% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % % The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth % % % This eBook is for the use of anyone anywhere at no cost and with % % almost no restrictions whatsoever. You may copy it, give it away or % % re-use it under the terms of the Project Gutenberg License included % % with this eBook or online at book.klll.cc % % % % % % Title: Plane Geometry % % % % Author: George Albert Wentworth % % % % Release Date: July 3, 2010 [EBook #33063] % % % % Language: English % % % % Character set encoding: ISO-8859-1 % % % % *** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY *** % % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \def\ebook{33063} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% %% %% Packages and substitutions: %% %% %% %% book: Required. %% %% inputenc: Standard DP encoding. Required. %% %% %% %% ifthen: Logical conditionals. Required. %% %% %% %% amsmath: AMS mathematics enhancements. Required. %% %% amssymb: Additional mathematical symbols. Required. %% %% %% %% alltt: Fixed-width font environment. Required. %% %% %% %% footmisc: Extended footnote capabilities. Optional. %% %% %% %% indentfirst: Indent first word of each sectional unit. Optional. %% %% %% %% fancyhdr: Enhanced running headers and footers. Required. %% %% graphicx: Standard interface for graphics inclusion. Required. %% %% %% %% geometry: Enhanced page layout package. Required. %% %% hyperref: Hypertext embellishments for pdf output. Required. %% %% %% %% enumerate: Configurable enumeration labels. Required. %% %% supertabular: The supertabular environment. Required. %% %% %% %% %% %% Producer's Comments: %% %% %% %% lacheck: ..................................... OK %% %% Numerous false positives %% %% %% %% PDF pages: 326 (if ForPrinting set to false) %% %% PDF page size: 6 x 8.5in (non-standard) %% %% PDF bookmarks: created, point to ToC entries %% %% PDF document info: filled in %% %% Images: 313 pdf diagrams %% %% %% %% Summary of log file: %% %% * One harmless underfull hbox on title page. %% %% %% %% Compile History: %% %% %% %% Jeremy Weatherford (PG xidus) %% %% ca. 2005 %% %% Fedora Core 3 Linux %% %% tetex-2.0.2-21 RPM %% %% tetex-latex-2.0.2-21 RPM %% %% %% %% Andrew D. Hwang (DP:adhere) %% %% July 2010 %% %% texlive2007, GNU/Linux %% %% %% %% Majority of figures contributed by: %% %% Owen Whitby (PG nilrem) %% %% %% %% Command block: %% %% %% %% pdflatex x2 %% %% %% %% %% %% July 2010: pglatex. %% %% Compile this project with: %% %% pdflatex 33063-t.tex ..... TWO times %% %% %% %% pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) %% %% %% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \listfiles \documentclass[12pt,titlepage]{book}[2005/09/16] %%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \usepackage[latin1]{inputenc}[2006/05/05] \usepackage{ifthen}[2001/05/26] %% Logical conditionals \usepackage{amsmath}[2000/07/18] %% Displayed equations \usepackage{amssymb}[2002/01/22] %% and additional symbols \usepackage{alltt}[1997/06/16] %% boilerplate, credits, license %% extended footnote capabilities \IfFileExists{footmisc.sty}{% \usepackage[symbol]{footmisc}[2005/03/17] }{} \IfFileExists{indentfirst.sty}{% \usepackage{indentfirst}[1995/11/23] }{} \usepackage{fancyhdr} \usepackage{graphicx}[1999/02/16]%% For diagrams \usepackage{enumerate}[1999/03/05] \usepackage{supertabular}[2004/02/20] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% Interlude: Set up PRINTING (default) or SCREEN VIEWING %%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % ForPrinting=true (default) false % Asymmetric margins Symmetric margins % Black hyperlinks Blue hyperlinks % Start Preface, ToC, etc. recto No blank verso pages % % Chapter-like ``Sections'' start both recto and verso in the scanned % book. This behavior has been retained. \newboolean{ForPrinting} %% UNCOMMENT the next line for a PRINT-OPTIMIZED VERSION of the text %% %\setboolean{ForPrinting}{true} %% Initialize values to ForPrinting=false \newcommand{\Margins}{hmarginratio=1:1} % Symmetric margins \newcommand{\HLinkColor}{blue} % Hyperlink color \newcommand{\PDFPageLayout}{SinglePage} \newcommand{\TransNote}{Transcriber's Note---} \newcommand{\TransNoteCommon}{% Minor typographical corrections and presentational changes have been made without comment. \bigskip } \newcommand{\TransNoteText}{% \TransNoteCommon This PDF file is optimized for screen viewing, but may easily be recompiled for printing. Please see the preamble of the \LaTeX\ source file for instructions. } %% Re-set if ForPrinting=true \ifthenelse{\boolean{ForPrinting}}{% \renewcommand{\Margins}{hmarginratio=2:3} % Asymmetric margins \renewcommand{\HLinkColor}{black} % Hyperlink color \renewcommand{\PDFPageLayout}{TwoPageRight} \renewcommand{\TransNoteText}{% \TransNoteCommon This PDF file is optimized for printing, but may easily be recompiled for screen viewing. Please see the preamble of the \LaTeX\ source file for instructions. } }{% If ForPrinting=false, don't skip to recto \renewcommand{\cleardoublepage}{\clearpage} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% End of PRINTING/SCREEN VIEWING code; back to packages %%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \ifthenelse{\boolean{ForPrinting}}{% \setlength{\paperwidth}{8.5in}% \setlength{\paperheight}{11in}% \usepackage[body={5.5in,7.33in},\Margins]{geometry}[2002/07/08] }{% \setlength{\paperwidth}{6in}% \setlength{\paperheight}{8.5in}% \usepackage[body={5.5in,7.33in},\Margins,includeheadfoot]{geometry}[2002/07/08] } \providecommand{\ebook}{00000} % Overridden during white-washing \usepackage[pdftex, hyperref, hyperfootnotes=false, pdftitle={The Project Gutenberg eBook \#\ebook: Plane Geometry}, pdfauthor={George Albert Wentworth}, pdfkeywords={Jeremy Weatherford, Stan Goodman, Kevin Handy, Project Gutenberg Online Distributed Proofreading Team}, pdfstartview=Fit, % default value pdfstartpage=1, % default value pdfpagemode=UseNone, % default value bookmarks=true, % default value linktocpage=false, % default value pdfpagelayout=\PDFPageLayout, pdfdisplaydoctitle, pdfpagelabels=true, bookmarksopen=true, bookmarksopenlevel=0, colorlinks=true, linkcolor=\HLinkColor]{hyperref}[2007/02/07] % Re-crop screen-formatted version, accommodating wide displays \ifthenelse{\boolean{ForPrinting}} {} {\hypersetup{pdfpagescrop= 0 30 612 765}} %%%% Fixed-width environment to format PG boilerplate %%%% \newenvironment{PGtext}{% \begin{alltt} \fontsize{9.2}{10.5}\ttfamily\selectfont}% {\end{alltt}} \newlength{\TmpLen} % [** TN: The Note to Teachers makes verbal reference to a portion of the book] \newcommand{\PageName}[1]{% \ifthenelse{\equal{\pageref{page:PageName}}{17}}{% seventeen % Value at current text block size }{% \ifthenelse{\equal{\pageref{page:PageName}}{14}}{% fourteen }{% \ifthenelse{\equal{\pageref{page:PageName}}{15}}{% fifteen }{% \ifthenelse{\equal{\pageref{page:PageName}}{16}}{% sixteen }{% \ifthenelse{\equal{\pageref{page:PageName}}{18}}{% eighteen }{% \ifthenelse{\equal{\pageref{page:PageName}}{19}}{% nineteen }{% \pageref{page:PageName} % Punt }% }% }% }% }% }% } \newcommand{\DittoMark}{``}%'' \newcommand{\Ditto}{\quad\DittoMark\quad} \DeclareInputText{176}{\ifmmode{{}^\circ}\else\textdegree\fi} \newcommand{\Headings}[1]{\textbf{\small#1}} %[** TN: Make all parentheses upright. % Code due to Will Robertson in comp.text.tex, Sept 3, 2008] \makeatletter \DeclareRobustCommand\IfMath{% \ifmmode \expandafter\@firstoftwo \else \expandafter\@secondoftwo \fi} \begingroup \global\let\origlparen=( \global\let\origrparen=) \catcode`\(=\active \catcode`\)=\active \AtBeginDocument{% \catcode`\(=\active \catcode`\)=\active \def({\IfMath{\origlparen}{\textup\origlparen}}% \def){\IfMath{\origrparen}{\textup\origrparen}}% } \endgroup \makeatother % tweak page layout \emergencystretch=11pt \raggedbottom \setlength{\parskip}{1mm} \makeatletter % convenience \newcommand{\hsp}{\hspace{\stretch{1}}} % Page separators \newcommand{\scanpage}[1]{} \newcommand{\indexbf}[1]{\textbf{#1}\label{#1}} \newcommand{\indexemph}[1]{\emph{#1}\label{#1}} % section heading styles in the text \renewcommand{\part}[2][]{% \def\mypartname{#2}% \clearpage \thispagestyle{plain}% \addcontentsline{toc}{part}{#2}% \def\firstchap{yes}% {\centering \normalfont \Huge #2\par}% } % chapter \renewcommand{\chapter}{ \ifthenelse{\equal{\firstchap}{yes}}{\vspace*{30\p@}}{\clearpage\thispagestyle{plain}}% \def\firstchap{no}% \global\@topnum\z@ \@afterindentfalse \secdef\@chapter\@schapter} \def\firstchap{yes} % modify chapter headings \renewcommand{\@makechapterhead}[1]{% {\centering \large \scshape #1\par\vskip 40\p@}} % section \def\section{\@startsection{section}{3}{\z@} {\baselineskip}{\parskip}% {\bfseries\centering}} % subsection \def\subsection{\@startsection{subsection}{4}{\z@} {\baselineskip}{\parskip}% {\bfseries\centering}} \newcommand{\proposition}[1]{\filbreak\bigskip\centerline{\textsc{\theproposition{} #1}}\medskip} % numbering depth and TOC depth \setcounter{tocdepth}{1} \setcounter{secnumdepth}{6} % section naming and numbering styles \def\partname{} \def\chaptername{} \def\sectionname{} \def\subsectionname{Proposition} \renewcommand{\thepart}{} \renewcommand{\thechapter}{} \renewcommand{\thesection}{} \renewcommand{\thesubsection}{} \newcounter{proposition}[chapter] \renewcommand{\theproposition}{\stepcounter{proposition}Proposition \Roman{proposition}.} % running headers \pagestyle{fancy} \renewcommand{\chaptermark}[1]{\markboth{BOOK \Roman{chapter}. \mypartname}{}} \renewcommand{\sectionmark}[1]{\markright{#1}} \newcommand{\SetRunningHeads}{% \fancyhf{} \ifthenelse{\boolean{ForPrinting}}{% \fancyhead[LE,RO]{\Headings{\thepage}} }{% \fancyhead[R]{\Headings{\thepage}} } \fancyhead[CO]{\Headings{\rightmark}} \fancyhead[CE]{\Headings{\leftmark}} } \renewcommand{\headrulewidth}{0.5pt} \renewcommand{\footrulewidth}{0.0pt} \addtolength{\headheight}{2.5pt} \fancypagestyle{plain}{ \fancyhead{} \renewcommand{\headrulewidth}{0pt} } \def\mypartname{} % numbered points \newcounter{point} \newcounter{exnum}[part] \newcommand{\pointno}{\stepcounter{point}\textbf{\thepoint{}.}} \newcommand{\exno}{\stepcounter{exnum}\textbf{\theexnum{}.}} \newenvironment{point}{\medskip\pointno~}{\filbreak} \newenvironment{proof}{\begin{point}}{\end{point}} \newenvironment{proofex}{\medskip\textbf{Ex.~\exno~}}{\filbreak} \newcommand{\pp}[1]{\medskip\pointno~#1\filbreak} \newcommand{\ex}[1]{\medskip\textbf{Ex.~\exno~}\ #1\filbreak} \newcommand{\defn}[1]{\textsc{Def.~}#1} \newcommand{\ax}[1]{\textbf{Axiom.~}\textit{#1}} \newcommand{\thm}[1]{\textbf{Theorem.~}\textit{#1}} % an outdented line heading several exercises % eg, ``To construct a triangle equivalent to:'' \newcommand{\exheader}[1]{\bigskip#1} \newcommand{\note}[2][foo]{\medskip\begin{small}\textsc{Note% \ifthenelse{\equal{#1}{foo}}{.~}{~#1.~}}% #2\end{small}\filbreak} \newcommand{\cor}[2][foo]{\textsc{Cor.~% \ifthenelse{\equal{#1}{foo}}{}{#1.~}}% \begin{itshape}#2\end{itshape}} % proof contents \newlength{\dentwidth}\setlength{\dentwidth}{\textwidth} \addtolength{\dentwidth}{-\parindent} \newcommand{\pnote}[1]{% \noindent\parbox{\linewidth}{\spaceskip0.5em plus 1em minus 0.25em\centering\small\textit{#1}}% } \newcommand{\obs}[1]{\textit{#1}} \newcommand{\fig}[1]{\fbox{figure here}\par\lett{#1}} \newcommand{\lett}[1]{\boldmath\textbf{#1}\unboldmath} \newcommand{\prove}[2][To prove that ]{\textit{#1#2}} \newcommand{\proveq}[3][\indent To prove that ]{\eq[\textit{#1}]{#2}{#3}{}} \newcommand{\qed}{\textsc{q.e.d.}} \newcommand{\qef}{\textsc{q.e.f.}} \newcommand{\step}[3][]{% \setlength{\TmpLen}{\parindent}% \noindent\parbox{\textwidth}{\setlength{\parindent}{\TmpLen}\rlap{#1}\hfill#2\hfill\llap{#3}}% } \newlength{\eqalign}\setlength{\eqalign}{.5\dentwidth} \newcommand{\eq}[4][]{% \setlength{\TmpLen}{\parindent}% \noindent\parbox{\textwidth}{\setlength{\parindent}{\TmpLen}\rlap{#1}\hfill~#2~#3~\hfill\llap{#4}}% } % math symbols \DeclareMathOperator{\arc}{arc} \DeclareMathOperator{\base}{base} \DeclareMathOperator{\rect}{rect.} % ugly parallelogram \def\Par{% \mathchoice {\@parallelogram\scriptsize}% {\@parallelogram\scriptsize}% {\@parallelogram\tiny}% {\@parallelogram\tiny}} \def\@parallelogram#1{% \textnormal{\setbox\z@\hbox{#1/}\dimen@\wd\z@ \@tempdima 2.45\dimen@ \vbox{\offinterlineskip \hbox{\kern.8\dimen@\vrule\@width\@tempdima\@height.4\p@}% \kern-.0\p@ \hbox to\@tempdima{#1/\hfil\rlap/}% \kern-.5\p@ \hbox{\kern.1\dimen@\vrule\@width\@tempdima\@height.4\p@}}}} \newcommand{\smallrule}{\medskip\centerline{\rule{10em}{.2ex}}\medskip} \newcommand{\figc}[2]{\centerline{\includegraphics{./images/#1.pdf}}% \ifthenelse{\equal{#2}{}}{}{\par\small\boldmath\textbf{#2}\unboldmath}} \newcommand{\figcc}[2]{\centerline{\includegraphics{./images/#1.pdf}\qquad\includegraphics{./images/#2.pdf}}} \newcommand{\figccc}[3]{\centerline{\includegraphics{./images/#1.pdf}\qquad% \includegraphics{./images/#2.pdf}\qquad% \includegraphics{./images/#3.pdf}}} \newcommand{\figcccc}[4]{\centerline{\includegraphics{./images/#1.pdf}\quad% \includegraphics{./images/#2.pdf}\quad% \includegraphics{./images/#3.pdf}\quad% \includegraphics{./images/#4.pdf}}} % lists \newcounter{Lcount} \newenvironment{myenum} {\setcounter{Lcount}{0}% \begin{list}{\arabic{Lcount}.} {\usecounter{Lcount}% \setlength{\itemsep}{0pt}% \setlength{\leftmargin}{2\parindent}}} {\end{list}} \makeatother %%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%% \begin{document} \pagestyle{empty} \pagenumbering{Alph} \phantomsection \pdfbookmark[-1]{FRONT MATTER.}{FRONT MATTER} %%%% PG BOILERPLATE %%%% \phantomsection \pdfbookmark[0]{PG BOILERPLATE.}{BOILERPLATE} \begin{center} \begin{minipage}{\textwidth} \small \begin{PGtext} The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at book.klll.cc Title: Plane Geometry Author: George Albert Wentworth Release Date: July 3, 2010 [EBook #33063] Language: English Character set encoding: ISO-8859-1 *** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY *** \end{PGtext} \end{minipage} \end{center} \clearpage %%%% Credits and transcriber's note %%%% \begin{center} \begin{minipage}{\textwidth} \begin{PGtext} Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy and the Online Distributed Proofreading Team at http://www.pgdp.net \end{PGtext} \end{minipage} \end{center} \vfill \begin{minipage}{0.85\textwidth} \small \phantomsection \pdfbookmark[0]{TRANSCRIBER'S NOTE.}{TRANSCRIBER'S NOTE} \subsection*{\centering\normalfont\scshape% \normalsize\MakeLowercase{\TransNote}}% \raggedright \TransNoteText \end{minipage} %%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% \frontmatter \pagestyle{fancy} \thispagestyle{empty} \begin{titlepage} \null\vfil \begin{center} {\LARGE PLANE GEOMETRY \par} \vspace{3em} {BY \par} {\large G.A. WENTWORTH \par} \textsc{Author of a Series of Text-Books in Mathematics} \vspace{3em} \emph{REVISED EDITION} \vspace{3em} \vfil\null {\large GINN \& COMPANY} BOSTON · NEW YORK · CHICAGO · LONDON \end{center} \end{titlepage} \newpage \thispagestyle{empty} \begin{center} \null\vfil Entered, according to Act of Congress, in the year 1888, by G.A. WENTWORTH in the Office Of the Librarian of Congress, at Washington \smallrule \textsc{Copyright, 1899} \textsc{By G.A. WENTWORTH} \smallrule {\small ALL RIGHTS RESERVED} {\small 67 10} \null\vfil The Athenæum Press \smallrule \settowidth{\TmpLen}{PRIETORS · BOSTON · U.S.A.}% \parbox{\TmpLen}{GINN \& COMPANY · PRO\-PRIETORS · BOSTON · U.S.A.} \end{center} \cleardoublepage \SetRunningHeads \section*{PREFACE.} \pdfbookmark[0]{PREFACE.}{PREFACE} Most persons do not possess, and do not easily acquire, the power of abstraction requisite for apprehending geometrical conceptions, and for keeping in mind the successive steps of a continuous argument. Hence, with a very large proportion of beginners in Geometry, it depends mainly upon the \emph{form} in which the subject is presented whether they pursue the study with indifference, not to say aversion, or with increasing interest and pleasure. Great care, therefore, has been taken to make the pages attractive. The figures have been carefully drawn and placed in the middle of the page, so that they fall directly under the eye in immediate connection with the text; and in no case is it necessary to turn the page in reading a demonstration. Full, long-dashed, and short-dashed lines of the figures indicate given, resulting, and auxiliary lines, respectively. Bold-faced, italic, and roman type has been skilfully used to distinguish the hypothesis, the conclusion to be proved, and the proof. As a further concession to the beginner, the reason for each statement in the early proofs is printed in small italics, immediately following the statement. This prevents the necessity of interrupting the logical train of thought by turning to a previous section, and compels the learner to become familiar with a large number of geometrical truths by constantly seeing and repeating them. This help is gradually discarded, and the pupil is left to depend upon the knowledge already acquired, or to find the reason for a step by turning to the given reference. It must not be inferred, because this is not a geometry of interrogation points, that the author has lost sight of the real object of the study. The training to be obtained from carefully following the logical steps of a complete proof has been provided for by the Propositions of the \scanpage{005.png}% Geometry, and the development of the power to grasp and prove new truths has been provided for by original exercises. The chief value of any Geometry consists in the happy combination of these two kinds of training. The exercises have been arranged according to the test of experience, and are so abundant that it is not expected that any one class will work them all out. The methods of attacking and proving original theorems are fully explained in the first Book, and illustrated by sufficient examples; and the methods of attacking and solving original problems are explained in the second Book, and illustrated by examples worked out in full. None but the very simplest exercises are inserted until the student has become familiar with geometrical methods, and is furnished with elementary but much needed instruction in the art of handling original propositions; and he is assisted by diagrams and hints as long as these helps are necessary to develop his mental powers sufficiently to enable him to carry on the work by himself. The law of converse theorems, the distinction between positive and negative quantities, and the principles of reciprocity and continuity have been briefly explained; but the application of these principles is left mainly to the discretion of teachers. The author desires to express his appreciation of the valuable suggestions and assistance which he has received from distinguished educators in all parts of the country. He also desires to acknowledge his obligation to Mr.~Charles Hamilton, the Superintendent of the composition room of the Athen\ae{}um Press, and to Mr.~I.~F. White, the compositor, for the excellent typography of the book. Criticisms and corrections will be thankfully received. \hfill G.~A. WENTWORTH. \textsc{Exeter}, N.H., June, 1899. \scanpage{006.png}% \clearpage \section*{NOTE TO TEACHERS.} \pdfbookmark[0]{NOTE TO TEACHERS.}{NOTE TO TEACHERS} It is intended to have the first \PageName\ pages of this book simply read in the class, with such running comment and discussion as may be useful to help the beginner catch the spirit of the subject-matter, and not leave him to the mere letter of dry definitions. In like manner, the definitions at the beginning of each Book should be read and discussed in the recitation room. There is a decided advantage in having the definitions for each Book in a single group so that they can be included in one survey and discussion. For a similar reason the theorems of limits are considered together. The subject of limits is exceedingly interesting in itself, and it was thought best to include in the theory of limits in the second Book every principle required for Plane and Solid Geometry. When the pupil is reading each Book for the first time, it will be well to let him write his proofs on the blackboard in his own language, care being taken that his language be the simplest possible, that the arrangement of work be vertical, and that the figures be accurately constructed. This method will furnish a valuable exercise as a language lesson, will cultivate the habit of neat and orderly arrangement of work, and will allow a brief interval for deliberating on each step. After a Book has been read in this way, the pupil should review the Book, and should be required to draw the figures free-hand. He should state and prove the propositions orally, using a pointer to indicate on the figure every line and angle named. He should be encouraged in reviewing each Book, to do the original exercises; to state the converse propositions, and determine whether they are true or false; and also to give well-considered answers to questions which may be asked him on many propositions. \scanpage{007.png}% The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus, a rectangle with a constant base $b$, and a variable altitude $x$, will afford an obvious illustration of the truth that the product of a constant and a variable is also a variable; and that the limit of the product of a constant and a variable is the product of the constant by the limit of the variable. If $x$ increases and approaches the altitude $a$ as a limit, the area of the rectangle increases and approaches the area of the rectangle $ab$ as a limit; if, however, $x$ decreases and approaches zero as a limit, the area of the rectangle decreases and approaches zero as a limit. An arithmetical illustration of this truth may be given by multiplying the approximate values of any repetend by a constant. If, for example, we take the repetend $0.3333$ etc., the approximate values of the repetend will be $\frac{3}{10}$, $\frac{33}{100}$, $\frac{333}{1000}$, $\frac{3333}{10000}$, etc., and these values multiplied by $60$ give the series $18$, $19.8$, $19.98$, $19.998$, etc., which evidently approaches $20$ as a limit; but the product of $60$ into $\frac{1}{3}$ (the limit of the repetend $0.333$ etc.) is also $20$. Again, if we multiply $60$ into the different values of the decreasing series $\frac{1}{30}$, $\frac{1}{300}$, $\frac{1}{3000}$, $\frac{1}{30000}$, etc., which approaches zero as a limit, we shall get the decreasing series $2$, $\frac{1}{5}$, $\frac{1}{50}$, $\frac{1}{500}$, etc.; and this series evidently approaches zero as a limit. The Teacher is likewise advised to give frequent written examinations. These should not be too difficult, and sufficient time should be allowed for accurately constructing the figures, for choosing the best language, and for determining the best arrangement. The time necessary for the reading of examination books will be diminished by more than one half, if the use of symbols is allowed. \textsc{Exeter}, N.H., 1899. \scanpage{008.png}% \clearpage \phantomsection\pdfbookmark[0]{TABLE OF CONTENTS.}{CONTENTS} \tableofcontents \mainmatter \scanpage{009.png}% \scanpage{010.png}% \phantomsection% \part{GEOMETRY.} \markboth{GEOMETRY.}{} % fake a chapter heading, no chapters yet \section{INTRODUCTION.} \begin{point}% If a block of wood or stone is cut in the shape represented in Fig.~1, it will have \emph{six flat faces}. Each face of the block is called a surface\label{surface}; and if the faces are made smooth by polishing, so that, when a straight edge is applied to any one of them, the straight edge in every part will touch the surface, the faces are called \textbf{plane surfaces}, or \indexbf{planes}. % [Illustration: Fig. 1.] \centering{\includegraphics{./images/woodcutsmall.pdf}} \centering{\small\textsc{Fig. 1.}} \end{point} \pp{The intersection\label{intersection} of any two of these surfaces is called a \indexbf{line}.} \pp{The intersection of any three of these lines is called a \indexbf{point}.} \begin{point}% The block extends in three principal directions: \begin{list}{}{\setlength{\itemsep}{0pt}} \item From left to right, $A$ to $B$. \item From front to back, $A$ to $C$. \item From top to bottom, $A$ to $D$. \end{list} These are called the \textbf{dimensions}\label{dimensions} of the block, and are named in the order given, \textbf{length}, \textbf{breadth} (or \textit{width}), and \textbf{thickness} (\textit{height} or \textit{depth}). \end{point} \scanpage{011.png}% \begin{point}% A \textbf{solid}, in common language, is a limited portion of space \textit{filled with matter}; but in Geometry we have nothing to do with \textit{the matter} of which a body is composed; we study simply its \textit{shape} and \textit{size}; that is, we regard a solid as a limited portion of space which may be occupied by a physical body, or marked out in some other way. Hence, \textit{A geometrical solid\label{geometrical1} is a limited portion of space.} \end{point} \begin{point}% The surface\label{surface2} of a solid is simply the boundary of the solid, that which separates it from surrounding space. The surface is no part of a solid and has no thickness. Hence, \textit{A surface has only two dimensions, length and breadth.} \end{point} \begin{point}% A line\label{line2} is simply a boundary of a surface, or the intersection of two surfaces. Since the surfaces have no thickness, a line has no thickness. Moreover, a line is no part of a surface and has no width. Hence, \textit{A line has only one dimension, length.} \end{point} \begin{point}% A point\label{point2} is simply the extremity of a line, or the intersection of two lines. A point, therefore, has no thickness, width, or length; therefore, no magnitude. Hence, \textit{A point has no dimension, but denotes position simply.} \end{point} \begin{point}% It must be distinctly understood at the outset that the points, lines, surfaces, and solids of Geometry are \textit{purely ideal}, though they are represented to the eye in a material way. Lines, for example, drawn on paper or on the blackboard, will have some width and some thickness, and will so far fail of being \textit{true lines}; yet, when they are used to help the mind in reasoning, it is assumed that they represent true lines, without breadth and without thickness. \end{point} \scanpage{012.png}% \figc{012aaZ10}{} \begin{point}% A point is \textit{represented} to the eye by a fine dot, and named by a letter, as $A$ (Fig.~2). A line is named by two letters, placed one at each end, as $BF$. A surface\label{surface3} is represented and named by the lines which bound it, as $BCDF$. A solid\label{geometrical2} is represented by the faces which bound it. \end{point} \pp{A point in space may be considered by itself, without reference to a line\label{line3}.} \pp{If a point moves in space, its path is a line. This line may be considered apart from the idea of a surface.} \pp{If a line moves in space, it generates, in general, a surface. A surface can then be considered apart from the idea of a solid.} \begin{point}% If a surface moves in space, it generates, in general, a solid. \filbreak \figc{012bbZ14}{} Thus, let the upright surface $ABCD$ (Fig.~3) move to the right to the position $EFGH$, the points $A$, $B$, $C$, and $D$ generating the lines $AE$, $BF$, $CG$, and $DH$, respectively. The lines $AB$, $BC$, $CD$, and $DA$ will generate the surfaces $AF$. $BG$, $CH$, and $DE$, respectively. The surface $ABCD$ will generate the solid $AG$. \end{point} \pp{\indexbf{Geometry} is the science which treats of \textit{position, form}, and \textit{magnitude}.} \pp{A \indexbf{geometrical figure} is a combination of points, lines, surfaces, or solids.} \begin{point}% {\indexbf{Plane Geometry} treats of figures all points of which are in the same plane.} \indexbf{Solid Geometry} treats of figures all points of which are not in the same plane. \end{point} \scanpage{013.png}% \section{GENERAL TERMS.} \begin{point}% A \indexbf{proof} is a course of reasoning by which the truth or falsity of any statement is logically established. \end{point} \begin{point}% An \textbf{axiom}\label{axiom} is a statement admitted to be true without proof. \end{point} \begin{point}% A \indexbf{theorem} is a statement to be proved. \end{point} \begin{point}% A \textbf{construction}\label{construction} is the representation of a required figure by means of points and lines. \end{point} \begin{point}% A \indexbf{postulate} is a construction admitted to be possible. \end{point} \begin{point}% A \textbf{problem} is a construction to be made so that it shall satisfy certain given conditions. \end{point} \begin{point}% A \indexbf{proposition} is an axiom, a theorem, a postulate, or a problem. \end{point} \begin{point}% A \textbf{corollary} is a truth that is easily deduced from known truths. \end{point} \begin{point}% A \indexbf{scholium} is a remark upon some particular feature of a proposition. \end{point} \begin{point}% The \textbf{solution of a problem} consists of four parts: 1. The \emph{analysis}\label{analysis}, or course of thought by which the construction of the required figure is discovered. 2. The \emph{construction} of the figure with the aid of ruler and compasses. 3. The \emph{proof} that the figure satisfies all the conditions. 4. The \emph{discussion} of the limitations, if any, within which the solution is possible. \end{point} \scanpage{014.png}% \begin{point}% A theorem consists of two parts: the \indexbf{hypothesis}, or that which is assumed; and the \indexbf{conclusion}, or that which is asserted to follow from the hypothesis. \end{point} \settowidth{\TmpLen}{\qquad\qquad Its contradictory:\quad}% \begin{point}% The \textbf{contradictory} of a theorem\label{contradictory} is a theorem which must be true if the given theorem is false, and must be false if the given theorem is true. Thus, \\[0.5\baselineskip] \parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\ \parbox{\TmpLen}{\qquad\qquad Its contradictory:} If $A$ is $B$, then $C$ is not $D$. \end{point} \begin{point}% The \indexbf{opposite} of a theorem is obtained by making \emph{both the hypothesis and the conclusion negative}. Thus, \\[0.5\baselineskip] \parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\ \parbox{\TmpLen}{\qquad\qquad Its opposite:} If $A$ is not $B$, then $C$ is not $D$. \end{point} \begin{point}% The \textbf{converse}\label{converse1} of a theorem is obtained by \emph{interchanging the hypothesis and conclusion}. Thus, \\[0.5\baselineskip] \parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\ \parbox{\TmpLen}{\qquad\qquad Its converse:} If $C$ is $D$, then $A$ is $B$. \end{point} \begin{point}% The converse of a truth is not \emph{necessarily} true. Thus, Every horse is a quadruped is true, but the converse, Every quad\-ru\-ped is a horse, is not true. \end{point} \begin{point}% \textit{If a direct proposition and its opposite are true, the converse proposition is true; and if a direct proposition and its converse are true, the opposite proposition is true}. Thus, if it were true that 1. If an animal is a horse, the animal is a quadruped; 2. If an animal is not a horse, the animal is not a quadruped;\\ it would follow that 3. If an animal is a quadruped, the animal is a horse. Moreover, if 1 and 3 were true, then 2 would be true. \end{point} \scanpage{015.png}% \section[GENERAL AXIOMS.]{\pointno\hsp GENERAL AXIOMS.\hsp\phantom{XX.}} \label{generalaxioms} 1. Magnitudes which are equal to the same magnitude, or equal magnitudes, are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals are taken from equals, the remainders are equal. 4. If equals are added to unequals, the sums are unequal in the same order; if unequals are added to unequals in the same order, the sums are unequal in that order. 5. If equals are taken from unequals, the remainders are unequal in the same order; if unequals are taken from equals, the remainders are unequal in the reverse order. 6. The doubles of the same magnitude, or of equal magnitudes are equal; and the doubles of unequals are unequal. 7. The halves of the same magnitude, or of equal magnitudes are equal; and the halves of unequals are unequal. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. \section[SYMBOLS AND ABBREVIATIONS.] {\pointno\hsp SYMBOLS AND ABBREVIATIONS. \hsp\phantom{XX.}} \label{abbr}\label{symbols}% \noindent\small\enlargethispage{8pt}% \begin{tabular*}{\dentwidth}{rl@{\extracolsep{\fill}}l@{\extracolsep{0pt}}@{\dots}l} $>$ & is (or are) greater than. & Def. & definition. \\ $<$ & is (or are) less than. & Ax. & axiom. \\ $\Bumpeq$ & is (or are) equivalent to. & Hyp. & hypothesis. \\ $\therefore$ & therefore. & Cor. & corollary. \\ $\perp$ & perpendicular. & Scho. & scholium. \\ $\perp_s$ & perpendiculars. & Ex. & exercise. \\ $\parallel$ & parallel.\qquad $\parallel_s$ parallels. & Adj. & adjacent. \\ $\angle$ & angle.\qquad $\angle_s$ angles. & Iden. & identical. \\ $\triangle$ & triangle.\qquad $\triangle_s$ triangles. & Const. & construction. \\ $\Par$ & parallelogram. & Sup. & supplementary. \\ $\Par_s$ & parallelograms. & Ext. & exterior. \\ $\odot$ & circle.\qquad $\odot_s$ circles. & Int. & interior. \\ rt. & right.\qquad st.\ straight. & Alt. & alternate. \\ \end{tabular*} \qed\ stands for quod erat demonstrandum, \emph{which was to be proved}. \qef\ stands for quod erat faciendum, \emph{which was to be done.} The signs $+$, $-$, $×$, $\div$, $=$, have the same meaning as in Algebra. \scanpage{016.png}% \normalsize \phantomsection% \part{PLANE GEOMETRY.} \chapter{BOOK I\@. RECTILINEAR FIGURES.} \section{DEFINITIONS.} \vspace{2ex} \figc{016aaZ37}{} \begin{point}% A \indexbf{straight line} is a line such that any part of it, however placed on any other part, will lie wholly in that part if its extremities lie in that part, as~$AB$. \end{point} \begin{point}% A \indexbf{curved line} is a line no part of which is straight, as~$CD$. \end{point} \begin{point}% A \textbf{broken line} is made up of different straight lines, as~$EF$. \end{point} \note{A straight line is often called simply a \emph{line}.} \begin{point}% A \textbf{plane surface}, or a \indexbf{plane}, is a surface in which, if any two points are taken, the straight line joining these points lies wholly in the surface. \end{point} \begin{point}% A \textbf{curved surface}\label{curvedsurf} is a surface no part of which is plane. \end{point} \begin{point}% A \indexbf{plane figure} is a figure all points of which are in the same plane. \end{point} \begin{point}% Plane figures which are bounded by straight lines are called \indexbf{rectilinear} figures; by curved lines, \indexbf{curvilinear} figures. \end{point} \begin{point}% Figures that have the \emph{same shape} are called \indexbf{similar}. Figures that have the \textit{same size but not the same shape} are called \textbf{equivalent}\label{equivalent1}. Figures that have the \textit{same shape and the same size} are called \textbf{equal}\label{equal} or \textbf{congruent}\label{congruent}. \end{point} \scanpage{017.png}% \section{THE STRAIGHT LINE.} \pp{\textbf{Postulate.} A straight line can be drawn from one point to another.} \pp{\textbf{Postulate.} A straight line can be produced indefinitely.} % footnote keeps this from being a normal \ax \pp{\textbf{Axiom.}\footnote{The general axioms on page \pageref{generalaxioms} apply to all magnitudes. Special geometrical axioms will be given when required.} \textit{Only one straight line can be drawn from one point to another.} Hence, two points \textit{determine} a straight line.} \pp{\cor[1]{Two straight lines which have two points in common coincide and form but one line.}} \pp{\cor[2]{Two straight lines can intersect in only one point.}} For if they had two points common, they would coincide and not intersect. Hence, two intersecting lines \textit{determine} a point. \pp{\ax{A straight line is the shortest line that can be drawn from one point to another.}\label{axiomstraight}} \pp{\defn{The \textbf{distance}\label{distance1} between two points is the length of the straight line that joins them.}} \pp{A straight line determined by two points may be considered as prolonged indefinitely.} \pp{If only the part of the line between two fixed points is considered, this part is called a \textbf{segment} of the line\label{lineseg}.} \pp{For brevity, we say ``the line $AB$,'' to designate a segment of a line limited by the points $A$ and $B$.} \pp{If a line is considered as extending from a fixed point, this point is called the \indexbf{origin} of the line.} \scanpage{018.png}% \figc{018aaZ55}{} \begin{point}% If any point, $C$, is taken in a given straight line, $AB$, the two parts $CA$ and $CB$ are said to have \emph{opposite directions} from the point $C$ (Fig.~5). Every straight line, as $AB$, may be considered as extending in either of two opposite directions, namely, from $A$ towards $B$, which is expressed by $AB$, and read segment $AB$; and from $B$ towards $A$, which is expressed by $BA$, and read segment $BA$. \end{point} \begin{point}% If the magnitude of a given line is changed, it becomes longer or shorter. Thus (Fig.~5), by prolonging $AC$ to $B$ we add $CB$ to $AC$, and $AB = AC+CB$. By diminishing $AB$ to $C$, we subtract $CB$ from $AB$, and $AC = AB-CB$. If a given line increases so that it is prolonged by its own magnitude several times in succession, the line is \emph{multiplied}, and the resulting line is called a \emph{multiple} of the given line. \figc{018bbZ56}{} Thus (Fig.~6), if $AB = BC = CD = DE$, then $AC = 2AB$, $AD = 3AB$, and $AE = 4AB$. Hence, \textit{Lines of given length may be added and subtracted; they may also be multiplied by a number.} \end{point} \section{THE PLANE ANGLE.} \label{angle} \figc{018ccZ57}{} \begin{point}% The \emph{opening} between two straight lines drawn from the same point is called a \textbf{plane angle}. The two lines, $ED$ and $EF$, are called the \textbf{sides}\label{anglesides}, and $E$, the point of meeting, is called the \indexbf{vertex} of the angle. The size of an angle depends upon the \emph{extent of opening} of its sides, and not upon the length of its sides. \end{point} \scanpage{019.png}% \begin{point}% If there is but one angle at a given vertex, the angle is designated by a capital letter placed at the vertex, and is read by simply naming the letter. \figcc{019aaZ58}{019bbZ58} If two or more angles have the same vertex, each angle is designated by three letters, and is read by naming the three letters, the one at the vertex between the others. Thus, $DAC$ (Fig.~8) is the angle formed by the sides $AD$ and $AC$. An angle is often designated by placing a small \emph{italic} letter between the sides and near the vertex, as in Fig.~9. \end{point} \begin{point}% \textbf{Postulate of Superposition.}\label{superposition} Any figure may be moved from one place to another without altering its size or shape. \end{point} \begin{point}% The \textbf{test of equality} of two geometrical magnitudes is that they may be made to coincide throughout their whole extent. Thus, Two straight lines are equal, if they can be placed one upon the other so that the points at their extremities coincide. Two angles are equal, if they can be placed one upon the other so that their vertices coincide and their sides coincide, each with each. \end{point} \begin{point}% A line or plane that divides a geometric magnitude into \emph{two equal parts} is called the \textbf{bisector}\label{bisector} of the magnitude. If the angles $BAD$ and $CAD$ (Fig.~8) are equal, $AD$ \emph{bisects} the angle $BAC$. \end{point} \begin{point}% Two angles are called \textbf{adjacent angles}\label{adjacent1} when they have the same vertex and a common side between them; as the angles $BOD$ and $AOD$ (Fig.~10). \end{point} \scanpage{020.png}% \figcc{019ccZ62}{020aaZ63} \begin{point}% When one straight line meets another straight line and makes the \emph{adjacent angles equal}, each of these angles is called a \textbf{right angle}\label{right}; as angles $DCA$ and $DCB$ (Fig.~11). \end{point} \begin{point}% A \indexbf{perpendicular} to a straight line is a straight line that makes a right angle with it. Thus, if the angle $DCA$ (Fig.~11) is a right angle, $DC$ is perpendicular to $AB$, and $AB$ is perpendicular to $DC$. \end{point} \begin{point}% The point (as $C$, Fig.~11) where a perpendicular meets another line is called the \indexbf{foot} of the perpendicular. \end{point} \begin{point}% When the sides of an angle extend in opposite directions, so as to be in the same straight line, the angle is called a \textbf{straight angle}.\label{straight} \figc{020bbZ66}{} Thus, the angle formed at $C$ (Fig.~12) with its sides $CA$ and $CB$ extending in opposite directions from $C$ is a straight angle. \end{point} \pp{\cor{A right angle is half a straight angle.}} \figcc{020ccZ68}{020ddZ69} \begin{point}% An angle less than a right angle is called an \textbf{acute angle}\label{acute}; as, angle $A$ (Fig.~13). \end{point} \begin{point}% An angle greater than a right angle and less than a straight angle is called an \textbf{obtuse angle}\label{obtuse}; as, angle $AOD$ (Fig.~14). \end{point} \begin{point}% An angle greater than a straight angle and less than two straight angles is called a \textbf{reflex angle}\label{reflex}; as, angle $DOA$, indicated by the dotted line (Fig.~14). \end{point} \scanpage{021.png}% \begin{point}% Angles that are neither right nor straight angles are called \textbf{oblique angles}\label{oblique}; and intersecting lines that are not perpendicular to each other are called \indexbf{oblique lines}. \end{point} \subsection{EXTENSION OF THE MEANING OF ANGLES.} \figc{021aaZ72}{} \begin{point}% Suppose the straight line $OC$ (Fig.~15) to move in the plane of the paper from coincidence with $OA$, about the point $O$ as a pivot, to the position $OC$; then the line $OC$ describes or generates \emph{the angle $AOC$}, and the magnitude of the angle $AOC$ depends upon the \emph{amount of rotation} of the line from the position $OA$ to the position $OC$. If the rotating line moves from the position $OA$ to the position $OB$, \emph{perpendicular} to $OA$, it generates the right angle $AOB$; if it moves to the position $OD$, it generates the obtuse angle $AOD$; if it moves to the position $OA'$, it generates the straight angle $AOA'$; if it moves to the position $OB'$ it generates the reflex angle $AOB'$, indicated by the dotted line; and if it moves to the position $OA$ again, it generates two straight angles. Hence, \end{point} \begin{point}% \textit{The angular magnitude about a point in a plane is equal to two straight angles, or four right angles; and the angular magnitude about a point on one side of a straight line drawn through the point is equal to a straight angle, or two right angles.} \end{point} \begin{point}% The whole angular magnitude about a point in a plane is called a \indexbf{perigon}; and two angles whose sum is a perigon are called \indexbf{conjugate angles}. \end{point} \note{This \emph{extension of the meaning of angles} is necessary in the applications of Geometry, as in Trigonometry, Mechanics, etc.} \scanpage{022.png}% \figccc{022aaZ75}{022bbZ76}{022ccZ77} \begin{point}% When two angles have the same vertex, and the sides of the one are prolongations of the sides of the other, they are called \indexbf{vertical angles}; as, angles $a$ and $b$, $c$ and $d$ (Fig.~16). \end{point} \begin{point}% Two angles are called \textbf{complementary}\label{complementary} when their sum is equal to a right angle; and each is called the \emph{complement}\label{complement} of the other; as, angles $DOB$ and $DOC$ (Fig.~17). \end{point} \begin{point}% Two angles are called \textbf{supplementary}\label{supplementary} when their sum is equal to a straight angle; and each is called the \indexemph{supplement} of the other; as, angles $DOB$ and $DOA$ (Fig.~18). \end{point} \subsection{UNIT OF ANGLES.} \begin{point}% By adopting a suitable unit for measuring angles we are able to express the magnitudes of angles by numbers. If we suppose $OC$ (Fig.~15) to turn about $O$ from coincidence with $OA$ until it makes \emph{one three hundred sixtieth} of a revolution, it generates an angle at $O$, which is taken as the unit for measuring angles. This unit is called a \emph{degree}. The degree is subdivided into sixty equal parts, called \emph{minutes}, and the minute into sixty equal parts, called \emph{seconds}. Degrees, minutes, and seconds are denoted by symbols. Thus, $5$~degrees $13$~minutes $12$~seconds is written $5°\ 13'\ 12''$. A right angle is generated when $OC$ has made \emph{one fourth} of a revolution and contains~$90°$; a straight angle, when $OC$ has made \emph{half} of a revolution and contains~$180°$; and a perigon, when $OC$ has made a complete revolution and contains~$360°$. \end{point} \note{The natural angular unit is one complete revolution. But this unit would require us to express the values of most angles by fractions. The advantage of using the degree as the unit consists in its convenient size, and in the fact that $360$~is divisible by so many different integral numbers.} \scanpage{023.png}% \figc{023aaZ79}{} \begin{point}% By the method of superposition we are able to compare magnitudes of the same kind. Suppose we have two angles, $ABC$ and $DEF$ (Fig.~19). Let the side $ED$ be placed on the side $BA$, so that the vertex $E$ shall fall on $B$; then, if the side $EF$ falls on $BC$, the angle $DEF$ equals the angle $ABC$; if the side $EF$ falls between $BC$ and $BA$ in the position shown by the dotted line $BG$, the angle $DEF$ is less than the angle $ABC$; but if the side $EF$ falls in the position shown by the dotted line $BH$, the angle $DEF$ is greater than the angle $ABC$. \end{point} \figc{023bcZ80}{} \begin{point}% If we have the angles $ABC$ and $DEF$ (Fig.~20), and place the vertex $E$ on $B$ and the side $ED$ on $BC$, so that the angle $DEF$ takes the position $CBH$, the angles $DEF$ and $ABC$ will together be equal to the angle $ABH$. If the vertex $E$ is placed on $B$, and the side $ED$ on $BA$, so that the angle $DEF$ takes the position $ABF$, the angle $FBC$ will be the difference between the angles $ABC$ and $DEF$. If an angle is increased by its own magnitude two or more times in succession, the angle is \emph{multiplied} by a number. Thus, if the angles $ABM$, $MBC$, $CBP$, $PBD$ (Fig.~21) are all equal, the angle $ABD$ is $4$~times the angle $ABM$. Therefore, \textit{Angles may be added and subtracted; they may also be multiplied by a number.} \label{page:PageName}% [** TN: For verbal ref. in Note to Teachers] \end{point} \scanpage{024.png}% \pagebreak \section{PERPENDICULAR AND OBLIQUE LINES.} \proposition{Theorem.} \begin{proof}% \obs{All straight angles are equal.} \figc{024aaZ81}{Let the angles $ACB$ and $DEF$ be any two straight angles.} \prove{$\angle ACB = \angle DEF$}. \textbf{Proof.} Place the $\angle ACB$ on the $\angle DEF$, so that the vertex $C$ shall fall on the vertex $E$, and the side $CB$ on the side $EF$. \step{Then $CA$ will fall on $ED$,}{§~47} \pnote{(because $ACB$ and $DEF$ are straight lines).} \step{$\therefore \angle ACB = \angle DEF$.}{§~60} \hfill\qed \end{proof} \pp{\cor[1]{All right angles are equal.}\hfill~Ax.~7} \begin{point}% \cor[2]{At a given point in a given line there can be but one perpendicular to the line.} \figc{024bbZ83}{} For, if there could be two $\perp_s$, we should have rt.~$\angle_s$ of different magnitudes; but this is impossible, §~82. \end{point} \pp{\cor[3]{The complements of the same angle or of equal angles are equal.}\hfill~Ax.~3} \pp{\cor[4]{The supplements of the same angle or of equal angles are equal.}\hfill~Ax.~3} \note{The beginner must not forget that in Plane Geometry all the points of a figure are in the same plane. Without this restriction in Cor.~2, an indefinite number of perpendiculars can be erected at a given point in a given line.} \scanpage{025.png}% \proposition{Theorem.} \begin{proof}% \obs{If two adjacent angles have their exterior sides in a straight line, these angles are supplementary.} \figc{025aaZ86}{Let the exterior sides $OA$ and $OB$ of the adjacent angles $AOD$ and $BOD$ be in the straight line $AB$.} \prove{$\angle_s AOD$ and $BOD$ are supplementary.} \step[\indent\textbf{Proof.}]{$AOB$ is a straight line.}{Hyp.} \step{$\therefore \angle AOB$ is a st.~$\angle$.}{§~66} \step[\indent But]{$\angle AOD + \angle BOD =$ the st.~$\angle AOB$.}{Ax.~9} \step{$\therefore$ the $\angle_s AOD$ and $BOD$ are supplementary.}{§~77} \hfill\qed \end{proof} \begin{point}% \defn{Adjacent angles that are supplements of each other are called \emph{supplementary-adjacent angles}\label{suppladj}.} Since the angular magnitude about a point is neither increased nor diminished by the number of lines which radiate from the point, it follows that, \end{point} \pp{\cor[1]{The sum of all the angles about a point in a plane is equal to a perigon, or two straight angles.}} \pp{\cor[2]{The sum of all the angles about a point in a plane, on the same side of a straight line passing through the point, is equal to a straight angle, or two right angles.}} \scanpage{026.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} If two adjacent angles are supplementary, their exterior sides are in the same straight line.} \figc{026aaZ90}{Let the adjacent angles $OCA$ and $OCB$ be supplementary.} \prove{$AC$ and $CB$ are in the same straight line.} \textbf{Proof.} Suppose $CF$ to be in the same line with $AC$. \step[\indent Then]{$\angle_s OCA$ and $OCF$ are supplementary,}{§~86} \pnote{(if two adjacent angles have their exterior sides in a straight line, these angles are supplementary).} \step[\indent But]{$\angle_s OCA$ and $OCB$ are supplementary.}{Hyp.} \step{$\therefore \angle_s OCF$ and $OCB$ have the same supplement.}{} \eq{$\therefore \angle OCF$}{$= \angle OCB$.}{§~85} \step{$\therefore CB$ and $CF$ coincide.}{§~60} \step{$\therefore AC$ and $CB$ are in the same straight line.}{\qed} Since Propositions II.\ and III.\ are true, their opposites are true. Hence,~\hfill§~33 \end{proof} \pp{\cor[1]{If the exterior sides of two adjacent angles are not in a straight line, these angles are not supplementary.}} \pp{\cor[2]{If two adjacent angles are not supplementary, their exterior sides are not in the same straight line.}} \scanpage{027.png}% \proposition{Theorem.} \begin{proof}% \obs{If one straight line intersects another straight line, the vertical angles are equal.} \figc{027aaZ93}{Let the lines $OP$ and $AB$ intersect at $C$.} \proveq{$\angle OCB$}{$= \angle ACP$.} \step[\indent\textbf{Proof.}]{$\angle OCA$ and $\angle OCB$ are supplementary.}{§~86} \step{$\angle OCA$ and $\angle ACP$ are supplementary,}{§~86} \pnote{(if two adjacent angles have their exterior sides in a straight line, these angles are supplementary).} \step{$\therefore \angle_s OCB$ and $ACP$ have the same supplement.}{} \eq{$\therefore \angle OCB$}{$= \angle ACP$.}{§~85} \eq[\indent In like manner,]{$\angle ACO$}{$= \angle PCB$.}{\qed} \end{proof} \pp{\cor{If one of the four angles formed by the intersection of two straight lines is a right angle, the other three angles are right angles.}} \ex{Find the complement and the supplement of an angle of~$49°$.} \ex{Find the number of degrees in an angle if it is double its complement; if it is one fourth of its complement.} \ex{Find the number of degrees in an angle if it is double its supplement; if it is one third of its supplement.} \scanpage{028.png}% \proposition{Theorem} \begin{proof}% \obs{Two straight lines drawn from a point in a perpendicular to a given line, cutting off on the given line equal segments from the foot of the perpendicular, are equal and make equal angles with the perpendicular.} \figc{028aaZ95}{Let $CF$ be a perpendicular to the line $AB$, and $CE$ and $CK$ two straight lines cutting off on $AB$ equal segments $FE$ and $FK$ from $F$.} \prove{$CE = CK$; and $\angle FCE = \angle FCK$.} \textbf{Proof.} Fold over $CFA$, on $CF$ as an axis, until it falls on the plane at the right of $CF$. \step{$FA$ will fall along $FB$,}{} \pnote{(since $\angle CFA = \angle CFB$, each being a rt.~$\angle$, by hyp.).} \step{Point $E$ will fall on point $K$,}{} \pnote{(since $FE = FK$, by hyp.).} \eq{$\therefore CE$}{$= CK$,}{§~60} \pnote{(their extremities being the same points);} \eq{and $\angle FCE$}{$= \angle FCK$,}{§~60} \pnote{(since their vertices coincide, and their sides coincide, each with each).} \hfill\qed \end{proof} \ex{Find the number of degrees in the angle included by the hands of a clock at $1$~o'clock. $3$~o'clock. $4$~o'clock. $6$~o'clock.} \scanpage{029.png}% \proposition{Theorem.} \begin{proof}% \obs{Only one perpendicular can be drawn to a given line from a given external point.} \figc{029aaZ96}{Let $AB$ be the given line, $P$ the given external point, $PC$ a perpendicular to $AB$ from $P$, and $PD$ any other line from $P$ to $AB$.} \proveq{$PD$ is not}{$\perp$ to $AB$.} \textbf{Proof.} Produce $PC$ to $P'$, making $CP'$ equal to $PC$. \step{Draw $DP'$.}{} \step{By construction, $PCP'$ is a straight line.}{} \step{$\therefore PDP'$ is not a straight line,}{§~46} \pnote{(only one straight line can be drawn from one point to another).} \step{Hence, $\angle PDP'$ is not a straight angle.}{} \step{Since $PC$ is $\perp$ to $DC$, and $PC = CP'$,}{} \step{$AC$ is $\perp$ to $PP'$ at its middle point.}{} \step{$\therefore \angle PDC = \angle P'DC$,}{§~95} \pnote{(two straight lines from a point in a $\perp$ to a line, cutting off on the line equal segments from the foot of the $\perp$, make equal $\angle_s$ with the $\perp$)} \step{Since $\angle PDP'$ is not a straight angle,}{} \step{$\angle PDC$, the half of $\angle PDP'$, is not a right angle.}{} \step{$\therefore PD$ is not $\perp$ to $AB$.}{\qed} \end{proof} \scanpage{030.png}% \proposition{Theorem.} \begin{proof}% \obs{The perpendicular is the shortest line that can be drawn to a straight line from an external point.} \figc{030aaZ97}{Let $AB$ be the given straight line, $P$ the given point, $PC$ the perpendicular, and $PD$ any other line drawn from $P$ to $AB$.} \proveq{$PC$}{$< PD$.} \textbf{Proof.} Produce $PC$ to $P'$, making $CP' = PC$. \step{Draw $DP'$.}{} \eq[\indent Then]{$PD$}{$= DP'$,}{§~95} \pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the line equal segments from the foot of the $\perp$, are equal).} \eq{$\therefore PD + DP'$}{$= 2PD$,}{} \eq[and]{$PC + CP'$}{$= 2PC$.}{Const.} \eq[\indent But]{$PC + CP'$}{$< PD + DP'$.}{§~49} \eq{$\therefore 2 PC$}{$< 2 PD$.}{} \eq{$\therefore PC$}{$< PD$.}{Ax.~7} \hfill\qed \end{proof} \pp{\cor{The shortest line that can be drawn from a point to a given line is perpendicular to the given line.}} \pp{\defn{The \textbf{distance}\label{distance2} of a point from a line is the length of the perpendicular from the point to the line.}} \scanpage{031.png}% \proposition{Theorem.} \begin{proof}% \obs{The sum of two lines drawn from a point to the extremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them.} \figc{031aa100}{Let $CA$ and $CB$ be two lines drawn from the point $C$ to the extremities of the straight line $AB$. Let $OA$ and $OB$ be two lines similarly drawn, but included by $CA$ and $CB$.} \proveq{$CA + CB$}{$> OA + OB$.} \textbf{Proof.} Produce $AO$ to meet the line $CB$ at $E$. \eq[\indent Then]{$CA + CE$}{$> OA + OE$,}{} \eq[and]{$BE + OE$}{$> OB$,}{§~49} \pnote{(a straight line is the shortest line from one point to another).} \step{Add these inequalities, and we have}{} \eq{$CA + CE + BE + OE$}{$> OA + OE + OB$.}{Ax.~4} \step{Substitute for $CE + BE$ its equal $CB$, then}{} \eq{$CA + CB + OE$}{$> OA + OE + OB$.}{} \step{Take away $OE$ from each side of the inequality.}{} \eq{$CA + CB$}{$> OA + OB$.}{Ax.~5} \hfill\qed \end{proof} \scanpage{032.png}% \proposition{Theorem.} \begin{proof}% \obs{Of two straight lines drawn from the same point in a perpendicular to a given line, cutting off on the line unequal segments from the foot of the perpendicular, the more remote is the greater.} \figc{032aa101}{Let $OC$ be perpendicular to $AB$, $OG$ and $OE$ two straight lines to $AB$, and $CE$ greater than $CG$.} \proveq{$OE$}{$> OG$.} \textbf{Proof.} Take $CF$ equal to $CG$, and draw $OF$. \eq[\indent Then]{$OF$}{$= OG$,}{§~95} \pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the line equal segments from the foot of the $\perp$, are equal).} \step{Produce $OC$ to $D$, making $CD = OC$.}{} \step{Draw $ED$ and $FD$.}{} \step[\indent Then]{$OE = ED$, and $OF = FD$.}{§~95} \eq[\indent But]{$OE + ED$}{$> OF + FD$,}{§~100} \eq{$\therefore 2OE > 2OF$, $OE$}{$> OF$, and $OE > OG$.}{\qed} \end{proof} \pp{\cor{Only two equal straight lines can be drawn from a point to a straight line; and of two unequal lines, the greater cuts off on the line the greater segment from the foot of the perpendicular.}} \scanpage{033.png}% \pagebreak \section{PARALLEL LINES.} \pp{\defn{Two \indexbf{parallel lines} are lines that lie in the same plane and cannot meet however far they are produced.}} \proposition{Theorem.} \begin{proof}% \obs{Two straight lines in the same plane perpendicular to the same straight line are parallel.} \figc{033aa104}{Let $AB$ and $CD$ be perpendicular to $AC$.} \prove{$AB$ and $CD$ are parallel.} \textbf{Proof.} If $AB$ and $CD$ are not parallel, they will meet if sufficiently prolonged; and we shall have two perpendicular lines from their point of meeting to the same straight line; but this is impossible,~\hfill§~96 \pnote{(only one perpendicular can be drawn to a given line from a given external point).} \step{$\therefore AB$ and $CD$ are parallel.}{\qed} \end{proof} \pp{\ax{Through a given point only one straight line can be drawn parallel to a given straight line.}\label{axiomparallel}} \begin{point}% \cor{Two straight lines in the same plane parallel to a third straight line are parallel to each other.} For if they could meet, we should have two straight lines from the point of meeting parallel to a straight line; but this is impossible.~\hfill§~105 \end{point} \scanpage{034.png}% \proposition{Theorem.} \begin{proof}% \obs{If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other also.} \figc{034aa107}{Let $AB$ and $EF$ be two parallel lines, and let $HK$ be perpendicular to $AB$, and cut $EF$ at $C$.} \proveq{$HK$}{is $\perp$ to $EF$.} \textbf{Proof.} Suppose $MN$ drawn through $C \perp$ to $HK$. \eq[\indent Then]{$MN$} {is $\parallel$ to $AB$.}{§~104} \eq[\indent But]{$EF$} {is $\parallel$ to $AB$.}{Hyp.} \step{$\therefore EF$ coincides with $MN$.}{§~105} \eq[\indent But]{$MN$} {is $\perp$ to $HK$.}{Const.} \eq{$\therefore EF$}{is $\perp$ to $HK$,}{} \eq[that is,]{$HK$}{is $\perp$ to $EF$.}{\qed} \end{proof} \pp{\defn{A straight line that cuts two or more straight lines is called a \indexbf{transversal} of those lines.}} \figc{034bb109}{} \begin{point}% If the transversal $EF$ cuts $AB$ and $CD$, the angles $a$, $d$, $g$, $f$ are called \emph{interior}\label{interior} angles; $b$, $c$, $h$, $e$ are called \emph{exterior}\label{exterior} angles. The angles $d$ and $f$, and $a$ and $g$, are called \emph{alternate-interior}\label{altint} angles; the angles $b$ and $h$, and $c$ and $e$, are called \emph{alternate-exterior}\label{altext} angles. The angles $b$ and $f$, $c$ and $g$, $e$ and $a$, $h$ and $d$, are called \emph{exterior-interior}\label{extint} angles. \end{point} \scanpage{035.png}% \proposition{Theorem.} \begin{proof}% \obs{If two parallel lines are cut by a transversal, the alternate-interior angles are equal.} \figc{035aa110}{Let $EF$ and $GH$ be two parallel lines cut by the transversal $BC$.} \proveq{$\angle EBC$}{$= \angle BCH$.} \textbf{Proof.} Through $O$, the middle point of $BC$, suppose $AD$ drawn $\perp$ to $GH$. \step[\indent Then]{$AD$ is likewise $\perp$ to $EF$,}{§~107} \pnote{(a straight line $\perp$ to one of two $\parallel_s$ is $\perp$ to the other),} \step[that is,]{$CD$ and $BA$ are both $\perp$ to $AD$.}{} Apply the figure $COD$ to the figure $BOA$, so that $OD$ shall fall along $OA$. \step[\indent Then]{$OC$ will fall along $OB$,}{§~93} \pnote{(since $\angle COD = \angle BOA$, being vertical $\angle_s$);} \step[and]{$C$ will fall on $B$,}{} \pnote{(since $OC = OB$, by construction).} \step[\indent Then]{the $\perp CD$ will fall along the $\perp BA$,}{§~96} \pnote{(only one $\perp$ can be drawn to a given line from a given external point).} \step{$\therefore \angle OCD$ coincides with $\angle OBA$, and is equal to it,}{§~60} \pnote{(two angles are equal, if their vertices coincide and their sides coincide, each with each).} \hfill\qed \end{proof} \scanpage{036.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} When two straight lines in the same plane are cut by a transversal, if the alternate-interior angles are equal, the two straight lines are parallel.} \figc{036aa111}{Let $EF$ cut the straight lines $AB$ and $CD$ in the points $H$ and $K$, and let the angles $AHK$ and $HKD$ be equal.} \proveq{$AB$ is}{$\parallel$ to $CD$.} \textbf{Proof}. Suppose $MN$ drawn through $H \parallel$ to $CD$. \eq[\indent Then]{$\angle MHK$}{$= \angle HKD$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).} \eq[\indent But]{$\angle AHK$}{$= \angle HKD$.}{Hyp.} \eq{$\therefore \angle MHK$}{$= \angle AHK$.}{Ax.~1} \step{$\therefore MN$ and $AB$ coincide.}{§~60} \eq[\indent But]{$MN$ is}{$\parallel$ to $CD$.}{Const.} \step{$\therefore AB$, which coincides with $MN$, is $\parallel$ to $CD$.}{\qed} \end{proof} \ex{Find the complement and the supplement of an angle that contains $37°\ 53'\ 49''$.} \ex{If the complement of an angle is one third of its supplement, how many degrees does the angle contain?} \scanpage{037.png}% \proposition{Theorem.} \begin{proof}% \obs{If two parallel lines are cut by a transversal, the exterior-interior angles are equal.} \figc{037aa112}{Let $AB$ and $CD$ be two parallel lines cut by the transversal $EF$, in the points $H$ and $K$.} \proveq{$\angle EHB$}{$= \angle HKD$.} \eq[\indent\textbf{Proof.}]{$\angle EHB$}{$= \angle AHK$,}{§~93} \pnote{(being vertical $\angle_s$).} \eq{$\angle AHK$}{$= \angle HKD$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).} \eq{$\therefore \angle EHB$}{$= \angle HKD$.}{Ax.~1} \eq[\indent In like manner]{$\angle EHA$}{$= \angle HKC$.}{\qed} \end{proof} \pp{\cor{The alternate-exterior angles $EHB$ and $CKF$, and also $AHE$ and $DKF$, are equal.}} \proposition{Theorem.} \begin{point}% \obs{\textsc{Conversely:} When two straight lines in a plane are cut by a transversal, if the exterior-interior angles are equal, these two straight lines are parallel.} (Proof similar to that in §~111.) \end{point} \scanpage{038.png}% \proposition{Theorem.} \begin{proof}% \obs{If two parallel lines are cut by a transversal, the two interior angles on the same side of the transversal are supplementary.} \figc{038aa115}{Let $AB$ and $CD$ be two parallel lines cut by the transversal $EF$ in the points $H$ and $K$.} \prove{$\angle_s BHK$ and $HKD$ are supplementary.} \step[\indent\textbf{Proof.}]{$\angle EHB + \angle BHK = \text{a st.\ }\angle$,}{§~86} \pnote{(being sup.-adj.~$\angle_s$).} \step[\indent But]{$\angle EHB = \angle HKD$,}{§~112} \pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).} \step{\( \therefore \angle BHK + \angle HKD = \text{a st.\ }\angle \).}{} \step{$\therefore \angle_s BHK$ and $HKD$ are supplementary.}{§~77} \hfill\qed \end{proof} \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} When two straight lines in a plane are cut by a transversal, if two interior angles on the same side of the transversal are supplementary, the two straight lines are parallel.} (Proof similar to that in §~111.) \end{proof} \scanpage{039.png}% \section{TRIANGLES.} \begin{point}% A \indexbf{triangle} is a portion of a plane bounded by three straight lines; as, $ABC$ (Fig.~1). \figcc{039aa117}{039bb118} The bounding lines are called the \textbf{sides}\label{trisides} of the triangle, and their sum is called its \indexbf{perimeter}; the angles included by the sides are called the \textbf{angles} of the triangle\label{anglestri}, and the vertices of these angles, the \textbf{vertices} of the triangle\label{trivertices}. \end{point} \begin{point}% \textbf{Adjacent angles}\label{adjacent2} of a rectilinear figure are two angles that have one side of the figure common; as, angles $A$ and $B$ (Fig.~2). \end{point} \begin{point}% An \textbf{exterior angle} of a triangle\label{exteriortri} is an angle included by one side and another side produced; as, $ACD$ (Fig.~2). The interior angle $ACB$ is adjacent to the exterior angle; the interior angles, $A$ and $B$, are called \textbf{opposite interior angles}. \end{point} \figc{039ce119}{} \begin{point}% A triangle is called a \indexbf{scalene triangle} when no two of its sides are equal; an \indexbf{isosceles triangle}, when two of its sides are equal; an \indexbf{equilateral triangle}, when its three sides are equal. \figc{039fi120}{} \end{point} \scanpage{040.png}% \begin{point}% A triangle is called a \indexbf{right triangle}, when one of its angles is a right angle; an \indexbf{obtuse triangle}, when one of its angles is an obtuse angle; an \textbf{acute triangle}, when all three of its angles are acute angles; an \indexbf{equiangular triangle}, when its three angles are equal. \end{point} \begin{point}% In a right triangle, the side opposite the right angle is called the \indexbf{hypotenuse}, and the other two sides the \indexbf{legs}. \end{point} \begin{point}% The side on which a triangle is supposed to stand is called the base\label{basetri} of the triangle. In the isosceles triangle, the equal sides are called the legs, and the other side, the base\label{baseiso}; in other triangles, any one of the sides may be taken as the base. \end{point} \begin{point}% The angle opposite the base of a triangle is called the \indexbf{vertical angle}, and its vertex, the \textbf{vertex} of the triangle\label{trivertex}. \end{point} \begin{point}% The \textbf{altitude} of a triangle\label{alttri} is the perpendicular from the vertex to the base, or to the base produced; as, $AD$ (Fig.~1). \end{point} \begin{point}% The three perpendiculars from the vertices of a triangle to the opposite sides (produced if necessary) are called the \textbf{altitudes} of the triangle; the three bisectors of the angles are called the \textbf{bisectors} of the triangle\label{tribisectors}; and the three lines from the vertices to the middle points of the opposite sides are called the \textbf{medians}\label{trimedians} of the triangle. \end{point} \begin{point}% If two triangles have the angles of the one equal, respectively, to the angles of the other, the equal angles are called \indexbf{homologous angles}, and the sides opposite the equal angles are called \indexbf{homologous sides}. \end{point} \begin{point}% Two triangles are equal in all respects if they can be made to coincide (§~60). The homologous sides of \indexemph{equal triangles} are equal, and the homologous angles are equal. \end{point} \scanpage{041.png}% \proposition{Theorem.} \begin{proof}% \obs{The sum of the three angles of a triangle is equal to two right angles.} \figc{041aa129}{Let $A$, $B$, and $BCA$ be the angles of the triangle $ABC$.} \prove{$\angle A+\angle B+\angle BCA = 2$ rt.~$\angle_s$.} \textbf{Proof.} Suppose $CE$ drawn $\parallel$ to $AS$, and prolong $AC$ to $F$. \step[\indent Then]{$\angle ECF + \angle ECB + \angle BCA = 2$ rt.~$\angle_s$,}{§~89} \pnote{(the sum of all the $\angle_s$ about a point on the same side of a straight line passing through the point is equal to $2$~rt.~$\angle_s$).} \eq[\indent But]{$\angle A$}{$= \angle ECF$,}{§~112} \pnote{(being ext.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$),} \eq[and]{$\angle B$}{$= \angle BCE$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$).} Put for the $\angle_s ECF$ and $BCE$ their equals, the $\angle_s A$ and $B$. \step[\indent Then]{$\angle A +\angle B + \angle BCA = 2$ rt.~$\angle_s$.}{\qed} \end{proof} \pp{\cor[1]{The sum of two angles of a triangle is less than two right angles.}} \pp{\cor[2]{If the sum of two angles of a triangle is taken from two right angles, the remainder is equal to the third angle.}} \pp{\cor[3]{If two triangles have two angles of the one equal to two angles of the other, the third angles are equal.}} \scanpage{042.png}% \pp{\cor[4]{If two right triangles have an acute angle of the one equal to an acute angle of the other, the other acute angles are equal.}} \pp{\cor[5]{In a triangle there can be but one right angle, or one obtuse angle.}} \pp{\cor[6]{In a right triangle the two acute angles are together equal to one right angle, or~$90°$.}} \pp{\cor[7]{In an equiangular triangle, each angle is one third of two right angles, or~$60°$.}} \pp{\cor[8]{An exterior angle of a triangle is equal to the sum of the two opposite interior angles, and therefore greater than either of them.}} \proposition{Theorem.} \begin{proof}% \obs{The sum of two sides of a triangle is greater than the third side, and their difference is less than the third side.} \figc{042aa138}{In the triangle $ABC$, let $AC$ be the longest side.} \prove{$AB + BC > AC$, and $AC - BC < AB$}. \step[\indent\textbf{Proof.}]{$AB + BC > AC$,}{§~49} \pnote{(a straight line is the shortest line from one point to another).} \step{Take away $BC$ from both sides.}{} \step[\indent Then]{$AB > AC - BC$,}{Ax.~5} \step[or]{$AC - BC < AB$.}{\qed} \end{proof} \scanpage{043.png}% \proposition{Theorem.} \begin{proof}% \obs{Two triangles are equal if two angles and the included side of the one are equal, respectively, to two angles and the included side of the other.} \figc{043ab139}{In the triangles $ABC$, $DEF$, let the angle $A$ be equal to the angle $D$, $B$ to $E$, and the side $AB$ to $DE$.} \proveq{$\triangle ABC$}{$= \triangle DEF$.} \textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall coincide with its equal, $DE$. \step[\indent Then]{$AC$ will fall along $DF$, and $BC$ along $EF$,}{} \pnote{(for $\angle A = \angle D$, and $\angle B = \angle E$, by hyp.).} \step{$\therefore C$ will fall on $F$,}{§~48} \pnote{(two straight lines can intersect in only one point).} \step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{§~60} \hfill\qed \end{proof} \pp{\cor[1]{Two triangles are equal if a side and any two angles of the one are equal to the homologous side and two angles of the other.}~\hfill§~132} \pp{\cor[2]{Two right triangles are equal if the hypotenuse and an acute angle of the one are equal, respectively, to the hypotenuse and an acute angle of the other.}~\hfill§~133} \pp{\cor[3]{Two right triangles are equal if a leg and an acute angle of the one are equal, respectively, to a leg and the homologous acute angle of the other.}~\hfill§~133} \scanpage{044.png}% \proposition{Theorem.} \begin{proof}% \obs{Two triangles are equal if two sides and the included angle of the one are equal, respectively, to two sides and the included angle of the other.} \figc{044ab143}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$, and the angle $A$ to the angle $D$.} \proveq{$\triangle ABC$}{$= \triangle DEF$.} \textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall coincide with its equal, $DE$. \step{Then $AC$ will fall along $DF$,}{} \pnote{(for $\angle A = \angle D$, by hyp.);} \step{and $C$ will fall on $F$,}{} \pnote{(for $AC = DF$, by hyp.).} \step{$\therefore CB = FE$,}{} \pnote{(their extremities being the same points).} \step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{\qed} \end{proof} \pp{\cor{Two right triangles are equal if their legs are equal, each to each.}} \note{In §~139 we have given two angles and the included side, in §~143 two sides and the included angle; hence, by interchanging the words \emph{sides} and \emph{angles}, either theorem is changed to the other. This is called the \emph{Principle of Duality}\label{princduality}, or the \emph{Principle of Reciprocity}\label{princreciprocity}. The reciprocal of a theorem is not always true, just as the converse of a theorem is not always true.} \scanpage{045.png}% \proposition{Theorem.} \begin{proof}% \textit{In an isosceles triangle the angles opposite the equal sides are equal.} \figc{045aa145}{Let $ABC$ be an isosceles triangle, having $AB$ and $AC$ equal.} \proveq{$\angle B$}{$= \angle C$.} \textbf{Proof.} Suppose $AD$ drawn so as to bisect the $\angle BAC$. \step{In the $\triangle_s ADB$ and $ADC$,}{} \eq{$AB$}{$=AC$,}{Hyp.} \eq{$AD$}{$=AD$,}{Iden.} \eq{and $\angle BAD$}{$= \angle CAD$.}{Const.} \eq{$\therefore \triangle ADB$}{$= \triangle ADC$,}{§~143} \pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$ of the one are equal, respectively, to two sides and the included $\angle$ of the other).} \eq{$\therefore \angle B$}{$= \angle C$,}{§~128} \pnote{(being homologous angles of equal triangles).} \hfill\qed \end{proof} \pp{\cor{An equilateral triangle is equiangular, and each angle is two thirds of a right angle.}} \ex{If the equal sides of an isosceles triangle are produced, the angles on the other side of the base are equal.} \scanpage{046.png}% \proposition{Theorem.} \begin{proof}% \obs{If two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles.} \figc{046aa147}{In the triangle $ABC$, let the angle $B$ be equal to the angle~$C$.} \proveq{$AB$}{$= AC$.} \step[\indent\textbf{Proof.}]{Suppose $AD$ drawn $\perp$ to $BC$.}{} In the rt.~$\triangle_s ADB$ and $ADC$, \eq{$AD$}{$= AD$,}{Iden.} \eq{and $\angle B$}{$= \angle C$.}{Hyp.} \eq{$\therefore$ rt.~$\triangle ADB$}{$=$ rt.~$\triangle ADC$,}{§~142} \pnote{(having a leg and an acute $\angle$ of the one equal, respectively, to a leg and the homologous acute $\angle$ of the other).} \eq{$\therefore AB$}{$= AC$,}{§~128} \pnote{(being homologous sides of equal $\triangle_s$).} \hfill\qed \end{proof} \pp{\cor[1]{An equiangular triangle is also equilateral.}} \pp{\cor[2]{The perpendicular from the vertex to the base of an isosceles triangle bisects the base, and bisects the vertical angle of the triangle.}} \scanpage{047.png}% \proposition{Theorem.} \begin{proof}% \obs{Two triangles are equal if the three sides of the one are equal, respectively, to the three sides of the other.} \figc{047ab150}{In the triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, $AC$ to $A'C'$, $BC$ to $B'C'$.} \proveq{$\triangle ABC$}{$= \triangle A'B'C'$.} \textbf{Proof.} Place $\triangle A'B'C'$ in the position $\triangle AB'C$ having its greatest side $\triangle A'C'$ in coincidence with its equal $\triangle AC$, and its vertex at $B'$, opposite $B$; and draw $BB'$. \eq[\indent Since]{$AB$}{$=AB'$}{Hyp.} \eq{$\angle ABB'$}{$= \angle AB'B$}{§~145} \pnote{(in an isosceles $\triangle$ the $\angle_s$ opposite the equal sides are equal).} \eq[\indent Since]{$CB$}{$= CB'$,}{Hyp.} \eq{$\angle CBB'$}{$= \angle CB'B$.}{§~145} \eq{$\therefore \angle ABB' + \angle CBB'$}{$= \angle AB'B + \angle CB'B$.}{Ax.~2} \eq[\indent Hence,]{$\angle ABC$}{$= \angle AB'C$.}{} \eq{$\therefore \triangle ABC$}{$= \triangle AB'C$,}{§~143} \pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$ of the one are equal, respectively, to two sides and the included $\angle$ of the other).} \eq{$\therefore \triangle ABC$}{$= \triangle A'B'C'$.}{\qed} \end{proof} \scanpage{048.png}% \proposition{Theorem.} \begin{proof}% \obs{Two right triangles are equal if a leg and the hypotenuse of the one are equal, respectively, to a leg and the hypotenuse of the other.} \figc{048ac151}{In the right triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, and $AC$ to $A'C'$.} \proveq{$\triangle ABC$}{$= \triangle A'B'C'$.} \textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle A'B'C'$, so that $AB$ shall coincide with $A'B'$, $A$ falling on $A'$, $B$ on $B'$, and $C$ and $C'$ on opposite sides of $A'B'$. \step[\indent Then]{$BC$ will fall along $C'B'$ produced,}{} \pnote{(for $\angle ABC = \angle A'B'C'$, each being a rt.~$\angle$.).} \eq[\indent Since]{$AC$}{$= A'C'$,}{Hyp.} \step{the $\triangle A'CC'$ is an isosceles triangle.}{§~120} \eq{$\therefore \angle C$}{$= \angle C'$,}{§~145} \pnote{($\angle_s$ opposite the equal sides of an isosceles $\triangle$ are equal).} \step{$\therefore \triangle_s ABC$ and $A'B'C'$ are equal,}{§~141} \pnote{(two right $\triangle_s$ are equal if they have the hypotenuse and an acute $\angle$ of, the one equal to the hypotenuse and an acute $\angle$ of the other).} \hfill\qed \end{proof} \ex{How many degrees are there in each of the acute angles of an isosceles right triangle?} \scanpage{049.png}% \proposition{Theorem.} \begin{proof}% \obs{If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.} \figc{049aa152}{In the triangle $ACB$, let $AB$ be greater than~$AC$.} \prove{$\angle ACB$ is greater than $\angle B$.} \step[\indent\textbf{Proof.}]{On $AB$ take $AE$ equal to $AC$.}{} \step{Draw $EC$.}{} \eq{$\angle AEC$}{$= \angle ACE$}{§~145} \pnote{(being $\angle_s$ opposite equal sides).} \step[\indent But]{$\angle AEC$ is greater than $\angle B$}{§~137} \pnote{(an exterior $\angle$ of a $\triangle$ is greater than either opposite interior $\angle$),} \step[and]{$\angle ACB$ is greater than $\angle ACE$.}{Ax.~8} \step{Substitute for $\angle ACE$ its equal $\angle AEC$,}{} \step[then]{$\angle ACB$ is greater than $\angle AEC$.}{} Since $\angle AEC$ is greater than $\angle B$, and $\angle ACB$ is greater than $\angle AEC$, \step{$\angle ACB$ is greater than $\angle B$.}{\qed} \end{proof} \ex{If any angle of an isosceles triangle is equal to two thirds of a right angle~($60°$), what is the value of each of the two remaining angles?} \ex{One angle of a triangle is~$34°$. Find the other angles, if one of them is twice the other.} \scanpage{050.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Reciprocally:} If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle.} \figc{050aa153}{In the triangle $ACB$, let the angle $C$ be greater than the angle~$B$.} \prove{$AB > AC$.} \step[\indent\textbf{Proof.}]{Now $AB=AC$, or $< AC$, or $>AC$.}{} \label{41} \step{But $AB$ is not equal to $AC$;}{} \step{for then the $\angle C$ would be equal to the $\angle B$,}{§~145} \pnote{(being $\angle_s$ opposite equal sides).} \step{And $AB$ is not less than $AC$;}{} \step{for then the $\angle C$ would be less than the $\angle B$.}{§~152} Both these conclusions are contrary to the hypothesis that the $\angle C$ is greater than the $\angle B$. \step{Hence, $AB$ cannot be equal to $AC$ or less than $AC$.}{} \step{$\therefore AB > AC$.}{\qed} \end{proof} \ex{If the vertical angle of an isosceles triangle is equal to~$30°$, find the exterior angle included by a side and the base produced.} \ex{If the vertical angle of an isosceles triangle is equal to~$36°$, find the angle included by the bisectors of the base angles.} \scanpage{051.png}% \proposition{Theorem.} \begin{proof}% \obs{If two triangles have two sides of the one equal, respectively, to two sides of the other, but the included angle of the first triangle greater than the included angle of the second, then the third side of the first is greater than the third side of the second.} \figc{051ac154}{In the triangles $ABC$ and $ABE$, let $AB$ be equal to $AB$, $BC$ to $BE$; but let the angle $ABC$ be greater than the angle $ABE$.} \proveq{$AC$}{$> AE$.} \textbf{Proof.} Place the $\triangle_s$ so that $AB$ of the one shall fall on $AB$ of the other, and $BE$ within the $\angle ABC$. Suppose $BF$ drawn to bisect the $\angle EBC$, and draw $EF$. The $\triangle_s EBF$ and $CBF$ are equal.~\hfill§~143 \eq[\indent For]{$BF$}{$= BF$,}{Iden.} \eq{$BE$}{$=BC$,}{Hyp.} \eq[and]{$\angle EBF$}{$=\angle CBF$.}{Const.} \eq{$\therefore EF$}{$=FC$.}{§~128} \eq[\indent Now]{$AF+FE$}{$> AE$.}{§~138} \eq{$\therefore AF+FC$}{$> AE$.}{} \eq{$\therefore AC$}{$> AE$.}{\qed} \end{proof} \scanpage{052.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely}: If two sides of a triangle are equal, respectively, to two sides of another, but the third side of the first triangle is greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle opposite the third side of the second.} \figc{052ab155}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$; but let $BC$ be greater than EF.} \prove{the $\angle A$ is greater than the $\angle D$.} \textbf{Proof}. Now the $\angle A$ is equal to the $\angle D$, or less than the $\angle D$, or greater than the $\angle D$. \step{But the $\angle A$ is not equal to the $\angle D$;}{} \step{for then the $\triangle ABC$ would be equal to the $\triangle DEF$,}{§~143} \pnote{(having two sides and the included $\angle$ of the one equal, respectively, to two sides and the included $\angle$ of the other),} \step{and $BC$ would be equal to $EF$.}{} And the $\angle A$ is not less than the $\angle D$, for then $BC$ would be less than $EF$.~\hfill§~154 Both these conclusions are contrary to the hypothesis that $BC$ is greater than $EF$. Since the $\angle A$ is not equal to the $\angle D$ or less than the $\angle D$, \step{the $\angle A$ is greater than the $\angle D$.}{\qed} \end{proof} \scanpage{053.png}% \section{LOCI OF POINTS.} \begin{point}% If it is required to find a point which shall fulfil a \emph{single} geometric condition, the point may have an \emph{unlimited number of positions}. If, however, all the points are in the same plane, the required point will be confined to a \emph{particular line}, or \emph{group of lines}. A point in a plane at a given distance from a fixed straight line of indefinite length in that plane, is evidently in one of two straight lines, so drawn as to be everywhere at the given distance from the fixed line, one on one side of the fixed line, and the other on the other side of it. A point in a plane equidistant from two parallel lines in that plane is evidently in a straight line drawn between the two given parallel lines and everywhere equidistant from them. \end{point} \begin{point}% All points in a plane that satisfy a single geometrical condition lie, in general, in a line or group of lines; and this line or group of lines is called the \textbf{locus} of the points that satisfy the given condition. \end{point} \begin{point}% To prove \emph{completely} that a certain line is the locus of points that fulfil a given condition, it is necessary to prove 1. \textit{Any point in the line satisfies the given condition; and any point not in the line does not satisfy the given condition.} Or, to prove 2. \textit{Any point that satisfies the given condition lies in the line; and any point in the line satisfies the given condition}. \end{point} \note{The word \emph{locus} (pronounced lo\'{ }kus) is a Latin word that signifies \emph{place}. The plural of locus is loci (pronounced lo\'{ }si).} \pp{\defn{A line which bisects a given line and is perpendicular to it is called the \textbf{perpendicular bisector} of the line.}} \scanpage{054.png}% \proposition{Theorem.} \begin{proof}% \obs{The perpendicular bisector\label{perpbisector} of a given line is the locus of points equidistant from the extremities of the line.} \figc{054aa160}{Let $PR$ be the perpendicular bisector of the line $AB$, $O$ any point in $PR$, and $C$ any point not in $PR$.} \step{Draw $OA$ and $OB$, $CA$ and $CB$.}{} \prove[To prove ]{$OA$ and $OB$ equal, $CA$ and $CB$ unequal.} \eq[\indent\textbf{Proof.}]{\textbf{1. }$\triangle OPA$}{$= \triangle OPB$,}{§~144} \step{for $PA = PB$ by hypothesis, and $OP$ is common,}{} \pnote{(two right $\triangle_s$ are equal if their legs are equal, each to each).} \eq{$\therefore OA$}{$= OB$.}{§~128} \textbf{2.~}Since $C$ is not in the $\perp$, $CA$ or $CB$ will cut the $\perp$. \step{Let $CA$ cut the $\perp$ at $D$, and draw $DB$.}{} Then, by the first part of the proof $DA = DB$. \step[\indent But]{$CB < CD + DB$.}{§~138} \step{$\therefore CB < CD + DA$.}{} \step[\indent That is, ]{$CB < CA$.}{} $\therefore PR$ is the locus of points equidistant from $A$ and $B$.~\hfill§~158,1 \hfill\qed \end{proof} \pp{\cor{Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line.}} \scanpage{055.png}% \proposition{Theorem.} \begin{proof}% \obs{The bisector of a given angle is the locus of points equidistant from the sides of the angle.} \vspace{-1ex} \figc{055aa162}{Let $O$ be any point equidistant from the sides of the angle $PAQ$.} \prove{$O$ is in the bisector of the $\angle PAQ$.} \step [\indent\textbf{Proof.}] {Draw $AO$.} {} \step {Suppose $OF$ drawn $\perp$ to $AP$ and $OG \perp$ to $AQ$.} {} \step {In the rt.~$\triangle_s AFO$ and $AGO$,} {} \eq {$AO $}{$= AO$,} {Iden.} \eq {$OF $}{$= OG$,} {Hyp.} \eq {$\therefore \triangle AFO $} {$= \triangle AGO$.} {§~151} \eq {$\therefore \angle FAO $} {$= \angle GAO$.} {§~128} \step {$\therefore O$ is in the bisector of the $\angle PAQ$.}{} \textbf{Let $\mathbf{D}$ be any point in the bisector of the angle $\mathbf{PAQ}$.} \prove {$D$ is equidistant from $AP$ and $AQ$.} \step[\indent\textbf{Proof.}]{Suppose $DB$ drawn $\perp$ to $AP$ and $DC \perp$ to~$AQ$.}{} \step {In the rt.~$\triangle_s ABD$ and $ACD$,} {} \eq {$AD $} {$= AD$,} {Iden.} \eq {$\angle DAB $} {$= \angle DAC$,} {Hyp.} \eq {$\therefore \triangle ABD $} {$= \triangle ACD$.} {§~141} \eq {$\therefore DB $} {$= DC$.} {§~128} \step{$\therefore D$ is equidistant from $AP$ and $AQ$.}{} \step{$\therefore$ the bisector of the $\angle PAQ$ is the locus of points that are equidistant from its sides.}{§~158, 2} \end{proof} \scanpage{056.png}% \section{QUADRILATERALS.} \begin{point}% A \indexbf{quadrilateral} is a portion of a plane bounded by four straight lines. The bounding lines are the \textbf{sides}, the angles formed by these sides are the \textbf{angles}, and the vertices of these angles are the \textbf{vertices}, of the quadrilateral. \end{point} \begin{point}% A \indexbf{trapezium} is a quadrilateral which has no two sides parallel. \end{point} \begin{point}% A \indexbf{trapezoid} is a quadrilateral which has two sides, and only two sides, parallel. \end{point} \begin{point}% A \indexbf{parallelogram} is a quadrilateral which has its opposite sides parallel. \end{point} \figc{056ac166}{} \begin{point}% A \indexbf{rectangle} is a parallelogram which has its angles right angles. \end{point} \begin{point}% A \indexbf{square} is a rectangle which has its sides equal. \end{point} \begin{point}% A \indexbf{rhomboid} is a parallelogram which has its angles oblique angles. \end{point} \begin{point}% A \indexbf{rhombus} is a rhomboid which has its sides equal. \end{point} \figc{056dg170}{} \begin{point}% The side upon which a parallelogram stands, and the opposite side, are called its lower and upper \emph{bases}\label{basepar}. \end{point} \scanpage{057.png}% \begin{point}% Two parallel sides of a trapezoid are called its \textbf{bases}\label{basetrap}, the other two sides its \textbf{legs}\label{legstrap}, and the line joining the middle points of the legs is called the \textbf{median} of the trapezoid\label{mediantrap}. \end{point} \figc{057aa174}{} \begin{point}% A trapezoid is called an \indexbf{isosceles trapezoid} if its legs are equal. \end{point} \begin{point}% The \textbf{altitude} of a parallelogram\label{altpar} or trapezoid\label{alttrap} is the perpendicular distance between its bases, as $PQ$. \end{point} \begin{point}% A \textbf{diagonal}\label{diagonal1} of a quadrilateral is a straight line joining two opposite vertices, as $AC$. \end{point} \proposition{Theorem.} \begin{proof}% \obs{Two angles whose sides are parallel, each to each, are either equal or supplementary.} \figc{057bb176}{Let $BA$ be parallel to $HD$, and $BC$ be parallel to $MN$.} \prove[To prove ]{$\angle_s a$, $a'$ and $c$ equal; $a$ and $c'$ supplementary.} \step[\indent\textbf{Proof.}]{Let $HD$ and $BC$ prolonged intersect at $x$.}{} \step[\indent Then]{$\angle a = \angle x$, and $\angle a' = \angle x$.}{§~112} \step{$\therefore \angle a = \angle a'$.}{Ax.~1} \step[\indent Also]{$\angle c = \angle a'$ (§~93). $\therefore \angle c = \angle a$.}{Ax.~1} \step[\indent Now]{$\angle a'$ and $\angle c'$ are supplementary.}{§~89} \step{Put $\angle a$ for its equal, $\angle a'$.}{} \step[\indent Then]{$\angle a$ and $\angle c'$ are supplementary.}{\qed} \end{proof} \pp{\cor{The opposite angles of a parallelogram are equal, and the adjacent angles are supplementary.}} \scanpage{058.png}% \proposition{Theorem.} \begin{proof}% \obs{The opposite sides of a parallelogram are equal.} \figc{058aa178}{Let the figure $ABCE$ be a parallelogram.} \prove[To prove ]{$BC = AE$, and $AB = EC$.} \step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{} \step{$\triangle ABC = \triangle CEA$.}{§~139} \step{For $AC$ is common,}{} \step{$\angle BAC = \angle ACE$, and $\angle ACB = \angle CAE$,}{§~110} \pnote{(being alt-int. $\angle_s$ of $\parallel$ lines).} \step{$\therefore BC = AE$, and $AB = CE$,}{§~128} \pnote{(being homologous sides of equal $\triangle_s$).} \hfill\qed \end{proof} \pp{\cor[1]{A diagonal divides a parallelogram into two equal triangles.}} \pp{\cor[2]{Parallel lines comprehended between parallel lines are equal.}} \figc{058bb181}{} \begin{point}% \cor[3]{Two parallel lines are everywhere equally distant.} For if $AB$ and $DC$ are parallel, $\perp_s$ dropped from \emph{any} points in $AB$ to $DC$, are equal, §~180. Hence, \emph{all} points in $AB$ are equidistant from~$DC$. \end{point} \scanpage{059.png}% \proposition{Theorem.} \begin{proof}% \obs{If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.} \figc{059aa182}{Let the figure $ABCE$ be a quadrilateral, having $BC$ equal to $AE$ and $AB$ to $EC$.} \prove{the figure $ABCE$ is a $\Par$.} \step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{} In the $\triangle_s ABC$ and $CEA$, \eq{$BC$}{$= AE$,}{Hyp.} \eq{$AB$}{$= CE$,}{Hyp.} \eq[and]{$AC$}{$= AC$,}{Iden.} \eq{$\therefore \triangle ABC$}{$= \triangle CEA$,}{§~150} \pnote{(having three sides of the one equal, respectively, to the three sides of the other).} \eq{$\therefore \angle ACB$}{$= \angle CAE$,}{§~128} \eq[and]{$\angle BAC$}{$= \angle ACE$,}{} \pnote{(being homologous $\angle_s$ of equal $\triangle_s$).} \eq{$\therefore BC$}{is $\parallel$ to $AE$,}{} \eq[and]{$AB$}{is $\parallel$ to $EC$,}{§~111} \pnote{(two lines in the same plane cut by a transversal are parallel, if the alt.-int.~$\angle_s$ are equal).} \step{$\therefore$ the figure $ACBE$ is a $\Par$,}{§~166} \pnote{(having its opposite sides parallel).}\hfill\llap{\qed} \end{proof} \scanpage{060.png}% \proposition{Theorem.} \begin{proof}% \obs{If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and parallel, and the figure is a parallelogram.} \figc{060aa183}{Let the figure $ABCE$ be a quadrilateral, having the side $AE$ equal and parallel to $BC$.} \prove{$AB$ is equal and parallel to $EC$.} \step[\indent\textbf{Proof.}]{Draw $AC$.}{} The $\triangle_s ABC$ and $CEA$ are equal,~\hfill§~143 \pnote{(having two sides and the included $\angle$ of each equal, respectively).} \step[\indent For]{$AC$ is common,}{} \eq{$BC$}{$=AE$}{Hyp.} \eq[and]{$\angle BCA$}{$= \angle CAE$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).} \eq{$\therefore AB$}{$=EC$,}{} \eq[and]{$\angle BAC$}{$= \angle ACE$,}{§~128} \pnote{(being homologous parts of equal $\triangle_s$).} \step{$\therefore AB$ is $\parallel$ to $EC$,}{§~111} \pnote{(two lines are $\parallel$, if the alt.-int. $\angle_s$ are equal).} \step{$\therefore$ the figure $ABCE$ is a $\Par$,}{§~166} \pnote{(the opposite sides being parallel).} \hfill\qed \end{proof} \scanpage{061.png}% \proposition{Theorem.} \begin{proof}% \obs{The diagonals of a parallelogram bisect each other.} \figc{061aa184}{Let the figure $ABCE$ be a parallelogram, and let the diagonals $AC$ and $BE$ cut each other at $O$.} \prove{$AO = OC$, and $BO = OE$.} \textbf{Proof.} In the $\triangle_s AOE$ and $COB$, \eq{$AE$}{$=BC$,}{§~178} \pnote{(being opposite sides of a $\Par$).} \eq{$\angle OAE$}{$=\angle OCB$,}{§~110} \eq{and $\angle OEA$}{$= \angle OBC$,}{} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).} \eq{$\therefore\triangle AOE$}{$=\triangle COB$,}{§~139} \pnote{(having two $\angle_s$ and the included side of the one equal, respectively, to two $\angle_s$ and the included side of the other).} \step{$\therefore AO=OC$, and $BO=OE$,}{§~128} \pnote{(being homologous sides of equal $\triangle_s$).} \hfill\qed \end{proof} \ex{The median from the vertex to the base of an isosceles triangle is perpendicular to the base, and bisects the vertical angle.} \ex{If two straight lines are cut by a transversal so that the alternate-exterior angles are equal, the two straight lines are parallel.} \ex{If two parallel lines are cut by a transversal, the two exterior angles on the same side of the transversal are supplementary.} \ex{If two straight lines are cut by a transversal so as to make the exterior angles on the same side of the transversal supplementary, the two lines are parallel.} \scanpage{062.png}% \proposition{Theorem.} \begin{proof}% \obs{Two parallelograms are equal, if two sides and the included angle of the one are equal, respectively, to two sides and the included angle of the other.} \figc{062ab185}{In the parallelograms $ABCD$ and $A'B'C'D'$, let $AB$ be equal to $A'B'$, $AD$ to $A'D'$, and angle $A$ to $A'$.} \prove{the $\Par_s$ are equal}. \textbf{Proof.} Place the $\Par$ $ABCD$ on the $\Par$ $A'B'C'D'$, so that $AD$ will fall on and coincide with its equal, $A'D'$. \step{Then $AB$ will fall on $A'B'$, and $B$ on $B'$;}{} \pnote{(for $\angle A = \angle A'$, and $AB = A'B'$, by hyp.)} Now, $BC$ and $B'C'$ are both $\parallel$ to $A'D'$ and drawn through $B'$. \step{$\therefore BC$ and $B'C'$ coincide,}{§~105} \pnote{(through a given point only one line can be drawn $\parallel$ to a given line).} Also $DC$ and $D'C'$ are $\parallel$ to $A'B'$ and drawn through $D'$. \step{$\therefore DC$ and $D'C'$ coincide.}{§~105} \step{$\therefore C$ falls on $C'$,}{§~48} \pnote{(two lines can intersect in only one point),} \step{$\therefore$ the two $\Par_s$ coincide, and are equal.}{\qed} \end{proof} \pp{\cor{Two rectangles having equal bases and altitudes are equal.}} \scanpage{063.png}% \proposition{Theorem.} \begin{proof}% \obs{If three or more parallels intercept equal parts on one transversal, they intercept equal parts on every transversal.} \figc{063aa187}{Let the parallels $AH$, $BK$, $CM$, $DP$ intercept equal parts $HK$, $KM$, $MP$ on the transversal $HP$.} \prove{they intercept equal parts $AB$, $BC$, $CD$ on the transversal~$AD$.} \textbf{Proof.} Suppose $AH$, $BF$, and $CG$ drawn $\parallel$ to $HP$. \step{$\angle_s$ $AEB$, $BFC$, etc.\ $=\angle_s$ $HKE$, $KMF$, etc., respectively.}{§~112} \step{But $\angle_s$ $HKE$, $KMF$, etc.\ are equal.}{§~112} \step{$\therefore \angle_s$ $AEB$, $BFC$, etc.\ are equal.}{Ax.~1} \step{Also $\angle_s$ $BAE$, $CBF$, etc.\ are equal.}{§~112} \step{Now $AE = HK$, $BF = KM$, $CG = MP$,}{§~180} \pnote{(parallels comprehended between parallels are equal).} \step{$\therefore AE = BF = CG$.}{Ax.~1} \step{$\therefore \triangle ABE = \triangle BCF = \triangle CDG$,}{§~139} \pnote{(having two $\angle_s$ and the included side of each respectively equal).} \step{$\therefore AB = BC = CD$.}{§~128} \hfill\qed \end{proof} \scanpage{064.png}% \figc{064aa188}{} \begin{point}% \cor[1]{If a line is parallel to the base of a triangle and bisects one side, it bisects the other side also.} Let $DE$ be $\parallel$ to $BC$ and bisect $AB$. Suppose a line is drawn through $A \parallel$ to $BC$. Then this line is $\parallel$ to $DE$, by §~106. The three parallels by hypothesis intercept equal parts on the transversal $AB$, and therefore, by §~187, they intercept equal parts on the transversal $AC$; that is, the line $DE$ bisects $AC$. \end{point} \begin{point}% \cor[2]{The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side.} A line drawn through $D$, the middle point of $AB$, $\parallel$ to $BC$, passes through $E$, the middle point of $AC$, by §~188. Therefore the line joining $D$ and $E$ coincides with this parallel and is $\parallel$ to $BC$. Also, since $EF$ drawn $\parallel$ to $AB$ bisects $AC$, it bisects $BC$, by §~188; that is, $BF=FC = \frac{1}{2}BC$. But $BDEF$ is a $\Par$ by §~166, and therefore $DE = BF = \frac{1}{2}BC$. \end{point} \figc{064bb190}{} \begin{point}% \cor[3]{The median of a trapezoid is parallel to the bases, and is equal to half the sum of the bases.} Draw the diagonal $DB$. In the $\triangle ADB$ join $E$, the middle point of $AD$, to $F$, the middle point of $DB$. Then, by §~189, $EF$ is $\parallel$ to $AB$ and $= \frac{1}{2}AB$. In the $\triangle DBC$ join $F$ to $G$, the middle point of $BC$. Then $FG$ is $\parallel$ to $DC$ and $=\frac{1}{2}DC$. $AB$ and $FG$, being $\parallel$ to $DC$, are $\parallel$ to each other. But only one line can be drawn through $F \parallel$ to $AB$ (§~105). Therefore $FG$ is the prolongation of $EF$. Hence, $EFG$ is parallel to $AB$ and $DC$, and equal to $\frac{1}{2} (AB + DC)$. \end{point} \scanpage{065.png}% \section{POLYGONS IN GENERAL.} \begin{point}% A \indexbf{polygon} is a portion of a plane bounded by straight lines. The bounding lines are the sides\label{polysides}, and their sum, the \textbf{perimeter}\label{perimeter2} of the polygon. The angles included by the adjacent sides are the \textbf{angles}\label{polyangles} of the polygon, and the vertices of these angles are the \textbf{vertices} of the polygon\label{polyvertices}. The number of sides of a polygon is evidently equal to the number of its angles. \end{point} \begin{point}% A \textbf{diagonal}\label{diagonal2} of a polygon is a line joining the vertices of two angles not adjacent; as, $AC$ (Fig.~1). \figc{065ac192}{} \end{point} \begin{point}% An \indexbf{equilateral polygon} is a polygon which has all its sides equal. \end{point} \begin{point}% An \indexbf{equiangular polygon} is a polygon which has all its angles equal. \end{point} \begin{point}% A \indexbf{convex polygon} is a polygon of which no side, when produced, will enter the polygon. \end{point} \begin{point}% A \indexbf{concave polygon} is a polygon of which two or more sides, if produced, will enter the polygon. \end{point} \begin{point}% Each angle of a convex polygon (Fig.~2) is called a \emph{salient}\label{salient} angle, and is less than a straight angle. \end{point} \begin{point}% The angle $EDF$ of the concave polygon (Fig.~3) is called a \emph{re-entrant} angle, and is greater than a straight angle. When the term polygon is used, a \emph{convex} polygon is meant. \end{point} \scanpage{066.png}% \begin{point}% Two polygons are \emph{equal} when they can be divided by diagonals into the same number of triangles, equal each to each, and similarly placed; for if the polygons are applied to each other, the corresponding triangles will coincide, and hence the polygons will coincide and be equal. \end{point} \begin{point}% Two polygons are \indexemph{mutually equiangular}, if the angles of the one are equal to the angles of the other, each to each, when taken in the same order. Figs.~1 and 2. \end{point} \begin{point}% The equal angles in mutually equiangular polygons are called \emph{homologous} angles\label{homangles}; and the sides which are included by homologous angles are called \emph{homologous} sides\label{homsides}. \end{point} \begin{point}% Two polygons are \indexemph{mutually equilateral}, if the sides of the one are equal to the sides of the other, each to each, when taken in the same order. Figs.~1 and 2. \figc{066ad203}{} \end{point} \begin{point}% Two polygons may be mutually equiangular without being mutually equilateral; as, Figs.~4 and 5. And, \emph{except in the case of triangles}, two polygons may be mutually equilateral without being mutually equiangular; as, Figs.~6 and 7. If two polygons are mutually equilateral and mutually equiangular \emph{they are equal}, for they can be made to coincide. \end{point} \begin{point}% A polygon of three sides is called a \emph{triangle}\label{triangle2}; one of four sides, a \emph{quadrilateral}\label{quadrilateral2}; one of five sides, a \indexemph{pentagon}; one of six sides, a \indexemph{hexagon}; one of seven sides, a \indexemph{heptagon}; one of eight sides, an \indexemph{octagon}; one of ten sides, a \indexemph{decagon}; one of twelve sides, a \indexemph{dodecagon}. \end{point} \scanpage{067.png}% \proposition{Theorem.} \begin{proof}% \obs{The sum of the interior angles of a polygon is equal to two right angles, taken as many times less two as the figure has sides.} \figc{067aa205}{Let the figure $ABCDEF$ be a polygon, having $n$ sides.} \prove{$\angle A + \angle B + \angle C$, etc.\ $= (n-2)2$ rt.~$\angle_s$.} \textbf{Proof.} From $A$ draw the diagonals $AC$, $AD$, and $AE$. The sum of the $\angle_s$ of the $\triangle_s$ is equal to the sum of the $\angle_s$ of the polygon. \step{Now, there are $(n-2)$~$\triangle_s$,}{} \step{and the sum of the $\angle_s$ of each $\triangle = 2$ rt.~$\angle_s$.}{§~129} $\therefore$ the sum of the $\angle_s$ of the $\triangle_s$, that is, the sum of the $\angle_s$ of the polygon is equal to $(n-2) 2$ rt.~$\triangle_s$.~\hfill\qed \end{proof} \pp{\cor{The sum of the angles of a quadrilateral equals 4 right angles; and if the angles are all equal, each is a right angle. In general, each angle of an equiangular polygon of $n$ sides is equal to $\displaystyle \frac{2(n-2)}{n}$ right angles.}} \ex{How many diagonals can be drawn in a polygon of $n$ sides?} \scanpage{068.png}% \proposition{Theorem.} \begin{proof}% \obs{The exterior angles of a polygon, made by producing each of its sides in succession, are together equal to four right angles.} \figc{068aa207}{Let the figure $ABCDE$ be a polygon, having its sides produced in succession.} \prove[To prove ]{the sum of the ext.~$\angle_s = 4$ rt.~$\angle_s$.} \textbf{Proof.} Denote the int.~$\angle_s$ of the polygon by $A$, $B$, $C$, $D$, $E$, and the corresponding ext.~$\angle_s$ by $a$, $b$, $c$, $d$, $e$. \eq{$\angle A + \angle a$}{$= 2$ rt.~$\angle_s$,}{§~89} \eq[and]{$\angle B + \angle b$}{$= 2$ rt.~$\angle_s$,}{} \pnote{(being sup.-adj.~$\angle_s$).} In like manner each pair of adj.~$\angle_s = 2$ rt.~$\angle_s$. $\therefore$ the sum of the interior and exterior $\angle_s$ of a polygon of $n$ sides is equal to $2n$ rt.~$\angle_s$. %[** TN: ad hoc visual formatting] But the sum of the interior~$\angle_s = (n-2) 2$ rt.~$\angle_s$\hfill§~205 \\ $\phantom{\text{\indent But the sum of the interior}~\angle_s} = 2 n$ rt.~$\angle_s - 4$ rt.~$\angle_s$. \step{$\therefore$ the sum of the exterior $\angle_s = 4$ rt.~$\angle_s$.}{\qed} \end{proof} \ex{How many sides has a polygon if the sum of its interior $\angle_s$ is twice the sum of its exterior $\angle_s$? ten times the sum of its exterior $\angle_s$?} \scanpage{069.png}% \section{SYMMETRY.} \label{symmetry} \begin{point}% Two points are said to be \textbf{symmetrical} with respect to a third point, called the \textbf{centre of symmetry}\label{centresym}, if this third point bisects the straight line which joins them. \figc{069ac208}{} Two points are said to be \emph{symmetrical} with respect to a straight line, called the \textbf{axis of symmetry}\label{axissym}, if this straight line bisects at right angles the straight line which joins them. Thus, $P$ and $P'$ are symmetrical with respect to $O$ as a centre, and $XX'$ as an axis, if $O$ bisects the line $PP'$, and if $XX'$ bisects $PP'$ at right angles. \end{point} \begin{point}% A figure is symmetrical with respect to a point as a centre of symmetry, if the point bisects every straight line drawn through it and terminated by the boundary of the figure. \end{point} \begin{point}% A figure is symmetrical with respect to a line as an axis of symmetry if one of the parts of the figure coincides, point for point, with the other part when it is folded over on that line as an axis. \end{point} \figc{069dd211}{} \begin{point}% Two figures are said to be symmetrical with respect to an axis if every point of one has a corresponding symmetrical point in the other. Thus, if every point in the figure $A'B'C'$ has a symmetrical point in $ABC$, with respect to $XX'$ as an axis, the figure $A'B'C'$ is symmetrical to $ABC$ with respect to $XX'$ as an axis. \end{point} \scanpage{070.png}% \proposition{Theorem.} \begin{proof}% \obs{A quadrilateral which has two adjacent sides equal, and the other two sides equal, is symmetrical with respect to the diagonal joining the vertices of the angles formed by the equal sides, and the diagonals are perpendicular to each other.} \figc{070aa212}{Let $ABCD$ be a quadrilateral, having $AB$ equal to $AD$, and $CB$ equal to $CD$, and having the diagonals $AC$ and $BD$.} \prove{the diagonal $AC$ is an axis of symmetry, and that it is $\perp$ to the diagonal $BD$.} \textbf{Proof.} In the $\triangle_s ABC$ and $ADC$, \step{$AB = AD$, and $BC = DC$,}{Hyp.} \eq[and]{$AC$}{$= AC$.}{Iden.} \eq{$\therefore \triangle ABC$}{$= \triangle ADC$.}{§~150} \step{$\therefore \angle BAC = \angle DAC$, and $\angle BCA = \angle DCA$.}{} Hence, if $ABC$ is turned on $AC$ as an axis until it falls on $ADC$, $AB$ will fall upon $AD$, $CB$ on $CD$, and $OB$ on $OD$. \step{$\therefore$ the $\triangle ABC$ will coincide with the $\triangle ADC$.}{} \step{$\therefore AC$ is an axis of symmetry (§~210) and is $\perp$ to $BD$.}{§~208} \hfill\qed \end{proof} \scanpage{071.png}% \proposition{Theorem.} \begin{proof}% \obs{If a figure is symmetrical with respect to two axes perpendicular to each other, it is symmetrical with respect to their intersection as a centre.} \figc{071aa213}{Let the figure $ABCDEFGH$ be symmetrical with respect to the two perpendicular axes $XX'$, $YY'$, which intersect at $O$.} \prove{$O$ is the centre of symmetry of the figure.} \textbf{Proof.} Let $N$ be any point in the perimeter. \step{Suppose $NMI$ drawn $\perp$ to $YY'$, $IKL \perp$ to $XX'$.}{} \step[\indent Then]{$NI$ is $\parallel$ to $XX'$ and $IL$ is $\parallel$ to $YY'$.}{§~104} \step{Draw $LO$, $ON$, and $KM$.}{} \eq[\indent Now]{$KI$}{$= KL$,}{§~208} \pnote{(the figure being symmetrical with respect to $XX'$).} \eq[\indent But]{$KI$}{$=OM$.}{§~180} \step{$\therefore KL=OM$, and $KLOM$ is a $\Par$.}{§~183} \step{$\therefore LO$ is equal and parallel to $KM$.}{§~183} \step{In like manner $ON$ is equal and parallel to $KM$.}{} \step{$\therefore LON$ is a straight line.}{§~105} $\therefore O$ bisects $LN$, \emph{any} straight line and therefore \emph{every} straight line drawn through $O$ and terminated by the perimeter. \step{$\therefore O$ is the centre of symmetry of the figure.}{\qed} \end{proof} \scanpage{072.png}% \subsection{REVIEW QUESTIONS ON BOOK I.} \begin{myenum} \item What is the subject-matter of Geometry? \item What is a geometric magnitude? \item What is an axiom? a theorem? a converse theorem? an opposite theorem? a contradictory theorem? \item Define a straight line; a curved line; a broken line; a plane surface; a curved surface. \item How many points are necessary to determine a straight line? \item How many straight lines are necessary to determine a point? \item On what does the magnitude of an angle depend? \item Define a straight angle; a right angle; an oblique angle. \item Define adjacent angles; complementary angles; supplementary angles; conjugate angles. \item Define parallel lines and give the axiom of parallels. \item If two lines in the same plane are parallel and cut by a transversal, what pairs of angles are equal? what pairs are supplementary? \item Define a right triangle; an isosceles triangle; a scalene triangle. \item To how many right angles is the sum of the angles of a triangle equal? the sum of the acute angles of a right triangle? \item To what angles is the exterior angle of a triangle equal? \item What is the test of equality of two geometric magnitudes? \item How does a reciprocal theorem differ from a converse theorem? \item State the three cases in which two triangles are equal. \item State the cases in which two right triangles are equal. \item What is meant by a locus of points? \item Where are the points located in a plane that are each equidistant from two given points? from two intersecting lines? \item Define a parallelogram; a trapezoid; an isosceles trapezoid. \item When is a figure symmetrical with respect to a centre? \item When is a figure symmetrical with respect to an axis? \item Must a triangle be equiangular if equilateral? must a triangle be equilateral if equiangular? \item When are two polygons said to be mutually equiangular? \item When are two polygons said to be mutually equilateral? \item Can two polygons of more than three sides be mutually equiangular without being mutually equilateral? mutually equilateral without being mutually equiangular? \item What line do two points each equidistant from the extremities of a given straight line determine? \end{myenum} \scanpage{073.png}% \subsection{METHODS OF PROVING THEOREMS.} \begin{point}% There are \emph{three} general methods of proving theorems, the \textbf{synthetic}, the \textbf{analytic}, and the \textbf{indirect} methods. The \emph{synthetic} method is the method employed in most of the theorems already given, and consists in putting together known truths in order to obtain a new truth. The \emph{analytic} method is the reverse of the synthetic method. It asserts that the conclusion is true if another proposition is true, and so on step by step, until a known truth is reached. Thus, proposition $A$ is true if proposition $B$ is true, and $B$ is true if $C$ is true; but $C$ \emph{is} true, hence $A$ and $B$ are true. If a known truth \emph{suggests} the required proof, it is best to use the synthetic form at once. If no proof occurs to the mind, it is necessary to use the analytic method to \emph{discover} the proof, and then the synthetic proof may be given. The \emph{indirect} method, or the method of \emph{reductio ad absurdum}, is illustrated on page \pageref{41}. It consists in proving a theorem to be true by proving its contradictory to be false. \end{point} \begin{point}% Generally auxiliary lines are required, as a line \emph{connecting two points}; a line \emph{parallel to or perpendicular to a given line}; a line \emph{produced by its own length}; a line \emph{making with another line an angle equal to a given angle.} \textbf{Two lines are proved equal} by proving them \emph{homologous sides of equal triangles}; or \emph{legs of an isosceles triangle}; or \emph{opposite sides of a parallelogram.} \textbf{Two angles are proved equal} by proving them \emph{alternate-interior angles or exterior-interior angles of parallel lines}; or \emph{homologous angles of equal triangles}; or \emph{base angles of an isosceles triangle}; or \emph{opposite angles of a parallelogram.} Two suggestions are of special importance to the beginner: \begin{myenum} \item \emph{Draw as accurate figures as possible.} \item \emph{Draw as general figures as possible.} \end{myenum} \end{point} \scanpage{074.png}% \section{EXERCISES.} \exheader{Prove by the analytic method:} \figc{074aaZ19}{} \begin{proofex}% \obs{A median of a triangle is less than half the sum of the two adjacent sides.} \prove[To prove ]{the median $AD < \frac{1}{2}(AB + AC)$.} \eq[\indent Now]{}{$AD < \frac{1}{2}(AB + AC)$,}{} \eq[if]{}{$2AD < AB + AC$.}{} This suggests producing $AD$ by its own length to $E$, and joining $BE$. \eq[\indent Then]{}{$AE=2AD$,}{} \step[and]{$2AD KB$. (Why?) Now $CD > BE$, if $KD$ is greater than or equal to $KE$. But suppose $KD < KE$. Lay off $KH = KD$ and $KG = KB$, join $HG$, and draw $GF \parallel$ to $BE$. $\triangle KDB = \triangle KHG$. (Why?) $\therefore \angle KHG = \angle KDB$. (Why?) $\therefore \angle KEC$ is greater than $\angle KHG$. (Why?) $\therefore GF > HE$. (Why?) $\angle GFC$ is greater than $\angle FCG$ ($\frac{1}{2}ACB$). $\therefore CG > GF$, and $> HE$. $\therefore KC - KG > KE - KH$, or $KC + KD > KB + KE$, or $CD > BE$. \end{proofex} \ex{State the converse theorem of Ex.~54. Is the converse theorem true?} \figc{079eeZ57}{} \begin{proofex}% The perpendiculars dropped from the middle point of the base upon the legs of an isosceles triangle are equal. \step{$\triangle BED = \triangle CFD$ (§~141).}{} \end{proofex} \begin{proofex}% State and prove the converse. \step{$\triangle BED = \triangle CFD$ (§~151).}{} \end{proofex} \scanpage{080.png}% \filbreak \begin{proofex}% The difference of the distances from any point in the base produced of an isosceles triangle to the equal sides of the triangle is constant. Rt. $\triangle DGC=$ rt.~$\triangle DFC$. (Why?) $\therefore DF = DG$. $\therefore DE - DF = DE - DG = EG$, the $\perp$ distance between the two $\parallel_s$, $BA$ and $CH$. \end{proofex} \figcc{080aaZ59}{080bbZ60} \begin{proofex}% The sum of the perpendiculars dropped from any point in the base of an isosceles triangle to the legs is constant, and equal to the altitude upon one of the legs. Let $PE$ and $PD$ be the $\perp_s$ and $BF$ the altitude. Draw $PG \perp$ to $BF$. $EPGF$ is a parallelogram. (Why?) $\therefore GF = PE$. It remains to prove $GB = PD$. The rt.~$\triangle PGB =$ the rt.~$\triangle BDP$. (Why?) \end{proofex} %\pagebreak \begin{proofex}% The sum of the perpendiculars dropped from any point within an equilateral triangle to the three sides is constant, and equal to the altitude. $AD$ is the altitude, $PE$, $PG$, and $PF$ the three perpendiculars. Through $P$ draw $HK \parallel$ to $BC$, meeting $AD$ at $M$. \eq[\indent Then]{$MD$}{$= PE$. (Why?)}{} \eq{$PG + PF$}{$= AM$ (Ex.~60).}{} \end{proofex} \figcc{080ccZ61}{080ddZ62} \ex{$ABC$ and $ABD$ are two triangles on the same base $AB$, and on the same side of it, the vertex of each triangle being without the other. If $AC$ equals $AD$, show that $BC$ cannot equal $BD$ (§~154).} \ex{The sum of the lines which join a point within a triangle to the three vertices is less than the perimeter, but greater than half the perimeter.} \figcc{080eeZ63}{080ffZ64} \ex{If from any point in the base of an isosceles triangle parallels to the legs are drawn, a parallelogram is formed whose perimeter is constant, being equal to the sum of the legs of the triangle.} \scanpage{081.png}% %\pagebreak \ex{The bisector of the vertical angle $A$ of a triangle $ABC$, and the bisectors of the exterior angles at the base formed by producing the sides $AB$ and $AC$, meet in a point which is equidistant from the base and the sides produced (§~162).} \figcc{081aaZ65}{081bbZ66} \begin{proofex}% If the bisectors of the base angles of a triangle are drawn, and through their point of intersection a line is drawn parallel to the base, the length of this parallel between the sides is equal to the sum of the segments of the sides between the parallel and the base. \step{$\angle EOB = \angle OBC = \angle OBE$. \quad $\therefore BE = EO$.}{} \end{proofex} \begin{proofex}% The bisector of the vertical angle of a triangle makes with the perpendicular from the vertex to the base an angle equal to half the difference of the base angles. Let $\angle B$ be greater than $\angle A$. \eq{$\angle DCE$}{$= 90° - \angle A - \angle ACD$.}{} \eq{$\angle ACD$}{$= 90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B$.}{} \step{$\therefore \angle DCE = 90° - \angle A - (90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B) = \frac{1}{2}\angle B - \frac{1}{2}\angle A$.}{} \end{proofex} \figcc{081ccZ67}{081ddZ68} \ex{If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Prove $\triangle AOB = \triangle COD$.} \figc{081eeZ69}{} \ex{The diagonals of a rectangle are equal. Prove $\triangle ABC = \triangle BAD$.} \ex{If the diagonals of a parallelogram are equal, the figure is a rectangle.} \ex{The diagonals of a rhombus are perpendicular to each other, and bisect the angles of the rhombus.} \ex{The diagonals of a square are perpendicular to each other, and bisect the angles of the square.} \figc{081ffZ73}{} \begin{proofex}% Lines from two opposite vertices of a parallelogram to the middle points of the opposite sides trisect the diagonal. \step{$EBFD$ is a $\Par$ (why?), and $DF$ is $\parallel$ to $EB$.}{} \step{$AM = MN$, and $MN = CN$ (§~188).}{} \end{proofex} \scanpage{082.png}% \begin{proofex}% The lines joining the middle points of the sides of any quadrilateral, taken in order, enclose a parallelogram. Prove $HG$ and $EF \parallel$ to $AC$; and $FG$ and $EH \parallel$ to $BD$ (§~189). Then $HG$ and $EF$ are each equal to $\frac{1}{2}AC$. \end{proofex} \figc{082adZ74}{} \ex{The lines joining the middle points of the sides of a rhombus, taken in order, enclose a rectangle. (Proof similar to that of Ex.~74.)} \ex{The lines joining the middle points of the sides of a rectangle (not a square), taken in order, enclose a rhombus.} \ex{The lines joining the middle points of the sides of a square, taken in order, enclose a square.} \begin{proofex}% The lines joining the middle points of the sides of an isosceles trapezoid, taken in order, enclose a rhombus or a square. $SHR$ and $QFP$ drawn $\perp$ to $AB$ are parallel. $\therefore PQSR$ is a $\Par$, and by Const.~is a rectangle or a square. $\therefore EFGH$ is a rhombus or a square (Exs.~76, 77). \figc{082ehZ78}{} \end{proofex} \ex{The bisectors of the angles of a rhomboid enclose a rectangle.} \ex{The bisectors of the angles of a rectangle enclose a square.} \ex{If two parallel lines are cut by a transversal, the bisectors of the interior angles form a rectangle.} \filbreak \figc{082iiZ82}{} \begin{proofex}% The median of a trapezoid passes through the middle points of the two diagonals. The median $EF$ is $\parallel$ to $AB$ and bisects $AD$ (§~190). $\therefore$ it bisects $DB$. Likewise $EF$ bisects $BC$ and $BD$. \end{proofex} \scanpage{083.png}% \begin{proofex}% The lines joining the middle points of the diagonals of a trapezoid is equal to half the difference of the bases. \step{$\triangle BFG = \triangle DFC$. (Why?) $\therefore EF = \frac{1}{2}AG$ (§~180).}{} \step{$CF=FG$, $DC=BG$.}{} \step{$\therefore AG=AB-DC$. $\therefore EF=\frac{1}{2}(AB-DC)$}{} \figc{083adZ83}{} \end{proofex} \begin{proofex}% In an isosceles trapezoid each base makes equal angles with the legs. Draw $CE \parallel$ to $DB$. $CE=DB$. (Why?) $\angle A = \angle CEA$, $\angle B = \angle CEA$, $\angle_s C$ and $D$ have equal supplements. \end{proofex} \ex{If the angles at the base of a trapezoid are equal, the other angles are equal, and the trapezoid is isosceles.} \begin{proofex}% In an isosceles trapezoid the opposite angles are supplementary: \step{$\angle C = \angle D$ (Ex.~84)}{} \end{proofex} \begin{proofex}% The diagonals on an isosceles trapezoidal are equal. Prove $\triangle ACD = \triangle BDC$. \end{proofex} %\pagebreak \begin{proofex}% If the diagonals of a trapezoid are equal, the trapezoid is isosceles. Draw $CE$ and $DF \perp$ to $AB$. \eq{$\triangle ADF$}{$= \triangle BCE$.}{(Why?)} \eq{$\therefore \angle ADF$}{$= \angle CBA$.}{} \eq{$\triangle ABC$}{$= \triangle BAD$.}{} \end{proofex} \figcc{083eeZ88}{083ffZ89} \begin{proofex}% If from the diagonal $DB$, of a square $ABCD$, $BE$ is cut off equal to $BC$, and $EF$ is drawn perpendicular to $BD$ meeting $DC$ at $F$, then $DE$ is equal to $EF$ and also to $FC$. $\angle EDF = 45°$, and $\angle DFE = 45°$; and $DE=DF$. Rt.~$\triangle BEF =$ rt.~$\triangle BCF$ (§~151); and $EF=FC$. \end{proofex} \ex{Two angles whose sides are so perpendicular, each to each, are either equal or supplementary.} \scanpage{084.png}% \chapter{BOOK II\@. THE CIRCLE.} \section{DEFINITIONS.} \begin{point}% A \textbf{circle}\label{circle} is a portion of a plane bounded by a curved line, all points of which are equally distant from a point within called the \textbf{centre}\label{centrecirc}. The bounding line is called the \textbf{circumference}\label{circumference} of the circle. \end{point} \begin{point}% A \textbf{radius} is a straight line from the centre to the circumference; and a \textbf{diameter}\label{diameter} is a straight line through the centre, with its ends in the circumference. By the definition of a circle, \emph{all its radii are equal}. All its diameters are equal, since a diameter is equal to two radii. \end{point} \begin{point}% \textbf{Postulate.} A circumference can be described from any point as a centre, with any given radius. \end{point} \begin{point}% A \indexbf{secant} is a straight line of unlimited length which intersects the circumference in two points; as, $AD$ (Fig.~1). \end{point} \figc{084aa220}{} \begin{point}% A \indexbf{tangent} is a straight line of unlimited length which has one point, and only one, in common with the circumference; as, $BC$ (Fig.~1). In this case the circle is said to be tangent to the straight line. The common point is called the \indexbf{point of contact}, or \indexbf{point of tangency}. \end{point} \begin{point}% Two \emph{circles} are tangent to each other, if both are tangent to a straight line at the same point; and are said to be tangent \emph{internally} or \emph{externally}, according as one circle lies wholly \emph{within} or \emph{without} the other. \end{point} \scanpage{085.png}% \pp{An \textbf{arc}\label{arc} is any part of the circumference; as, $BC$ (Fig.~3). Half a circumference is called a \indexbf{semicircumference}. Two arcs are called \textbf{conjugate arcs}, if their sum is a circumference.} \pp{A \textbf{chord}\label{chord} is a straight line that has its extremities in the circumference; as, the straight line $BC$ (Fig.~3).} \pp{A chord subtends two conjugate arcs. If the arcs are unequal, the less is called the \indexbf{minor} arc, and the greater the \indexbf{major} arc. A minor arc is generally called simply an arc.} \figc{085ac224}{} \pp{A \indexbf{segment} of a circle is a portion of the circle bounded by an arc and its chord (Fig.~2).} \pp{A \indexbf{semicircle} is a segment equal to half the circle (Fig.~2).} \pp{A \indexbf{sector} of a circle is a portion of the circle bounded by two radii and the arc which they intercept. The angle included by the radii is called the \emph{angle of the sector} (Fig.~2).} \pp{A \indexbf{quadrant} is a sector equal to a quarter of the circle (Fig.~2).} \pp{An angle is called a \textbf{central angle}\label{central}, if its vertex is at the centre and its sides are radii of the circle; as, $\angle AOD$ (Fig.~2).} \begin{point}% An angle is called an \textbf{inscribed angle}\label{inscribedcirc}, if its vertex is in the circumference and its sides are chords; as, $\angle ABC$ (Fig.~3). An angle is \emph{inscribed in a segment}\label{inscribedseg}, if its vertex is in the arc of the segment and its sides pass through the extremities of the arc. \end{point} \scanpage{086.png}% \pp{A polygon is \emph{inscribed in a circle}\label{polyinscribed}, if its sides are chords; and a circle is \emph{circumscribed about a polygon}\label{circcircumscribed}, if all the vertices of the polygon are in the circumference (Fig.~3).} \pp{A circle is \emph{inscribed in a polygon}\label{circinscribed}, if the sides of the polygon are tangent to the circle; and a polygon is \emph{circumscribed about}\label{polycircumscribed} a circle if its sides are tangents (Fig.~4).} \begin{point}% \emph{Two circles are equal, if they have equal radii.} For they will coincide, if their centres are made to coincide. \textsc{Conversely:} \emph{Two equal circles have equal radii.} \end{point} \pp{\emph{Two circles are concentric}\label{concentric}, if they have the same centre.} \filbreak \section{ARCS, CHORDS, AND TANGENTS.} \proposition{Theorem.} \begin{proof}% \obs{A straight line cannot meet the circumference of a circle in more than two points}. \figc{086aa235}{Let $HK$ be any line meeting the circumference $HKM$ in $H$ and $K$.} \prove{$HK$ cannot meet the circumference in any other point}. \textbf{Proof.} If possible, let $HK$ meet the circumference in $P$. \step{Then the radii $OH$, $OP$, and $OK$ are equal.}{§~217} \step{$\therefore P$ does not lie in the straight line $HK$.}{§~102} \step{$\therefore HK$ meets the circumference in only two points.}{\llap{\qed}} \end{proof} \scanpage{087.png}% \proposition{Theorem.} \begin{proof}% \obs{In the same circle or in equal circles, equal central angles intercept equal arcs; and of two unequal central angles the greater intercepts the greater arc.} \figc{087ab236}{In the equal circles whose centres are $O$ and $O'$, let the angles $AOB$ and $A'O'B'$ be equal, and angle $AOC$ be greater than angle $A'O'C'$.} \prove{1. $\arc AB = \arc A'B'$;} \prove[\phantom{To prove that~}]{2. $\arc AC > \arc A'B'$.} \textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that the $\angle A'O'B'$ shall coincide with its equal, the $\angle AOB$. \step{Then $A'$ falls on $A$, and $B'$ on $B$.}{§~233} \step{$\therefore$ $\arc A'B'$ coincides with $\arc AB$.}{§~216} \textbf{2.~} Since the $\angle AOC$ is greater than the $\angle A'O'B'$, it is greater than the $\angle AOB$, the equal of the $\angle A'O'B'$. \step{Therefore, $OC$ falls without the $\angle AOB$.}{} \step{$\therefore$ $\arc AC > \arc AB$.}{Ax.~8} \step{$\therefore$ $\arc AC > \arc A'B'$, the equal of $\arc AB$.}{\qed} \end{proof} \begin{proof}% \obs{\textsc{Conversely:} In the same circle or in equal circles, equal arcs subtend equal central angles; and of two unequal arcs the greater subtends the greater central angle.} \scanpage{088.png}% \proveq{\textup{1.} $\angle AOB$}{$=\angle A'O'B'$;} \proveq[\indent]{\settowidth{\TmpLen}{\textit{To prove that}}\rule{\TmpLen}{0pt}\textup{2.} $\angle AOC$}{is greater than $\angle A'O'B'$.} \textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that $O'A$ shall fall on its equal $OA$, and the arc $A'B'$ on its equal $AB$. \step{Then $O'B'$ will coincide with $OB$.}{§~47} \step{$\therefore \angle A'O'B'=\angle AOB$.}{§~60} \textbf{2.~}Since $\arc AC>A'B'$, it is greater than $\arc AB$, the equal of $A'B'$, and $OB$ will fall within the $\angle AOC$. \eq{}{$\therefore \angle AOC$ is greater than $\angle AOB$.}{Ax.~8} \eq{}{$\therefore \angle AOC$ is greater than $\angle A'O'B'$.}{\qed} \end{proof} \pp{\cor[1]{In the same circle or in equal circles, two sectors that have equal angles are equal; two sectors that have unequal angles are unequal, and the greater sector has the greater angle.}} \pp{\cor[2]{In the same circle or in equal circles, equal sectors have equal angles; and of two unequal sectors the greater has the greater angle.}} \begin{point}% \textbf{Law of Converse Theorems.}\label{converse2} It was stated in §~32 that the converse of a theorem is not necessarily true. If, however, a theorem is in fact a group of three theorems, and if \emph{one of the hypotheses} of the group \emph{must} be true, and \emph{no two of the conclusions can be true at the same time}, then the converse of the theorem is \emph{necessarily} true. Proposition II. is a group of three theorems. It asserts that the arc $AB$ is equal to the arc $A'B'$, if the angle $AOB$ is equal to the angle $A'O'B'$; that the arc $AB$ is greater than the arc $A'B'$, if the angle $AOB$ is greater than the angle $A'O'B'$; that the arc $AB$ is less than the arc $A'B'$, if the angle $AOB$ is less than the angle $A'O'B'$. One of these hypotheses must be true; for the angle $AOB$ must be equal to, greater than, or less than, the angle $A'O'B'$. No two of the conclusions can be true at the same time, for the arc $AB$ cannot be both equal to and greater than the arc $A'B'$; nor can it be both equal to and less than the arc $A'B'$; nor both greater than and less than the arc $A'B'$. In such a case, the converse theorem is \emph{necessarily} true, and no proof like that given in the text is required to establish it. \end{point} \scanpage{089.png}% \proposition{Theorem.} \begin{proof}% \obs{In the same circle or in equal circles, equal arcs are subtended by equal chords; and of two unequal arcs the greater is subtended by the greater chord.} \figc{089ab241}{In the equal circles whose centres are $O$ and $O'$, let the arcs $AB$ and $A'B'$ be equal, and the arc $AF$ greater than arc $A'B'$.} \proveq{\textup{1. }chord $AB$}{$=$ chord $A'B'$;} \proveq[]{\textup{2. }chord $AF$}{$>$ chord $A'B'$.} \step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{} \step[\indent 1.]{The $\triangle_s AOB$ and $A'O'B'$ are equal.}{§~143} \step{For $OA = O'A'$, and $OB = O'B'$,}{§~233} \pnote{(radii of equal circles),} \step{and $\angle AOB = \angle A'O'B'$,}{§~237} \pnote{(in equal $\odot_s$ equal arcs subtend equal central $\angle_s$).} \step{$\therefore$ chord $AB = $ chord $A'B'$.}{§~128} \step{}{} \step[\indent 2.]{In the $\triangle_s AOF$ and $A'O'B'$,}{} \step{$OA = O'A'$, and $OF = O'B'$.}{§~233} \step{But the $\angle AOF$ is greater than the $\angle A'O'B'$,}{§~237} \pnote{(in equal $\odot_s$, the greater of two unequal arcs subtends the greater $\angle$).} \step{$\therefore$ chord $AF >$ chord $A'B'$.}{§~154} \hfill\qed \end{proof} \pp{\cor{In the same circle or in equal circles, the greater of two unequal major arcs is subtended by the less chord.}} \scanpage{090.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} In the same circle or in equal circles, equal chords subtend equal arcs; and of two unequal chords the greater subtends the greater arc.} \figc{090ab243}{In the equal circles whose centres are $O$ and $O'$, let the chords $AB$ and $A'B'$ be equal, and the chord $AF$ greater than $A'B'$.} \prove{\quad\upshape{1.} $\arc AB = \arc A'B'$;} \prove[\phantom{To prove that~}]{\quad\upshape{2.} $\arc AF > \arc A'B'$.} \step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{} \step[\indent 1.]{The $\triangle_s OAB$ and $O'A'B'$ are equal.}{§~150} \step{For $OA = O'A'$, and $OB = O'B'$,}{§~233} \step{and chord $AB =$ chord $A'B'$.}{Hyp.} \step{$\therefore \angle AOB = \angle A'O'B'$.}{§~128} \step{$\therefore \arc AB = \arc A'B'$,}{§~236} \pnote{(in equal $\odot_s$ equal central $\angle_s$ intercept equal arcs).} \step[\indent 2.]{In the $\triangle_s OAF and O'A'B'$,}{} \step{$OA = O'A'$ and $OF = O'B'$.}{§~233} \step{But chord $AF >$ chord $A'B'$.}{Hyp.} \step{$\therefore$ the $\angle AOF$ is greater than the $\angle A'O'B'$.}{§~155} \step{$\therefore \arc AF > \arc A'B'$,}{§~236} \pnote{(in equal $\odot_s$ the greater central $\angle$ intercepts the greater arc).} \hfill\qed \end{proof} \pp{\cor{In the same circle or in equal circles, the greater of two unequal chords subtends the less major arc.}} \scanpage{091.png}% \proposition{Theorem.} \begin{proof}% \obs{A diameter perpendicular to a chord bisects the chord and the arcs subtended by it.} \figc{091aa245}{Let $ES$ be a diameter perpendicular to the chord $AB$ at $M$.} \prove{$AM = BM$, $AS = BS$, and $AE = BE$.} \textbf{Proof.} Draw $OA$ and $OB$ from $O$, the centre of the circle. \step{The rt.~$\triangle_s OAM$ and $OBM$ are equal.}{§~151} \eq[\indent For]{$OM$}{$= OM$,}{Iden.} \eq[and]{$OA$}{$= OB$.}{§~217} \step{$\therefore AM = BM$, and $\angle AOS = \angle BOS$.}{§~128} \eq[\indent Likewise]{$\angle AOE$}{$= \angle BOE$.}{§~85} \step{$\therefore AS = BS$, and $AE = BE$.}{§~236} \hfill\qed \end{proof} \pp{\cor[1]{A diameter bisects the circumference and the circle.}} \pp{\cor[2]{A diameter which bisects a chord is perpendicular to it.}} \pp{\cor[3]{The perpendicular bisector of a chord passes through the centre of the circle, and bisects the arcs of the chord.}} \scanpage{092.png}% \proposition{Theorem.} \begin{proof}% \obs{In the same circle or in equal circles, equal chords are equally distant from the centre. \textsc{Conversely:} Chords equally distant from the centre are equal.} \figc{092aa249}{Let $AB$ and $CF$ be equal chords of the circle $ABFC$.} \prove{$AB$ and $CF$ are equidistant from the centre $O$.} \textbf{Proof.} Draw $OP \perp$ to $AB$, $OH \perp$ to $CF$, and join $OA$ and $OC$. \step{$OP$ bisects $AB$, and $OH$ bisects $CF$.}{§~245} \step{The rt.~$\triangle_s OPA$ and $OHC$ are equal.}{§~151} \eq{$AP$}{$= CH$,}{Ax.~7} \eq[and]{$OA$}{$= OC$,}{§~217} \eq[\indent Hence,]{$OP$}{$= OH$.}{§~128} \step{$\therefore AB$ and $CF$ are equidistant from $O$.}{} \step{}{} \eq[\indent \textsc{Conversely:}]{\textbf{Let }$OP$}{$=OH$.}{} \proveq[\indent To prove]{$AB$}{$= CF$.} \textbf{Proof.} The rt.~$\triangle_s OPA$ and $OHC$ are equal.~\hfill§~151 \eq[\indent For]{$OA$}{$= OC$,}{§~217} \eq[and]{$OP$}{$= OH$,}{Hyp.} \eq[\indent Hence,]{$AP$}{$= CH$.}{§~128} \eq{$\therefore AB$}{$= CF$.}{Ax.~6} \hfill\qed \end{proof} \scanpage{093.png}% \proposition{Theorem.} \begin{proof}% \obs{In the same circle or in equal circles, if two chords are unequal, they are unequally distant from the centre; and the greater chord is at the less distance.} \figc{093aa250}{In the circle whose centre is $O$, let the chords $AB$ and $CD$ be unequal, and $AB$ the greater; and let $OE$ be perpendicular to $AB$ and $OF$ perpendicular to $CD$.} \proveq{$OE$}{$< OF$.} \textbf{Proof.} Suppose $AG$ drawn equal to $CD$, and $OH \perp$ to $AG$. \step{Draw $EH$.}{} \step{$OE$ bisects $AB$, and $OH$ bisects $AG$.}{§~245} \eq[By hypothesis,]{$AB$}{$> CD$.}{} \step{$\therefore AB > AG$, the equal of $CD$.}{} \eq{$\therefore AE$}{$> AH$.}{Ax.~7} \step{$\therefore \angle AHE$ is greater than $\angle AEH$.}{§~152} $\therefore \angle OHE$, the complement of $\angle AHE$, is less than $\angle OEH$, the complement of $\angle AEH$.\hfill~Ax.~5 \eq{$\therefore OE$}{$< OH$.}{§~153} \eq[\indent But]{$OH$}{$=OF$.}{§~249} \eq{$\therefore OE$}{$< OF$.}{\qed} \end{proof} \ex{The perpendicular bisectors of the sides of an inscribed polygon are concurrent (pass through the same point).} \scanpage{094.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} In the same circle or in equal circles, if two chords are unequally distant from the centre, they are unequal; and the chord at the less distance is the greater.} \figc{094aa251}{In the circle whose centre is $O$, let $AB$ and $CD$ be unequally distant from $O$; and let $OE$, the perpendicular to $AB$, be less than $OF$, the perpendicular to $CD$.} \proveq{$AB$}{$> CD$.} \textbf{Proof.} Suppose $AG$ drawn equal to $CD$, and $OH \perp$ to $AG$. \eq[\indent Then]{$OH$}{$= OF$}{§~249} \eq[\indent Hence,]{$OE$}{$< OH$.}{} \step{Draw $EH$.}{} \step{$\angle OHE$ is less than $\angle OEH$.}{§~152} $\therefore \angle AHE$, the complement of $\angle OHE$, is greater than $\angle AEH$, the complement of $\angle OEH$.\hfill~Ax.~5 \eq{$\therefore AE$}{$> AH$.}{§~153} \step[\indent But]{$AE = \frac{1}{2}AB$, and $AH = \frac{1}{2}AG$.}{§~245} \eq{$\therefore AB$}{$> AG$.}{Ax.~6} \eq[\indent But]{$CD$}{$= AG$.}{Const.} \eq{$\therefore AB$}{$> CD$.}{\qed} \end{proof} \pp{\cor{A diameter of a circle is greater than any other chord.}} \scanpage{095.png}% \proposition{Theorem.} \begin{proof}% \obs{A straight line perpendicular to a radius at its extremity is a tangent to the circle.} \figc{095aa253}{Let $MB$ be perpendicular to the radius $OA$ at $A$.} \prove{$MB$ is a tangent to the circle.} \textbf{Proof.} From $O$ draw any other line to $MB$, as $OH$. \eq[\indent Then]{$OH$}{$> OA$.}{§~97} \step{$\therefore$ the point $H$ is without the circle.}{§~216} Hence, \emph{every point}, except $A$, of the line $MB$ is without the circle, and therefore $MB$ is a tangent to the circle at $A$.~\hfill§~220 \hfill\qed \end{proof} \begin{point}% \cor[1]{A tangent to a circle is perpendicular to the radius drawn to the point of contact.} For $OA$ is the shortest line from $O$ to $MB$, and is therefore $\perp$ to $MB$ (§~98); that is, $MB$ is $\perp$ to $OA$. \end{point} \begin{point}% \cor[2]{A perpendicular to a tangent at the point of contact passes through the centre of the circle.} For a radius is $\perp$ to a tangent at the point of contact, and therefore a $\perp$ erected at the point of contact coincides with this radius and passes through the centre. \end{point} \pp{\cor[3]{A perpendicular from the centre of a circle to a tangent passes through the point of contact.}} \scanpage{096.png}% \proposition{Theorem.} \begin{proof}% \obs{Parallels intercept equal arcs on a circumference.} \figc{096ac257}{\textnormal{\textsc{Case 1.~}} Let $AB$ \textnormal{(Fig.~1)} be a tangent at $F$ parallel to $CD$, a secant.} \proveq{$\arc CF$}{$= \arc DF$.} \step[\indent\textbf{Proof.}]{Suppose $FF'$ drawn $\perp$ to $AB$.}{} \step{Then $FF'$ is a diameter of the circle.}{§~255} \step{And $FF'$ is also $\perp$ to $CD$.}{§~107} \step{$\therefore CF = DF$, and $CF' = DF'$.}{§~245} \step{}{} \textsc{Case 2.~}\textbf{Let $AB$ and $CD$} (Fig.~2) \textbf{be parallel secants.} \proveq{$\arc AC$}{$= \arc BD$.} \step[\indent\textbf{Proof.}]{Suppose $EF \parallel$ to $CD$ and tangent to the circle at $M$.}{} \eq[\indent Then]{$\arc AM$}{$= \arc BM$,}{Case~1} \eq[and]{$\arc CM$}{$= \arc DM$.}{} %proofrule \eq{$\therefore \arc AC$}{$= \arc BD$.}{Ax.~3} \step{}{} \textsc{Case 3.~}\textbf{Let $AB$ and $CD$} (Fig.~3) \textbf{be parallel tangents at $E$ and $F$.} \proveq{$\arc EGF$}{$= \arc EHF$.} \step[\indent\textbf{Proof.}]{Suppose $GH$ drawn $\parallel$ to $AB$.}{} \eq[\indent Then]{$\arc EG$}{$= \arc EH$,}{Case~1} \eq[and]{$\arc GF$}{$= \arc HF$.}{} %proofrule \eq{$\therefore \arc EGF$}{$= \arc EHF$.}{Ax.~2} \hfill\qed \end{proof} \scanpage{097.png}% \proposition{Theorem.} \begin{proof}% \obs{Through three points not in a straight line one circumference, and only one, can be drawn.} \figc{097aa258}{Let $A$, $B$, $C$ be three points not in a straight line.} \prove{one circumference, and only one, can be drawn through $A$, $B$, and~$C$.} \step[\indent\textbf{Proof.}]{Draw $AB$ and $BC$.}{} At the middle points of $AB$ and $BC$ suppose $\perp_s$ erected. These $\perp_s$ will intersect at some point $O$, since $AB$ and $BC$ are not in the same straight line. The point $O$ is in the perpendicular bisector of $AB$, and is therefore equidistant from $A$ and $B$; the point $O$ is also in the perpendicular bisector of $BC$, and is therefore equidistant from $B$ and $C$.~\hfill§~160 Therefore, $O$ is equidistant from $A$, $B$, and $C$; and a circumference described from $O$ as a centre, with a radius $OA$, will pass through the three given points. The centre of a circumference passing through the three points must be in both perpendiculars, and hence at their intersection. As two straight lines can intersect in only one point, $O$ is the centre of the only circumference that can pass through the three given points.~\hfill\qed \end{proof} \pp{\cor{Two circumferences can intersect in only two points. \textup{For, if two circumferences have three points common, they coincide and form one circumference.}}} \scanpage{098.png}% \pp{\defn{A \textbf{tangent from an external point to a circle}\label{tangent2} is the part of the tangent between the external point and the point of contact.}} \proposition{Theorem.} \begin{proof}% \obs{The tangents to a circle drawn from an external point are equal, and make equal angles with the line joining the point to the centre.} \figc{098aa261}{Let $AB$ and $AC$ be tangents from $A$ to the circle whose centre is $O$, and let $AO$ be the line joining $A$ to the centre $O$.} \prove{$AB = AC$, and $\angle BAO = \angle CAO$.} \step[\indent\textbf{Proof.}]{Draw $OB$ and $OC$.}{} \step{$AB$ is $\perp$ to $OB$, and $AC \perp$ to $OC$,}{§~254} \pnote{(a tangent to a circle is $\perp$ to the radius drawn to the point of contact).} \step{The rt.~$\triangle_s OAB$ and $OAC$ are equal.}{§~151} For $OA$ is common, and the radii $OB$ and $OC$ are equal.~\hfill§~217 \step{$\therefore AB=AC$, and $\angle BAO = \angle CAO$.}{§~128} \hfill\qed \end{proof} \pp{\defn{The line joining the centres of two circles is called the \indexbf{line of centres}.}} \pp{\defn{A tangent to two circles is called a \indexbf{common external tangent} if it does not cut the line of centres, and a \indexbf{common internal tangent} if it cuts the line of centres.}} \scanpage{099.png}% \proposition{Theorem.} \begin{proof}% \obs{If two circles intersect each other, the line of centres is perpendicular to their common chord at its middle point.} \figc{099aa264}{Let $C$ and $C'$ be the centres of the two circles, $AB$ the common chord, and $CC'$ the line of centres.} \prove{$CC'$ is $\perp$ to $AB$ at its middle point.} \step[\indent\textbf{Proof.}]{Draw $CA$, $CB$, $C'A$, and $C'B$.}{} \step{$CA = CB$, and $C'A = C'B$.}{§~217} \step{$\therefore C$ and $C'$ are two points, each equidistant from $A$ and $B$.}{} \step{$\therefore CC'$ is the perpendicular bisector of $AB$.}{§~161} \hfill\qed \end{proof} \begin{proofex}% Describe the relative position of two circles if the line of centres: \begin{myenum} \item is greater than the sum of the radii; \item is equal to the sum of the radii; \item is less than the sum but greater than the difference of the radii; \item is equal to the difference of the radii; \item is less than the difference of the radii. \end{myenum} Illustrate each case by a figure. \end{proofex} \ex{The straight line drawn from the middle point of a chord to the middle point of its subtended arc is perpendicular to the chord.} \ex{The line which passes through the middle points of two parallel chords passes through the centre of the circle.} \scanpage{100.png}% \proposition{Theorem.} \begin{proof}% \obs{If two circles are tangent to each other, the line of centres passes through the point of contact.} \figc{100aa265}{Let the two circles, whose centres are $C$ and $C'$, be tangent to the straight line $AB$ at $Q$, and $CC'$ the line of centres.} \prove{$O$ is in the straight line $CC'$.} \textbf{Proof.} A $\perp$ to $AB$, drawn through the point $O$, passes through the centres $C$ and $C'$,~\hfill§~255 \pnote{(a $\perp$ to a tangent at the point of contact passes through the centre of the circle).} $\therefore$ the line $CC'$, having two points in common with this $\perp$ must coincide with it.~\hfill§~47 \step{$\therefore O$ is in the straight line $CC'$.}{\qed} \end{proof} \begin{proofex}% Describe the relative position of two circles if they may have: \begin{myenum} \item two common external and two common internal tangents; \item two common external tangents and one common internal tangent; \item two common external tangents and no common internal tangent; \item one common external and no common internal tangent; \item no common tangent. \end{myenum} Illustrate each case by a figure. \end{proofex} \ex{The line drawn from the centre of a circle to the point of intersection of the two tangents is the perpendicular bisector of the chord joining the points of contact.} \scanpage{101.png}% \section{MEASUREMENT.} \begin{point}% To \textbf{measure} a quantity of any kind is to find \emph{the number of times} it contains a known quantity of the \emph{same kind}, called the \textbf{unit of measure}. The \emph{number} which shows the number of times a quantity contains the unit of measure is called the \indexbf{numerical measure} of that quantity. \end{point} \begin{point}% No quantity is great or small except by comparison with another quantity of the \emph{same kind}. This comparison is made by finding the numerical measures of the two quantities in terms of a common unit, and then dividing one of the measures by the other. The quotient is called their \indexbf{ratio}. In other words the ratio of two quantities of the same kind is the \emph{ratio} of their \emph{numerical measures} expressed in terms of a common unit. The ratio of $a$ to $b$ is written $a : b$, or $\dfrac{a}{b}$. \end{point} \begin{point}% Two quantities that can be expressed in \emph{integers} in terms of a common unit are said to be \textbf{commensurable}\label{commensurable}, and the exact value of their ratio can be found. The common unit is called their \emph{common measure}, and each quantity is called a \emph{multiple} of this common measure. Thus, a common measure of $2\frac{1}{2}$~feet and $3\frac{2}{3}$~feet is $\frac{1}{6}$~of a foot, which is contained $15$~times in $2\frac{1}{2}$~feet, and $22$~times in $3\frac{2}{3}$~feet. Hence, $2\frac{1}{2}$~feet and $3\frac{2}{3}$~feet are multiples of $\frac{1}{6}$~of a foot, since $2\frac{1}{2}$~feet may be obtained by taking $\frac{1}{6}$~of a foot $15$~times, and $3\frac{2}{3}$~feet by taking $\frac{1}{6}$~of a foot $22$~times. The ratio of $2\frac{1}{2}$~feet to $3\frac{2}{3}$~feet is expressed by the fraction~$\frac{15}{22}$. \end{point} \begin{point}% Two quantities of the same kind that cannot \emph{both} be expressed in \emph{integers} in terms of a common unit, are said to be \textbf{incommensurable}, and the \emph{exact value} of their ratio cannot be found. But by taking the unit sufficiently small, an \emph{approximate value} can be found that shall differ from the true value of the ratio by less than any assigned value, however small. \scanpage{102.png}% Thus, suppose the ratio, $\dfrac{a}{b} = \sqrt{2}$. Now $\sqrt{2} = 1.41421356\cdots$, a value greater than $1.414213$, but less than $1.414214$. If, then, a \emph{millionth part} of $b$ is taken as the unit of measure, the value of $\dfrac{a}{b}$ lies between $1.414213$ and $1.414214$, and therefore differs from either of these values by less than $0.000001$. By carrying the decimal further, an approximate value may be found that will differ from the true value of the ratio by less than \emph{a billionth, a trillionth, or any other assigned value}. In general, if $\dfrac{a}{b} > \dfrac{m}{n}$ but $< \dfrac{m+1}{n}$, then the error in taking either of these values for $\dfrac{a}{b}$ is less than $\dfrac{1}{n}$, the difference between these two fractions. But by increasing $n$ indefinitely, $\dfrac{1}{n}$ can be decreased indefinitely, and a value of the ratio can be found within any required degree of accuracy. \end{point} \pp{The ratio of two incommensurable quantities is called an \indexbf{incommensurable ratio}; and is a \emph{fixed value} which its successive approximate values constantly approach.} \section[THEORY OF LIMITS.]{THE THEORY OF LIMITS.} \begin{point}% When a quantity is regarded as having a \emph{fixed} value throughout the same discussion, it is called a \indexbf{constant}; but when it is regarded, under the conditions imposed upon it, as having \emph{different successive} values, it is called a \indexbf{variable}. If a variable, by having different successive values, can be made to differ from a given constant by less than any assigned value, however small, but cannot be made absolutely equal to the constant, that constant is called the \indexbf{limit} of the variable, and the variable is said to \textbf{approach the constant as its limit}. \end{point} \scanpage{103.png}% \figc{103aa272}{} \begin{point}% Suppose a point to move from $A$ toward $B$, under the conditions that the first second it shall move one half the distance from $A$ to $B$, that is, to $M$; the next second, one half the remaining distance, that is, to $M'$; and so on indefinitely. Then it is evident that the moving point \emph{may approach as near to $B$ as we choose, but will never arrive at $B$}. For, however near it may be to $B$ at any instant, the next second it will pass over half the distance still remaining; it must, therefore, approach nearer to $B$, since \emph{half} the distance still remaining is \emph{some} distance, but will not reach $B$, since \emph{half} the distance still remaining is not the \emph{whole} distance. Hence, the distance from $A$ to the moving point is an increasing variable, which indefinitely approaches the constant $AB$ as its \emph{limit}; and the distance from the moving point to $B$ is a decreasing variable, which indefinitely approaches the \emph{constant zero} as its \emph{limit}. \end{point} \figc{103bb273}{} \begin{point}% Again, suppose a square $ABCD$ inscribed in a circle, and $E$, $F$, $H$, $K$ the middle points of the arcs subtended by the sides of the square. If we draw the lines $AE$, $EB$, $BF$, etc., we shall have an inscribed polygon of double the number of sides of the square. The length of the perimeter of this polygon, represented by the dotted lines, is greater than that of the square, since two sides replace each side of the square and form with it a triangle, and two sides of a triangle are together greater than the third side; but less than the length of the circumference, for it is made up of straight lines, each one of which is less than the part of the circumference between its extremities. \end{point} \scanpage{104.png}% By continually doubling the number of sides of each resulting inscribed figure, the length of the perimeter will increase with the increase of the number of sides, but will not become equal to the length of the circumference. The difference between the perimeter of the inscribed polygon and the circumference of the circle can be made less than any assigned value, but cannot be made equal to zero. The length of the circumference is, therefore, the \emph{limit} of the length of the perimeter as the \emph{number of sides} of the inscribed figure is \emph{indefinitely increased.}~\hfill§~271 \begin{point}% Consider the decimal $0.333 \cdots$ which may be written \centerline{\( \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots \)} The value of each fraction after the first is one tenth of the preceding fraction, and by continuing the series we shall reach a fraction less than \emph{any} assigned value, that is, the values of the successive fractions \emph{approach zero as a limit.} The \emph{sum} of these fractions is less than $\frac{1}{3}$; but the more terms we take, the nearer does the sum \emph{approach $\frac{1}{3}$ as a limit.} \end{point} \begin{point}% \textbf{Test for a limit.} In order to prove that a variable approaches a constant as a limit, it is necessary to prove that the difference between the variable and the constant: \begin{myenum} \item \emph{Can be made less than any assigned value, however small.} \item \emph{Cannot be made absolutely equal to zero.} \end{myenum} \end{point} \begin{point}% \thm{If the limit of a variable $x$ is zero, then the limit of $kx$, the product of the variable by any finite constant $k$, is zero.} 1. Let $q$ be any assigned quantity, however small. Then $\dfrac{q}{k}$ is not~$0$. Hence $x$, which may differ as little as we please from~$0$, may be taken less than $\dfrac{q}{k}$, and then $kx$ will be less than $q$. 2. Since $x$ cannot be~$0$, $kx$ cannot be~$0$. \step{Therefore, the limit of $kx=0$}{§~275} \end{point} \scanpage{105.png}% \begin{point}% \cor{If the limit of a variable~$x$ is zero, then the limit of the quotient of the variable by any finite constant~$k$, is also zero.} For $\dfrac{x}{k} = \dfrac{1}{k} × x$, which by §~276 can be made less than any assigned value, however small, but cannot be made equal to zero. \end{point} \begin{point}% \thm{The limit of the sum of a finite number of variables \mbox{$x$, $y$, $z$, $\cdots$} is equal to the sum of their respective limits \mbox{$a$, $b$, $c$, $\cdots$.}} \sloppy Let \mbox{$d$, $d'$, $d''$, $\cdots$} denote the differences between \mbox{$x$, $y$, $z$, $\cdots$} and \mbox{$a$, $b$, $c$, $\cdots$,} respectively. Then \mbox{$d+d'+d''+\cdots$} can be made less than any assigned quantity $q$. \fussy For, if \mbox{$d$, $d'$, $d''$, $\cdots$} are $n$ in number and $d$ is the largest, \step{$d+d'+d''+\cdots < nd$.}{(1)} Since $d$ may be diminished at pleasure, we may make $d$ so small that \step{$ d < \dfrac{q}{n}$; and therefore $nd < q$.}{} But by (1), $d+d'+d''+\cdots < nd$, and therefore $< q$. Therefore, the difference between ($x + y + z + \cdots$) and $(a + b + c + \cdots)$ can be made less than any assigned quantity, but not zero. Therefore, the limit of $(x + y + z + \cdots) = a + b + c + \cdots$.~\hfill§~275 \end{point} \begin{point}% \thm{If the limit of a variable~$x$ is not zero, and if $k$~is any finite constant, the limit of the product~$kx$ is equal to the limit of~$x$ multiplied by~$k$.} 1. If $a$ denotes the limit of $x$, then $x$ cannot be equal to $a$.~\hfill§~271 \step{Therefore, $kx$ cannot be equal to $ka$.}{} 2. The limit of $(a - x) = 0$. Hence, the limit of $ka - kx=0$.~\hfill§~276 \step{Therefore, the limit of $kx = ka$.}{§~275} \end{point} \begin{point}% \cor{The limit of the quotient of a variable $x$ by any finite constant $k$ is the limit of $x$ divided by $k$.} For $\dfrac{x}{k} = \dfrac{1}{k} × x$, and $\dfrac{\text{the limit of }x}{k} = \dfrac{1}{k}$ $×$ the limit of $x$. \end{point} \scanpage{106.png}% \begin{point}% \thm{The limit of the product of two or more variables is the product of their respective limits, provided no one of these limits is zero.} If $x$ and $y$ are variables, $a$ and $b$ their respective limits, we may put $x = a - d$, $y = b - d'$; then $d$ and $d'$ are variables which can be made less than any assigned quantity, but not zero.~\hfill§~275 Now, %[**TN: ad hoc visual formatting] \vspace*{-1.5\baselineskip} \begin{align*} xy &= (a - d) (b - d') \\ &= ab - ad' - bd + dd' \\ \therefore ab - xy &= ad' + bd - dd'. \end{align*} Since every term on the right contains $d$ or $d'$, the whole right member can be made less than any assigned quantity, but not zero.~\hfill§~278 Hence, $ab - xy$ can be made less than any assigned quantity, but not zero. \step{Therefore, the limit of $xy = ab$.}{§~275} \step{Similarly, for three or more variables.}{} \end{point} \begin{point}% \cor[1]{The limit of the nth power of a variable is the nth power of its limit.} For the limit of the product of the variables $x$, $y$, $z$, $\cdots$ to $n$ factors is the product of their respective limits, the constants $a$, $b$, $c$, $\cdots$ to $n$ factors~(§~281). If the $n$ factors $xyz\cdots$ are each equal to $x$, and the $n$ factors $abc\cdots$ are each equal to $a$, we have $xyz\cdots = x^n$, and $abc\cdots = a^n$. \step{Therefore, the limit of $x^n = a^n$.}{} \end{point} \begin{point}% \cor[2]{The limit of the nth root of a variable is the nth root of its limit.} For if the limit of $x = a$, we may put this in the following form, \step{the limit of \( \sqrt[n]{x^n} = \sqrt[n]{a^n} \);}{} \noindent that is, the limit of \( \sqrt[n]{xxx\cdots\text{ to $n$~factors }} \) is \( \sqrt[n]{aaa\cdots\text{ to $n$~factors}} \). Now, $xxx\cdots$ is a variable since each factor is a variable, and $aaa\cdots$ is a constant since each factor is a constant. If we denote $xxx\cdots$ to $n$~factors by the variable~$y$, and $aaa\cdots$ to $n$ factors by the constant~$b$, we have % the vphantoms make the two sqrt boxes the same height, to line up better \step{the limit of \( \displaystyle \sqrt[n]{\vphantom{b}y} = \sqrt[n]{\vphantom{y}b} \).}{} \end{point} \scanpage{107.png}% \begin{point}% \thm{If two variables are constantly equal, and each approaches a limit, the limits are equal.} Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits, $d$ and $d'$ the respective differences between the variables and their limits. Then, if the variables are \emph{increasing} toward their limits \step{$a = x + d$, and $b = y + d'$.}{} Since the equation $x = y$ is always true, we have by subtraction \step{$a - b = d - d'$.}{} Since $a$ and $b$ are constants, $a - b$ is a constant; therefore, $d - d'$, which is equal to $a - b$, is a constant. But the only constant which is less than any assigned value is~$0$. Therefore, $d - d' = 0$. Therefore, $a - b = 0$, and $a = b$. If the variables $x$ and $y$ are \emph{decreasing} toward their limits $a$ and $b$, respectively, then \step{$a = x - d$ and $b = y - d'$.}{} Therefore, by subtraction \step{$a - b = d' - d$.}{} Therefore, by the same proof as for increasing variables \step{$a = b$.}{} \end{point} \begin{point}% \thm{If two variables have a constant ratio, and each approaches a limit that is not zero, the limits have the same ratio.} Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits. \eq[\indent Let]{$\dfrac{x}{y}$}{$= r$, a constant; then $x = ry$.}{} Since $x$ and $ry$ are two variables that are always equal, \eq{the limit of $x$}{$=$ the limit of $ry$.}{§~284} \eq[\indent Now,]{the limit of $ry$}{$= r$ $×$ limit of $y$.}{§~279} But the limit of $x$ is $a$, and the limit of $y$ is $b$. \eq[\indent Therefore,]{$a$}{$= rb$; that is, $\dfrac{a}{b} = r$.}{} \end{point} \scanpage{108.png}% \proposition{Problem.} \begin{proof}% \obs{To find the ratio of two straight lines.} \figc{108aa286}{Let $AB$ and $CD$ be two straight lines.} \prove[To find ]{the ratio of $AB$ and $CD$.} \step{Apply $CD$ to $AB$ as many times as possible.}{} \step{Suppose twice, with a remainder $EB$.}{} \step{Then apply $EB$ to $CD$ as many times as possible.}{} \step{Suppose three times, with a remainder $FD$.}{} \step{Then apply $FD$ to $EB$ as many times as possible.}{} \step{Suppose once, with a remainder $HB$.}{} \step{Then apply $HB$ to $FD$ as many times as possible.}{} \step{Suppose once, with a remainder $KD$.}{} \step{Then apply $KD$ to $HB$ as many times as possible.}{} \step{Suppose $KD$ is contained just twice in $HB$.}{} \setlength{\eqalign}{.33\dentwidth} \eq[\indent Then]{$HB$}{$= 2 KD$;}{} \eq{$FD$}{$= HB + KD = 3 KD$;}{} \eq{$EB$}{$= FD + HB = 5 KD$;}{} \eq{$CD$}{$= 3 EB + FD = 18 KD$;}{} \eq{$AB$}{$= 2 CD + EB = 41 KD$;}{} \eq{$\therefore \dfrac{AB}{CD}$}{$= \dfrac{41 KD}{18 KD} = \dfrac{41}{18}$.}{\qef} \setlength{\eqalign}{.5\dentwidth} \end{proof} \note{By the same process the ratio of two arcs of the same circle or of equal circles can be found. \par If the lines or arcs are incommensurable, an approximate value of the ratio can be found by the same method.} \scanpage{109.png}% \clearpage \section{MEASURE OF ANGLES.} \proposition{Theorem.} \begin{proof}% \obs{In the same circle or in equal circles, two central angles have the same ratio as their intercepted arcs.} \figc{109ac287}{In the equal circles whose centres are $C$ and $C'$, let $ACB$ and $A'C'B'$ be the angles, $AB$ and $A'B'$ the intercepted arcs.} \proveq{$\dfrac{\angle A'C'B'}{\angle ACB}$} {$=\dfrac{\arc A'B'}{\arc AB}$.} \textsc{Case~1.} \emph{When the arcs are commensurable} (Figs.~1 and 2). \textbf{Proof.} Let the arc~$m$ be a common measure of $A'B'$ and $AB$. \step{Suppose $m$ to be contained $4$~times in $A'B'$,}{} \step{and $7$~times in $AB$.}{} \eq[\indent Then]{$\dfrac{\arc A'B'}{\arc AB}$}{$=\dfrac{4}{7}$.}{} At the several points of division on $AB$ and $A'B'$ draw radii. These radii will divide $\angle ACB$ into $7$~parts, and $\angle A'C'B'$ into $4$~parts, equal each to each,~\hfill§~237 \pnote{(in the same $\odot$, or equal $\odot_s$, equal arcs subtend equal central $\angle_s$).} \eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$} {$=\dfrac{4}{7}$.}{} \eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$} {$=\dfrac{\arc A'B'}{\arc AB}$.}{Ax.~1} \scanpage{110.png}% \step{}{} \filbreak \textsc{Case 2.} \emph{When the arcs are incommensurable} (Figs.~2 and 3). \textbf{Proof.} Divide $AB$ into any number of equal parts, and apply one of these parts to $A'B'$ as many times as $A'B'$ will contain it. Since $AB$ and $A'B'$ are incommensurable, a certain number of these parts will extend from $A'$ to some point, as $D$, leaving a remainder $DB'$ less than one of these parts. Draw $C'D$. \step{By construction $AB$ and $A'D$ are commensurable.}{} \eq{$\therefore \dfrac{\angle A'C'D}{\angle ACB}$} {$=\dfrac{\arc A'D}{\arc AB}$.}{Case~1} By increasing the \emph{number} of equal parts into which $AB$ is divided we can diminish at pleasure the \emph{length} of each part, and therefore make $DB'$ less than any assigned value, however small, since $DB'$ is always less than one of the equal parts into which $AB$ is divided. We cannot make $DB'$ equal to zero, since, by hypothesis, $AB$ and $A'B'$ are incommensurable.~\hfill§~269 Hence, $DB'$ approaches zero as a limit, if the number of parts of $AB$ is indefinitely increased.~\hfill§~275 And the corresponding angle $DC'B'$ approaches zero as a limit. Therefore, the arc $A'D$ approaches the arc $A'B'$ as a limit,~\hfill§~271\linebreak and the $\angle A'C'D$ approaches the $\angle A'C'B'$ as a limit. \step[\indent Therefore,]{$\dfrac{\arc A'D}{\arc AB}$ approaches $\dfrac{\arc A'B'}{\arc AB}$ as a limit,}{§~280} \step[and]{$\dfrac{\angle A'C'D}{\angle ACB}$ approaches $\dfrac{\angle A'C'B'}{\angle ACB}$ as a limit.}{§~280} \step[\indent But]{$\dfrac{\angle A'C'D}{\angle ACB}$ is constantly equal to $\dfrac{\arc A'D}{\arc AB}$,}{} \step{as $A'D$ varies in value and approaches $A'B'$ as a limit.}{} \step{$\therefore \dfrac{\angle A'C'B'}{\angle ACB} = \dfrac{\arc A'B'}{\arc AB}$.}{§~284} \hfill\qed \end{proof} \scanpage{111.png}% \begin{point}% A circumference is divided into $360$~equal parts, called \emph{degrees}; and therefore a unit angle at the centre intercepts a unit arc on the circumference. Hence, the \emph{numerical measure of a central angle} expressed in terms of the unit angle is equal to the \emph{numerical measure of its intercepted arc} expressed in terms of the unit arc. This must be understood to be the meaning when it is said that \emph{A central angle is measured by its intercepted arc.} \end{point} \proposition{Theorem.} \begin{proof}% \obs{An inscribed angle is measured by half the arc intercepted between its sides.} \figc{111ac289}{1. Let the centre $C$ \textnormal{(Fig.~1)} be in one of the sides of the angle.} \prove{the $\angle B$ is measured by $\frac{1}{2}$ the arc $PA$.} \step[\indent\textbf{Proof.}]{Draw $CA$.}{} \eq{$CA$}{$= CB$.}{§~217} \eq{$\therefore \angle B$}{$= \angle A$.}{§~145} \eq[\indent But]{$\angle PCA$}{$= \angle B + \angle A$.}{§~137} \eq{$\therefore \angle PCA$}{$= 2 \angle B$.}{} \step[\indent But]{$\angle PCA$ is measured by $\arc PA$,}{§~288} \pnote{(a central $\angle$ is measured by its intercepted arc).} \step{$\therefore \angle B$ is measured by $\frac{1}{2} \arc PA$.}{} \scanpage{112.png}% \step{}{} \textbf{2. Let the centre $C$ \textnormal{(Fig.~2)} fall within the angle $PBA$.} \prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.} \step[\indent\textbf{Proof.}]{Draw the diameter $BCE$.}{} \setlength{\eqalign}{.33\dentwidth} \eq[\indent Then]{$\angle EBA$}{is measured by $\frac{1}{2} \arc AE$,}{} \eq[and]{$\angle EBP$}{is measured by $\frac{1}{2} \arc EP$.}{Case~1} \eq{$\therefore \angle EBA + \angle EBP$}{is measured by $\frac{1}{2}(\arc AE +\arc EP)$,}{} \eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{} \step{}{} \textbf{3. Let the centre $C$} (Fig.~3) \textbf{fall without the angle $PBA$.} \prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.} \step[\indent\textbf{Proof.}]{Draw the diameter $BCF$.}{} \eq[\indent Then]{$\angle FBA$}{is measured by $\frac{1}{2} \arc FA$,}{} \eq[and]{$\angle FBP$}{is measured by $\frac{1}{2} \arc FP$.}{Case~1} \eq{$\therefore \angle FBA - \angle FBP$}{is measured by $\frac{1}{2}(\arc FA - \arc FP)$,}{} \eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{\qed} \setlength{\eqalign}{.5\dentwidth} \end{proof} \figc{112ac290}{} \pp{\cor[1]{An angle inscribed in a semicircle is a right angle. \textup{For it is measured by half a semicircumference (Fig.~4).}}} \pp{\cor[2]{An angle inscribed in a segment greater than a semicircle is an acute angle. \textup{For it is measured by an arc less than half a semicircumference; as, $\angle CAD$ (Fig.~5).}}} \pp{\cor[3]{An angle inscribed in a segment less than a semicircle is an obtuse angle. \textup{For it is measured by an arc greater than half a semicircumference; as, $\angle CBD$ (Fig.~5).}}} \pp{\cor[4]{Angles inscribed in the same segment or in equal segments are equal \textup{(Fig.~6).}}} \scanpage{113.png}% \proposition{Theorem.} \begin{proof}% \obs{An angle formed by two chords intersecting within the circumference is measured by half the sum of the intercepted arcs.} \figc{113aa294}{Let the angle $COD$ be formed by the chords $AC$ and $BD$.} \prove{the $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$.} \step[\indent\textbf{Proof.}]{Suppose $AE$ drawn $\parallel$ to $BD$.}{} \eq{}{Then $\arc AB = \arc DE$,}{§~257} \pnote{(parallels intercept equal arcs on a circumference).} \eq{}{Also $\angle COD = \angle CAE$,}{§~112} \pnote{(ext.-int.~$\angle_s$ of $\parallel_s$).} \step{But $\angle CAE$ is measured by $\frac{1}{2}(CD + DE)$,}{§~289} \pnote{(an inscribed $\angle$ is measured by half its intercepted arc).} Put $\angle COD$ for its equal, the $\angle CAE$, and $\arc AB$ for its equal, the arc $DE$; then $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$. \hfill\qed \end{proof} \ex{The opposite angles of an inscribed quadrilateral are supplementary.} \ex{If through a point within a circle two perpendicular chords are drawn, the sum of either pair of the opposite arcs which they intercept is equal to a semicircumference.} \ex{The line joining the centre of the square described upon the hypotenuse of a right triangle to the vertex of the right angle bisects the right angle.} \scanpage{114.png}% \proposition{Theorem.} \begin{proof}% \obs{An angle included by a tangent and a chord drawn from the point of contact is measured by half the intercepted arc.} \figc{114aa295}{Let $MAH$ be the angle included by the tangent $MO$ to the circle at $A$ and the chord $AH$.} \prove{the $\angle MAH$ is measured by $\frac{1}{2}$ the arc $AEH$.} \step[\indent\textbf{Proof.}]{Suppose $HF$ drawn $\parallel$ to $MO$.}{} \step{Then $\arc AF = \arc AEH$,}{§~257} \pnote{(parallels intercept equal arcs on a circumference).} \step{Also $\angle MAH = \angle AHF$,}{§~110} \pnote{(alt.-int.~$\angle_s$ of $\parallel_s$).} \step{$\angle AHF$ is measured by $\frac{1}{2} AF$,}{§~289} \pnote{(an inscribed $\angle$ is measured by half its intercepted arc).} Put $\angle MAH$ for its equal, the $\angle AHF$, and $\arc AEH$ for its equal, the arc $AF$; then $\angle MAH$ is measured by $\frac{1}{2} \arc AEH$. Likewise, the $\angle OAH$, the supplement of the $\angle MAH$, is measured by half the arc $AFH$, the conjugate of the arc $AEH$. \hfill\qed \end{proof} \ex{Two circles are tangent externally at $A$, and a common external tangent touches them at $B$ and $C$, respectively. Show that angle $BAC$ is a right angle.} \scanpage{115.png}% \proposition{Theorem.} \begin{proof}% \obs{An angle formed by two secants, two tangents, or a tangent and a secant, drawn to a circle from an external point, is measured by half the difference of the intercepted arcs.} \figc{115ac296}{} The proof of this theorem is left as an exercise for the student. \end{proof} \figcc{115dd297}{115ee297} \begin{point}% \textbf{Positive and Negative Quantities.} In measurements it is convenient to mark the distinction between two quantities that are measured in \emph{opposite directions}, by calling one of them \indexbf{positive} and the other \indexbf{negative}. Thus, if $OA$ is considered positive, then $OC$ may be considered negative, and if $OR$ is considered positive, then $OD$ may be considered negative. When this distinction is applied to angles, an angle is considered to be \emph{positive}, if the rotating line that describes it moves in the opposite direction to the hands of a clock (counter clockwise), and to be \emph{negative}, if the rotating line moves in the same direction as the hands of a clock (clockwise). Arcs corresponding to positive angles are considered \emph{positive}, and arcs corresponding to negative angles are considered \emph{negative}. Thus, the angle $ACB$ described by a line rotating about $C$ from $CA$ to $CB$ is positive, and the arc $AB$ is positive; the angle $ACB'$ described by the line rotating about $C$ from $CA$ to $CB'$ is negative, and the arc $AB'$ is negative. \end{point} \scanpage{116.png}% \pp{\textbf{The Principle of Continuity.}\label{princcont} By marking the distinction between quantities measured in opposite directions, a theorem may often be so stated as to include two or more particular theorems.} The following theorem furnishes a good illustration: \begin{point}% \textit{The angle included between two lines of unlimited length that cut or touch a circumference is measured by half the sum of the intercepted arcs.} Here the word \emph{sum} means the algebraic sum and includes both the arithmetical sum and the arithmetical difference of two quantities. \figc{116ag299}{} 1. If the lines intersect at the centre, the two intercepted arcs are equal, and half the sum will be one of the arcs (§~288). 2. If the lines intersect between the centre and the circumference, the angle is measured by half the sum of the arcs (§~294). 3. If the lines intersect on the circumference, one of the arcs becomes zero and we have an inscribed angle (§~289), or an angle formed by a tangent and a chord (§~295). In each case the angle is measured by half the sum of the intercepted arcs. 4. If the lines intersect without the circumference, then the arc $ab$ is negative and the algebraic sum is the arithmetical difference of the included arcs. When the reasoning employed to prove a theorem is continued in the manner just illustrated to include two or more theorems, we are said to reason by the \emph{Principle of Continuity.} \end{point} \subsection{REVIEW QUESTIONS ON BOOK II.} \begin{myenum} \item What do we call the locus of points in a plane that are equidistant from a fixed point in the plane? \item What does the chord of a segment become when the segment is a semicircle? \item What kind of an angle do the radii of a sector include when the sector is a semicircle? \item What is the difference between a chord and a secant? \item What part of a tangent is meant by a tangent to a circle from an external point? \scanpage{117.png}% \item Two chords are equal in equal circles under either of two conditions. What are the two conditions? \item Points that lie in a straight line are called \emph{collinear}; points that lie in a circumference are called \emph{concyclic}. How many collinear points can be concyclic? \item What is meant by the statement that a central angle is measured by the arc intercepted between its sides? \item What is an inscribed angle? What is its measure? \item What kind of an angle is the angle inscribed in a semicircle? in a segment less than a semicircle? in a segment greater than a semicircle? \item What is the measure of an angle included by two intersecting chords? by two secants intersecting without the circle? \item What is the measure of an angle included by a tangent and a chord drawn to the point of contact? \item When are two quantities of the same kind incommensurable? \item When are two quantities of the same kind commensurable? \item Define a variable and the limit of a variable. \item Does the series $\frac{1}{2}$ in., $\frac{3}{4}$ in., $\frac{7}{8}$ in., $\frac{15}{16}$ in., etc., constitute a variable? Is the variable increasing or decreasing? \item What is the limit of this variable? \item What is the test of a limit? \end{myenum} %\section{EXERCISES.} \subsection{THEOREMS.} \ex{An angle formed by a tangent and a chord is equal to the angle inscribed in the opposite segment.} \ex{Two chords drawn perpendicular to a third chord at its extremities are equal.} \ex{The sum of two opposite sides of a circumscribed quadrilateral is equal to the sum of the other two sides.} \figcc{117aa104}{117bb105} \begin{proofex}% If the sum of two opposite angles of a quadrilateral is equal to two right angles, a circle may be circumscribed about the quadrilateral. Let $\angle A + \angle C = 2 \text{ rt.\ } \angle_s$. Pass a circumference through $D$, $A$, and $B$, and prove that this circumference passes through $C$. \end{proofex} \begin{proofex}% The shortest line that can be drawn from a point within a circle to the circumference is the shorter segment of the diameter through that point. Let $A$ be the given point. Prove $AB$ shorter than any other line $AD$ from $A$ to the circumference. \end{proofex} \scanpage{118.png}% \ex{The longest line that can be drawn from a point within a circle to the circumference is the longer segment of the diameter through that point.} \figc{118aa107}{} \ex{The shortest line that can be drawn from a point without a circle to the circumference will pass through the centre of the circle if produced.} \ex{The longest line that can be drawn from a point without a circle to the concave arc of the circumference passes through the centre of the circle.} \figc{118be108}{} \ex{The shortest chord that can be drawn through a point within a circle is perpendicular to the diameter at that point.} \begin{proofex}% If two intersecting chords make equal angles with the diameter drawn through the point of intersection, the two chords are equal. % special case, two centered columns \hspace{\stretch{1}}Rt.~$\triangle COM =$ rt.~$\triangle CON$. \hspace{\stretch{1}}$\therefore OM = ON$. \hspace{\stretch{1}} \end{proofex} \ex{The angles subtended at the centre of a circle by any two opposite sides of a circumscribed quadrilateral are supplementary.} \begin{proofex}% The radius of a circle inscribed in an equilateral triangle is equal to one third the altitude of the triangle. $\triangle OEF$ is equiangular and equilateral; $\angle FEA = \angle FAE$. % special case, two centered columns \hspace{\stretch{1}}$\therefore AF = EF$. \hspace{\stretch{1}}$\therefore AF = FO = OD$. \hspace{\stretch{1}} \end{proofex} \ex{The radius of a circle circumscribed about an equilateral triangle is equal to two thirds the altitude of the triangle (Ex.~27).} \ex{A parallelogram inscribed in a circle is a rectangle.} \ex{A trapezoid inscribed in a circle is an isosceles trapezoid.} \ex{All chords of a circle which touch an interior concentric circle are equal, and are bisected at the point of contact.} \ex{If the inscribed and circumscribed circles of a triangle are concentric, the triangle is equilateral (Ex.~116).} \scanpage{119.png}% \ex{If two circles are tangent to each other the tangents to them from any point of the common internal tangent are equal.} \ex{If two circles touch each other and a line is drawn through the point of contact terminated by the circumferences, the tangents at its ends are parallel.} \figcc{119aa120}{119bb121} \begin{proofex}% If two circles touch each other and two lines are drawn through the point of contact terminated by the circumferences, the chords joining the ends of these lines are parallel. $\angle A = \angle MPC$ and $\angle B = \angle NPD$. \quad $\therefore \angle A = \angle B$. \end{proofex} \ex{If two circles intersect and a line is drawn through each point of intersection terminated by the circumferences, the chords joining the ends of these lines are parallel.} \figcc{119cc122}{119dd123} \begin{proofex}% Through one of the points of intersection of two circles a diameter of each circle is drawn. Prove that the line joining the ends of the diameters passes through the other point of intersection. \step{$\angle ABC = \angle ABD = 90°$}{§~290} \end{proofex} \ex{If two common external tangents or two common internal tangents are drawn to two circles, the segments intercepted between the points of contact are equal.} \ex{The diameter of the circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse.} \figcc{119ee125}{119ff126} \begin{proofex}% If one leg of a right triangle is the diameter of a circle, the tangent at the point where the circumference cuts the hypotenuse bisects the other leg. \begin{center} $\angle BOE = \angle DOE$.\qquad $\angle BOE = \angle OAD$. $\therefore OE$ and $AC$ are $\parallel$.\qquad $\therefore BE = EC$ (§~188). \end{center} \end{proofex} \ex{If, from any point in the circumference of a circle, a chord and a tangent are drawn, the perpendiculars dropped on them from the middle point of the subtended arc are equal. $\angle BAM = \angle CAM$.} \ex{The median of a trapezoid circumscribed about a circle is equal to one fourth the perimeter of the trapezoid (Ex.~103).} \scanpage{120.png}% \ex{Two fixed circles touch each other externally and a circle of variable radius touches both externally. Show that the difference of the distances from the centre of the variable circle to the centres of the fixed circles is constant.} \ex{If two fixed circles intersect, and circles are drawn to touch both, show that either the sum or the difference of the distances of their centres from the centres of the fixed circles is constant, according as they touch (i)~one internally and one externally, (ii)~both internally or both externally.} \figcc{120aa130}{120bb131} \begin{proofex}% If two straight lines are drawn through any point in a diagonal of a square parallel to the sides of the square, the points where these lines meet the sides lie on the circumference of a circle whose centre is the point of intersection of the diagonals. $\triangle POE = \triangle POF$ (§~143). \quad $\therefore OE = OF$. \quad $\triangle POE' = \triangle POF'$. \end{proofex} \begin{proofex}% If $ABC$ is an inscribed equilateral triangle and $P$ is any point in the arc $BC$, then $PA = PB + PC$. Take $PM = PB$. $\triangle ABM = \triangle CBP$ (§~143) and $AM = PC$. \end{proofex} \ex{The tangents drawn through the vertices of an inscribed rectangle, which is not a square, enclose a rhombus.} \figc{120cd132}{} \begin{proofex}% The bisectors of the angles included by the opposite sides (produced) of an inscribed quadrilateral intersect at right angles. \eq{Arc $AF - \arc BM$}{$= \arc DF - \arc CM$}{} \eq{and $\arc AH - \arc DN$}{$= \arc BH - \arc CN$.}{} \eq{$\therefore \arc FH + \arc MN$}{$= \arc HM + \arc FN$.}{} \eq{$\therefore \angle FIH$}{$= \angle HIM$.}{} \textbf{Discussion.} This problem is impossible, if any two sides of the quadrilateral are parallel. \end{proofex} \scanpage{121.png}% \section{PROBLEMS OF CONSTRUCTION.} \note{Hitherto we have supposed the figures constructed. We now proceed to explain the methods of constructing simple problems, and afterwards to apply these methods to the solution of more difficult problem.} \proposition{Problem.} \begin{proof}% \obs{To let fall a perpendicular upon a given line from a given external point.} \figc{121aa300}{Let $AB$ be the given straight line, and $C$ the given external point.} \prove[]{To let fall a $\perp$ to the line $AB$ from the point $C$}. From $C$ as a centre, with a radius sufficiently great, describe an arc cutting $AB$ in two points, $H$ and $K$. From $H$ and $K$ as centres, with equal radii greater than $\frac{1}{2}HK$, \step{describe two arcs intersecting at $O$.}{} \step{Draw $CO$,}{} \step{and produce it to meet $AB$ at $M$.}{} \step{$CM$ is the $\perp$ required.}{} \textbf{Proof.} Since $C$ and $O$ are two points each equidistant from $H$ and $K$, they determine a $\perp$ to $HK$ at its middle point.~\hfill§~161 \hfill\qef \end{proof} \note{\emph{Given} lines of the figures are represented by full lines, \emph{resulting} lines by long-dashed, and \emph{auxiliary} lines by short-dashed lines.} \scanpage{122.png}% \proposition{Problem.} \begin{proof}% \obs{At a given point in a straight line, to erect a perpendicular to that line.} \figc{122ab301}{1. Let $O$ be the given point in $AC$. \textnormal{Fig.~1.}} \step{Take $OH$ and $OB$ equal.}{} From $H$ and $B$ as centres, with equal radii greater than $OB$, describe two arcs intersecting at $R$. Join $OR$. Then the line $OR$ is the $\perp$ required. \textbf{Proof.} $O$ and $R$, two points each equidistant from $H$ and $B$, determine the perpendicular bisector of $HB$.~\hfill§~161 \lett{2. Let $B$ be the given point. \textnormal{Fig.~2.}} Take any point $C$ without $AB$; and from $C$ as a centre, with the distance $CB$ as a radius, describe an arc intersecting $AB$ at $E$. Draw $EC$, and prolong it to meet the arc again at $D$. Join $BD$, and $BD$ is the $\perp$ required. \step[\indent\textbf{Proof.}]{The $\angle B$ is a right angle.}{§~290} \step{$\therefore BD$ is $\perp$ to $AB$.}{\qef} \textbf{Discussion.} The point $C$ must be so taken that it will not be in the required perpendicular. \end{proof} \scanpage{123.png}% \proposition{Problem.} \begin{proof}% \obs{To bisect a given straight line.} \figc{123aa302}{To bisect the given straight line $AB$.} From $A$ and $B$ as centres, with equal radii greater than $\frac{1}{2} AB$, describe arcs intersecting at $C$ and $E$. \step{Join $CE$.}{} \step{Then $CE$ bisects $AB$.}{§~161} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To bisect a given arc.} \figc{123bb303}{To bisect the given arc $AB$.} \step{Draw the chord $AB$.}{} From $A$ and $B$ as centres, with equal radii greater than $\frac{1}{2} AB$, describe arcs intersecting at $D$ and $E$. \scanpage{124.png}% \step{Draw $DE$.}{} \step{Then $DE$ is the $\perp$ bisector of the chord $AB$.}{§~161} \step{$\therefore DE$ bisects the arc $ACB$.}{§~248} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To bisect a given angle.} \figc{124aa304}{Let $AEB$ be the given angle.} From $E$ as a centre, with any radius, as $EA$, describe an arc cutting the sides of the $\angle E$ at $A$ and $B$. From $A$ and $B$ as centres, with equal radii greater than half the distance from $A$ to $B$, describe two arcs intersecting at $D$. \step{Draw $DE$.}{} \step{Then $DE$ bisects the arc $AB$ at $C$.}{§~303} \step{$\therefore DE$ bisects the angle $E$.}{§~237} \hfill\qef \end{proof} \ex{To construct an angle of $45°$; of $135°$.} \ex{To construct an equilateral triangle, having given one side.} \ex{To construct an angle of $60°$; of $150°$.} \ex{To trisect a right angle.} \scanpage{125.png}% \proposition{Problem.} \begin{proof}% \obs{At a given point in a given straight line, to construct an angle equal to a given angle.} \figc{125ab305}{At $C$ in the line $CM$, construct an angle equal to the given angle $A$.} From $A$ as a centre, with any radius, $AE$, describe an arc cutting the sides of the $\angle A$ at $E$ and $F$. \step{From $C$ as a centre, with a radius equal to $AE$,}{} \step{describe an arc $HG$ cutting $CM$ at $H$.}{} \step{From $H$ as a centre, with a radius equal to the chord $EF$,}{} \step{describe an arc intersecting the arc $HG$ at $O$.}{} \step{Draw $CO$, and $\angle$ $HCO$ is the required angle.}{Why?} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To draw a straight line parallel to a given straight line through a given external point.} \figc{125cc306}{Let $AB$ be the given line, and $C$ the given point.} \scanpage{126.png}% \step{Draw $ECD$, making any convenient $\angle EDB$.}{} \step{At the point $C$ construct $\angle ECF$ equal to $\angle EDB$.}{§~305} \step{Then the line $HCF$ is $\parallel$ to $AB$.}{Why?} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To divide a given straight line into a given number of equal parts.} \figc{126aa307}{Let $AB$ be the given straight line.} From $A$ draw the line $AO$, making any convenient angle with $AB$. Take any convenient length, and apply it to $AO$ as many times as the line $AB$ is to be divided into parts. From $C$, the last point thus found on $AO$, draw $CB$. Through the points of division on $AO$ draw parallels to the line $CB$.~\hfill§~306 \step{These lines will divide $AB$ into equal parts.}{§~187} \hfill\qef \end{proof} \ex{To construct an equilateral triangle, having given the perimeter.} \ex{To divide a line into four equal parts by two different methods.} \begin{proofex}% Through a given point to draw a line which shall make equal angles with the two sides of a given angle. Through the given point draw a $\perp$ to the bisector of the given $\angle$. \end{proofex} \ex{To draw a line through a given point, so that it shall form with the sides of a given angle an isosceles triangle (Ex.~140).} \scanpage{127.png}% \proposition{Problem.} \begin{proof}% \obs{To find the third angle of a triangle when two of the angles are given.} \figc{127ab308}{Let $A$ and $B$ be the two given angles.} \step{At any point $H$ in any line $EF$,}{} \step{construct $\angle a$ equal to $\angle A$, and $\angle b$ equal to $\angle B$.}{§~305} \step[\indent Then]{$\angle c$ is the $\angle$ required.}{Why?} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To construct a triangle when two sides and the included angle are given.} \figc{127cd309}{Let $b$ and $c$ be the two sides of the triangle and $E$ the included angle.} \step{Take $AB$ equal to the side $c$.}{} At $A$, construct $\angle BAD$ equal to the given $\angle E$.~\hfill§~305 \scanpage{128.png}% \step{On $AD$ take $AC$ equal to $b$, and draw $CB$.}{} \step{Then $\triangle ACB$ is the $\triangle$ required.}{\qef} \end{proof} \proposition{Problem.} \begin{proof}% \obs{To construct a triangle when a side and two angles of the triangle are given.} \figc{128ab310}{Let $c$ be the given side, $A$ and $B$ the given angles.} \step{Take $EC$ equal to the side $c$.}{} \step{At $E$ construct the $\angle CEH$ equal to $\angle A$.}{§~305} \step{At $C$ construct the $\angle ECK$ equal to $\angle B$.}{} \step{Produce $EH$ and $CK$ until they intersect at $O$.}{} \step{Then $\triangle COE$ is the $\triangle$ required.}{\qef} \textsc{Remark.} If one of the given angles is opposite to the given side, find the third angle by §~308, and proceed as above. \textbf{Discussion.} The problem is impossible when the two given angles are together equal to or greater than two right angles. \end{proof} \ex{To construct an equilateral triangle, having given the altitude.} \exheader{To construct an isosceles triangle, having given:} \ex{The base and the altitude.} \ex{The altitude and one of the legs.} \ex{The angle at the vertex and the altitude.} \scanpage{129.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a triangle when two sides and the angle opposite one of them are given.} \figc{129ab311}{Let $a$ and $b$ be the given sides, and $A$ the angle opposite $a$.} \textsc{Case 1.} \emph{If $a$~is less than~$b$.} \step{Construct $\angle DAE$ equal to the given $\angle A$}{§~305} \step{On $AD$ take $AB$ equal to $b$.}{} \step{From $B$ as a centre, with a radius equal to $a$,}{} \step{describe an arc intersecting the line $AE$ at $C$ and $C'$.}{} \step{Draw $BC$ and $BC'$.}{} Then both the $\triangle_s ABC$ and $ABC'$ fulfil the conditions, and hence we have two constructions. This is called the \emph{ambiguous} case. \figcc{129cc311}{129dd311} \textbf{Discussion.} If the side $a$ is equal to the $\perp BH$, the arc described from $B$ will touch $AE$, and there will be but one construction, the right $\triangle ABH$. If the given side $a$ is less than the $\perp$ from $B$, the arc described from $B$ will not intersect or touch $AE$, and hence the problem is impossible. \scanpage{130.png}% If the $\angle A$ is right or obtuse, the problem is impossible; for the side opposite a right or obtuse angle is the greatest side.~\hfill§~153 \filbreak \textsc{Case 2.} \emph{If $a$ is equal to $b$.} \figc{130aa311}{} If the $\angle A$ is acute, and $a = b$, the arc described from $B$ as a centre, and with a radius equal to $a$, will cut the line $AE$ at the points $A$ and $C$. There is therefore but one solution: the isosceles $\triangle ABC$. \textbf{Discussion.} If the $\angle A$ is right or obtuse, the problem is impossible; for equal sides of a $\triangle$ have equal $\angle_s$ opposite them, and a $\triangle$ cannot have two right $\angle_s$ or two obtuse $\angle_s$. \figccc{130bb311}{130cc311}{130dd311} \textsc{Case 3.} \emph{If $a$ is greater than $b$.} If the given $\angle A$ is acute, the arc described from $B$ will cut the line $ED$ on opposite sides of $A$, at $C$ and $C'$. The $\triangle ABC$ answers the required conditions, but the $\triangle$ $ABC'$ does not, for it does not contain the acute $\angle A$. There is then only one solution; namely, the $\triangle ABC$. If the $\angle A$ is right, the arc described from $B$ cuts the line $ED$ on opposite sides of $A$, and we have two \emph{equal} right $\triangle_s$ which fulfil the required conditions. If the $\angle A$ is obtuse, the arc described from $B$ cuts the line $ED$ on opposite sides of $A$, at the points $C$ and $C'$. The $\triangle ABC$ answers the required conditions, but the $\triangle ABC'$ does not, for it does not contain the obtuse $\angle A$. There is then only one solution; namely, the $\triangle ABC$.~\hfill\qef \end{proof} \scanpage{131.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a triangle when the three sides of the triangle are given.} \figc{131aa312}{Let the three sides be $c$, $a$, and $b$.} Take $AB$ equal to $c$. From $A$ as a centre, with a radius equal to $b$, describe an arc. From $B$ as a centre, with a radius equal to $a$, describe an arc, intersecting the other arc at $C$. \step{Draw $CA$ and $CB$.}{} \step{$\triangle CAB$ is the $\triangle$ required.}{\qef} \textbf{Discussion.} The problem is impossible when one side is equal to or greater than the sum of the other two sides. \end{proof} \proposition{Problem.} \begin{proof}% \obs{To construct a parallelogram when two sides and the included angle are given}. \figc{131bb313}{Let $m$ and $o$ be the two sides, and $C$ the included angle.} \step{Take $AB$ equal to $o$.}{} \step{At $A$ construct $\angle BAD$ equal to $\angle C$.}{§~305} \scanpage{132.png}% Take $AH$ equal to $m$. From $H$ as a centre, with a radius equal to $o$, describe an arc, and from $B$ as a centre, with a radius equal to $m$, describe an arc, intersecting the other arc at $E$; and draw $EH$ and $EB$. \step{The quadrilateral $ABEH$ is the $\Par$ required.}{§~182} \hfill\qef \end{proof} \proposition{Problem.} \begin{proof}% \obs{To circumscribe a circle about a given triangle.} \figc{132aa314}{Let $ABC$ be the given triangle.} \step{Bisect $AB$ and $BC$.}{§~302} \step{At $E$ and $D$, the points of bisection, erect $\perp_s$.}{§~301} Since $BC$ is not the prolongation of $AB$, these $\perp_s$ will intersect at some point $O$. From $O$, with a radius equal to $OB$, describe a circle. \step{The $\odot ABC$ is the $\odot$ required.}{} \step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$,}{} \step{and also is equidistant from $B$ and $C$.}{§~160} \step{$\therefore$ the point $O$ is equidistant from $A$, $B$, and $C$,}{} \noindent and a $\odot$ described from $O$ as a centre, with a radius equal to $OB$, will pass through the vertices $A$, $B$, and $C$.~\hfill\qef \end{proof} The same construction serves to describe a circumference which shall pass through three points not in the same straight line; also to find the centre of a given circle or of a given arc. \note{The point $O$ is called the \emph{circum-centre}\label{circum-centre} of the triangle.} \scanpage{133.png}% \proposition{Problem.} \begin{proof}% \obs{To inscribe a circle in a given triangle.} \figc{133aa315}{Let $ABC$ be the given triangle.} \step{Bisect the $\angle_s A$ and $C$.}{§~304} \step{From $E$, the intersection of the bisectors,}{} \step{draw $EH \perp$ to the side $AC$.}{§~300} \step{From $E$ as centre, with radius $EH$, describe the $\odot KHM$.}{} \step{The $\odot KHM$ is the $\odot$ required.}{} \textbf{Proof.} Since $E$ is in the bisector of the $\angle A$, it is equidistant from the sides $AB$ and $AC$; and since $E$ is in the bisector of the $\angle C$, it is equidistant from the sides $AC$ and $BC$.~\hfill§~162 $\therefore$ a $\odot$ described from $E$ as centre, with a radius equal to $EH$, will touch the sides of the $\triangle$ and be inscribed in it.~\hfill\qef \end{proof} \note{The point $E$ is called the \indexemph{in-centre} of the triangle.} \figc{133bb316}{} \begin{point}% The intersections of the bisectors of the exterior angles of a triangle are the centres of three circles, each of which will touch one side of the triangle, and the two other sides produced. These three circles are called \emph{escribed}\label{escribed} circles; and their centres are called the \indexemph{ex-centres} of the triangle. \end{point} \scanpage{134.png}% \proposition{Problem.} \begin{proof}% \obs{Through a given point, to draw a tangent to a given circle.} \textsc{Case 1.} \emph{When the given point is on the circumference.} \figc{134aa317}{Let $C$ be the given point on the circumference whose centre is~$O$.} \step{From the centre $O$ draw the radius $OC$.}{} \step{Through $C$ draw $AM \perp$ to $OC$.}{§~301} \step{Then $AM$ is the tangent required.}{§~253} \textsc{Case 2.} \emph{When the given point is without the circle.} \lett{Let $O$ be the centre of the given circle, $E$ the given point.} \step{Draw $OE$.}{} On $OE$ as a diameter, describe a circumference intersecting the given circumference at the points $M$ and $H$. \step{Draw $OM$ and $EM$.}{} \step{Then $EM$ is the tangent required.}{} \step[\indent\textbf{Proof.}]{$\angle OME$ is a right angle.}{§~290} \step{$\therefore EM$ is tangent to the circle at $M$.}{§~253} In like manner, we may prove $EH$ tangent to the given $\odot$. \hfill\qef \end{proof} \ex{To draw a tangent to a given circle, so that it shall be parallel to a given straight line.} \scanpage{135.png}% \proposition{Problem.} \begin{proof}% \obs{Upon a given straight line, to describe a segment of a circle in which a given angle may be inscribed.} \figc{135aa318}{Let $AB$ be the given line, and $M$ the given angle.} \step{Construct the $\angle ABE$ equal to the $\angle M$.}{§~305} \step{Bisect the line $AB$ by the $\perp OF$.}{§~302} \step{From the point $B$ draw $BO \perp$ to $EB$.}{§~301} From $O$, the point of intersection of $FO$ and $BO$, as a centre with a radius equal to $OB$, describe a circumference. \step{The segment $AKB$ is the segment required.}{} \step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$.}{§~160} \step{$\therefore$ the circumference will pass through $A$.}{} \step{But $BE$ is $\perp$ to $OB$.}{Const.} \step{$\therefore BE$ is tangent to the $\odot$,}{§~253} \pnote{(a straight line $\perp$ to a radius at its extremity is tangent to the $\odot$).} \step{$\therefore \angle ABE$ is measured by $\frac{1}{2} \arc AB$,}{§~295} \pnote{(being an $\angle$ formed by a tangent and a chord).} But any $\angle$ as $\angle K$ inscribed in the segment $AKB$ is measured by $\frac{1}{2} \arc AB$.\hfill§~289 $\therefore$ the $\angle M$ may be inscribed in the segment $AKB$. \hfill\qef \end{proof} \scanpage{136.png}% \subsection{SOLUTION OF PROBLEMS.} \pp{If a problem is so simple that the solution is obvious from a known theorem, we have only to make the construction according to the theorem, and then give a synthetic proof, if a proof is necessary, that the construction is correct, as in the examples of the fundamental problems already given.} \begin{point}% But problems are usually of a more difficult type. The application of known theorems to their solution is not immediate, and often far from obvious. To discover the mode of application is the first and most difficult part of the solution. The best way to attack such problems is by a method resembling the analytic proof of a theorem, called the \textbf{analysis} of the problem. 1. \textbf{Suppose the construction made}, and let the figure represent all parts concerned, both given and required. 2. Study the relations among the parts with the aid of known theorems, and try to find some relation that will suggest the construction. 3. If this attempt fails, introduce new relations by drawing auxiliary lines, and study the new relations. If this attempt fails, make a new trial, and so on till a clue to the right construction is found. \end{point} \pp{A problem is \emph{determinate} if it has a \emph{definite} number of solutions, \emph{indeterminate} if it has an \emph{indefinite} number of solutions, and \emph{impossible} if it has \emph{no} solution. A problem is sometimes determinate for certain relative positions or magnitudes of the given parts, and indeterminate for other positions or magnitudes of the given parts.} \pp{The \textbf{discussion} of a problem consists in examining the problem with reference to all possible conditions, and in determining the conditions necessary for its solution.} \scanpage{137.png}% \begin{proofex}% \obs{\textsc{Problem.} To construct a circle that shall pass through a given point and cut chords of a given length from two parallels.} \figc{137aa147}{} \textbf{Analysis.} Suppose the problem solved. Let $A$ be the given point, $BC$ and $DE$ the given parallels, $MN$ the given length, and $O$ the centre of the required circle. Since the circle cuts equal chords from two parallels its centre must be equidistant from them. Therefore, one locus for $O$ is $FG \parallel$ to $BC$ and equidistant from $BC$ and $DE$. Draw the $\perp$ bisector of $MN$, cutting $FG$ in $P$. $PM$ is the radius of the circle required. With $A$ as centre and radius $PM$ describe an arc cutting $FG$ at $O$. Then $O$ is the centre of the required circle. \textbf{Discussion.} The problem is impossible if the distance from $A$ to $FG$ is greater than $PM$. \end{proofex} \figc{137bb148}{} \begin{proofex}% \obs{\textsc{Problem.} To construct a triangle, having given the perimeter, one angle, and the altitude from the vertex of the given angle.} \textbf{Analysis.} Suppose the problem solved, and let $ABC$ be the $\triangle$ required, $ACB$ the given $\angle$, and $CD$ the given altitude. Produce $AB$ both ways, and take $AE=AC$, and $BF=BC$, then $EF=$ the given perimeter. Join $CE$ and $CF$, forming the isosceles $\triangle_s CAE$ and $CBF$. In the $\triangle ECF$, $\angle E+\angle F+\angle ECF=180°$ (why?), but $\angle ECF=\angle ECA+\angle FCB+\angle ACB$. Since $\angle E=\angle ECA$ and $\angle F=\angle FCB$, we have $\angle ECF=\angle E+\angle F+\angle ACB$.\quad $\therefore 2\angle E+2\angle F+\angle ACB = 180°$. \( \therefore \angle E+\angle F+\frac{1}{2}\angle ACB = 90° \), and\ \( \angle E+\angle F = 90° - \frac{1}{2}\angle ACB \). By substitution, \(\angle ECF = 90° + \frac{1}{2}\angle ACB \). $\therefore \angle ECF$ is known. \end{proofex} \textbf{Construction.} To find the point $C$, construct on $EF$ a segment that will contain the $\angle ECF$ (§~318), and draw a parallel to $EF$ at the distance $CD$, the given altitude. To find the points $A$ and $B$, draw the $\perp$ bisectors of the lines $CE$ and $CF$, and the points $A$ and $B$ will be vertices of the required $\triangle$. Why? \scanpage{138.png}% \section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.} \ex{Find the locus of a point at a given distance from a given circumference.} \exheader{Find the locus of the centre of a circle:} \ex{Which has a given radius $r$ and passes through a given point $P$.} \ex{Which has a given radius $r$ and touches a given line $AB$.} \ex{Which passes through two given points $P$ and $Q$.} \ex{Which touches a given straight line $AB$ at a given point $P$.} \ex{Which touches each of two given parallels.} \ex{Which touches each of two given intersecting lines.} \begin{proofex}% To find in a given line a point $X$ which is equidistant from two given points. The required point is the intersection of the given line with the perpendicular bisector of the line joining the two given points (§~160). \end{proofex} \ex{To find a point $X$ equidistant from three given points.} \figc{138aa158}{} \ex{To find a point $X$ equidistant from two given points and at a given distance from a third given point.} \ex{To construct a circle which has a given radius and passes through two given points.} \ex{To find a point $X$ at given distances from two given points.} \ex{To construct a circle which has its centre in a given line and passes through two given points.} \ex{To find a point $X$ equidistant from two given points and also equidistant from two given intersecting lines (§§~160 and 162).} \ex{To find a point $X$ equidistant from two given points and also equidistant from two given parallel lines.} \ex{To find a point $X$ equidistant from two given intersecting lines and also equidistant from two given parallels.} \figc{138bb165}{} \ex{To find a point $X$ equidistant from two given intersecting lines and at a given distance from a given point.} \ex{To find a point $X$ which lies in one side of a given triangle and is equidistant from the other two sides.} \scanpage{139.png}% % 169 = c % 174 = d % 175 = e % 177 = f % 176 = g \filbreak \figcc{139aa167}{139bb168} \ex{A straight railway passes two miles from a town. A place is four miles from the town and one mile from the railway. To find by construction the places that answer this description.} \ex{In a triangle $ABC$, to draw $DE$ parallel to the base $BC$, cutting the sides of the triangle in $D$ and $E$, so that $DE$ shall equal $DB + EC$ (§~162).} \figc{139cc169}{} \begin{proofex}% To draw through two sides of a triangle a line parallel to the third side so that the part intercepted between the sides shall have a given length. \null \step{Take $BD = d$.}{} \end{proofex} \ex{Prove that the locus of the vertex of a right triangle, having a given hypotenuse as base, is the circumference described upon the given hypotenuse as diameter (§~290). } \ex{Prove that the locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms with the base a segment capable of containing the given angle (§~318).} \ex{Find the locus of the middle point of a chord of a given length that can be drawn in a given circle.} \ex{Find the locus of the middle point of a chord drawn from a given point in a given circumference.} \figccc{139dd174}{139ee175}{139gg176} \ex{Find the locus of the middle point of a straight line drawn from a given exterior point to a given circumference.} \ex{A straight line moves so that it remains parallel to a given line, and touches at one end a given circumference. Find the locus of the other end.} \ex{A straight rod moves so that its ends constantly touch two fixed rods which are perpendicular to each other. Find the locus of its middle point. } \scanpage{140.png}% \ex{In a given circle let $AOB$ be a diameter, $OC$ any radius, $CD$ the perpendicular from $C$ to $AB$. Upon $OC$ take $OM$ equal to $CD$. Find the locus of the point $M$ as $OC$ turns about $O$.} \figcc{139ff177}{140aa178} \ex{To construct an equilateral triangle, having given the radius of the circumscribed circle.} \exheader{To construct on isosceles triangle, having given:} \ex{The angle at the vertex and the base (§~160 and §~318).} \ex{The base and the radius of the circumscribed circle.} \ex{The base and the radius of the inscribed circle.} \figc{140bb182}{} \begin{proofex}% The perimeter and the altitude. Let $ABC$ be the $\triangle$ required, $EF$ the given perimeter. The altitude $CD$ passes through the middle of $EF$, and the $\triangle_s AEC$, $BFC$ are isosceles. \end{proofex} \exheader{To construct a right triangle, having given:} \ex{The hypotenuse and one leg.} \ex{One leg and the altitude upon the hypotenuse.} \ex{The median and the altitude drawn from the vertex of the right angle.} \ex{The hypotenuse and the altitude upon the hypotenuse.} \ex{The radius of the inscribed circle and one leg.} \ex{The radius of the inscribed circle and an acute angle.} \ex{An acute angle and the sum of the legs.} \ex{An acute angle and the difference of the legs.} \figc{140cc191}{} \ex{To construct an equilateral triangle, having given the radius of the inscribed circle.} \exheader{To construct a triangle, having given:} \ex{The base, the altitude, and an angle at the base.} \ex{The base, the altitude, and the $\angle$ at the vertex.} \ex{The base, the corresponding median, and the $\angle$ at the vertex.} \ex{The perimeter and the angles.} \ex{One side, an adjacent $\angle$, and the sum of the other sides.} \scanpage{141.png}% \exheader{To construct a triangle, having given:} \ex{One side, an adjacent $\angle$, and the difference of the other sides.} \ex{The sum of two sides and the angles.} \ex{One side, an adjacent $\angle$, and the radius of the circumscribed circle.} \ex{The angles and the radius of the circumscribed circle.} \ex{The angles and the radius of the inscribed circle. } \ex{An angle, and the bisector and the altitude drawn from the vertex of the given angle.} \ex{Two sides and the median corresponding to the other side.} \ex{The three medians.} \exheader{To construct a square, having given:} \ex{The diagonal.} \begin{proofex}% The sum of the diagonal and one side. Let $ABCD$ be the square required, $CA$ the diagonal. Produce $CA$, making $AE = AB$. $\triangle_s ABC$ and $ABE$ are isosceles and $\angle BAC = \angle BCA = 45°$. \end{proofex} \figcc{141aa206}{141bb207} \ex{Given two perpendiculars, $AB$ and $CD$, intersecting in $O$, and a straight line intersecting these perpendiculars in $E$ and $F$; to construct a square, one of whose angles shall coincide with one of the right angles at $O$, and the vertex of the opposite angle of the square shall lie in $EF$. (Two solutions.)} \exheader{To construct a rectangle, having given:} \ex{One side and the angle between the diagonals.} \ex{The perimeter and the diagonal.} \ex{The perimeter and the angle between the diagonals.} \ex{The difference of two adjacent sides and the angle between the diagonals.} \exheader{To construct a rhombus, having given:} \ex{The two diagonals.} \ex{One side and the radius of the inscribed circle.} \scanpage{142.png}% \ex{One angle and the radius of the inscribed circle.} \ex{One angle and one of the diagonals.} \exheader{To construct a rhomboid, having given:} \ex{One side and the two diagonals.} \ex{The diagonals and the angle between them.} \ex{One side, one angle, and one diagonal.} \ex{The base, the altitude, and one angle.} \exheader{To construct an isosceles trapezoid, having given:} \ex{The bases and one angle.} \ex{The bases and the altitude.} \ex{The bases and the diagonal.} \figc{142aa223}{} \begin{proofex}% The bases and the radius of the circumscribed circle. Let $ABCD$ be the isosceles trapezoid required, $O$ the centre of the circumscribed $\odot$. A diameter $\perp$ to $AB$ is $\perp$ to $CD$, and bisects both $AB$ and $CD$. Draw $CG$ $\parallel$ to $FE$. Then $EG = FC = \frac{1}{2}DC$. \end{proofex} \exheader{To construct a trapezoid, having given:} \ex{The four sides.} \ex{The two bases and the two diagonals.} \ex{The bases, one diagonal, and the $\angle$ between the diagonals.} \exheader{To construct a circle which has the radius $r$ and which also:} \ex{Touches each of two intersecting lines $AB$ and $CD$.} \ex{Touches a given line $AB$ and a given circle~$K$.} \ex{Passes through a given point $P$ and touches a given line~$AB$.} \ex{Passes through a given point~$P$ and touches a given circle~$K$.} \exheader{To construct a circle which shall:} \ex{Touch two given parallels and pass through a given point~$P$.} \ex{Touch three given lines two of which are parallel.} \ex{Touch a given line $AB$ at $P$ and pass through a given point~$Q$.} \ex{Touch a given circle at $P$ and pass through a given point~$Q$.} \ex{Touch two given lines and touch one of them at a given point~$P$.} \scanpage{143.png}% \ex{Touch a given line and touch a given circle at a point $P$.} \ex{Touch a given line $AB$ at $P$ and also touch a given circle.} \ex{To inscribe a circle in a given sector.} \ex{To construct within a given circle three equal circles, so that each shall touch the other two and also the given circle.} \ex{To describe circles about the vertices of a given triangle as centres, so that each shall touch the two others.} \figc{143aa241}{} \begin{proofex}% To bisect the angle formed by two lines, without producing the lines to their point of intersection. Draw any line $EF \parallel$ to $BA$. Take $EG = EH$, and produce $GH$ to meet $BA$ at $I$. Draw the $\perp$ bisector of $GI$. \figccc{143bb242}{143cc243}{143dd244} \end{proofex} \ex{To draw through a given point $P$ between the sides of an angle $BAC$ a line terminated by the sides of the angle and bisected at $P$.} \begin{proofex}% Given two points $P$, $Q$, and a line $AB$; to draw lines from $P$ and $Q$ which shall meet on $AB$ and make equal angles with $AB$. Make use of the point which forms with $P$ a pair of points symmetrical with respect to $AB$. \end{proofex} \ex{To find the shortest path from $P$ to $Q$ which shall touch a line~$AB$.} \figc{143ee245}{} \begin{proofex}% To draw a common tangent to two given circles. Let $r$ and $r'$ denote the radii of the circles, $O$ and $O'$ their centres. With centre $O$ and radius $r-r'$ describe a $\odot$. From $O'$ draw the tangents $O'M$, $O'N$. Produce $OM$ and $ON$ to meet the circumference at $A$ and $C$. Draw the radii $O'B$ and $O'D \parallel$, respectively, to $OA$ and $OC$. Draw $AB$ and $CD$. To draw the internal tangents use an auxiliary $\odot$ of radius $r + r'$. \end{proofex} \scanpage{144.png}% \chapter{BOOK III\@. PROPORTION\@. SIMILAR POLYGONS.} \section[THEORY OF PROPORTION.]{THE THEORY OF PROPORTION.} \pp{A \indexbf{proportion} is an expression of equality between two equal ratios; and is written in one of the following forms: \[ a:b = c:d; \qquad a:b::c:d; \qquad \frac{a}{b}=\frac{c}{d}. \] This proportion is read, ``$a$ is to $b$ as $c$ is to $d$''; or ``the ratio of $a$ to $b$ is equal to the ratio of $c$ to $d$.''} \begin{point}% The \indexbf{terms} of a proportion are the four quantities compared; the \emph{first} and \indexemph{third} terms are called the \indexbf{antecedents}, the \emph{second} and \indexemph{fourth} terms, the \indexbf{consequents}; the \emph{first} and \emph{fourth} terms, the \indexbf{extremes}, the \emph{second} and \emph{third} terms, the \indexbf{means}. Thus, in the proportion $a : b = c : d$; $a$ and $c$ are the antecedents, $b$ and $d$ the consequents, $a$ and $d$ the extremes, $b$ and $c$ the means. \end{point} \begin{point}% The fourth proportional to three given quantities is the fourth term of the proportion which has for its first three terms the three given quantities \emph{taken in order.} Thus, $d$ is the fourth proportional to $a$, $b$, and $c$ in the proportion \[ a:b = c:d. \] \end{point} \begin{point}% The quantities $a$, $b$, $c$, $d$, $e$, are said to be in \textbf{continued proportion}\label{continuedprop}, if $a:b = b:c = c:d = d:e$. If three quantities are in continued proportion, the second is called the \indexbf{mean proportional} between the other two, and the third is called the \textbf{third proportional} to the other two. Thus, in the proportion $a:b = b:c$; $b$ is the mean proportional between $a$ and $c$; and $c$ is the third proportional to $a$ and $b$. \end{point} \scanpage{145.png}% \proposition{Theorem.} \begin{proof}% \obs{In every proportion the product of the extremes is equal to the product of the means.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{§~323} \eq[\indent Whence]{$ad$}{$= bc$.}{\qed} \end{proof} \proposition{Theorem.} \begin{proof}% \obs{The mean proportional between two quantities is equal to the square root of their product.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= b:c}$.}{} \eq[\indent Then]{$b^2$}{$= ac$.}{§~327} Whence, extracting the square root, \eq{$b$}{$= \sqrt{ac}$.}{\qed} \end{proof} \proposition{Theorem.} \begin{proof}% \obs{If the product of two quantities is equal to the product of two others, either two may be made the extremes of the proportion in which the other two are made the means.} \eq[\indent\textbf{Let}]{$\mathbf{ad}$}{$\mathbf{= bc}$.}{} \proveq{$a:b$}{$= c:d$} Divide both members of the given equation by $bd$. \eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{} \eq[\indent Or]{$a:b$}{$= c:d$.}{\qed} \end{proof} \scanpage{146.png}% \proposition{Theorem.} \begin{proof}% \obs{If four quantities are in proportion, they are in proportion by \textnormal{\textbf{alternation}}\label{alternation}; that is, the first term is to the third as the second is to the fourth.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \proveq{$a:c$}{$=b:d$.} \eq[\indent Now]{$ad$}{$=bc$.}{§~327} Divide each member of the equation by $cd$. \eq[\indent Then]{$\dfrac{a}{c}$}{$= \dfrac{b}{d}$.}{} \eq[\indent Or]{$a:c$}{$= b:d$.}{\qed} \end{proof} \proposition{Theorem.} \begin{proof}% \obs{If four quantities are in proportion, they are in proportion by \textnormal{\indexbf{inversion}}; that is, the second term is to the first as the fourth is to the third.} \eq[\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \proveq{$b:a$}{$= d:c$.} \eq[\indent Now]{$bc$}{$= ad$.}{§~327} Divide each member of the equation by $ac$. \eq[\indent Then]{$\dfrac{b}{a}$}{$=\dfrac{d}{c}$.}{} \eq[\indent Or]{$b:a$}{$= d:c$.}{\qed} \end{proof} \scanpage{147.png}% \proposition{Theorem.} \begin{proof} \obs{If four quantities are in proportion, they are in proportion by \textnormal{\textbf{composition}}\label{composition} that is, the sum of the first two terms is to the second term as the sum of the last two terms is to the fourth term.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \proveq{$a+b:b$}{$= c+d:d$.} \eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{} \eq[\indent Then]{$\dfrac{a}{b}+1$}{$= \dfrac{c}{d}+1$;}{} \eq[that is,]{$\dfrac{a+b}{b}$}{$=\dfrac{c+d}{d}$.}{} \eq[\indent Or]{$a+b:b$}{$= c+d:d$.}{} \eq[\indent In like manner]{$a+b:a$}{$= c+d:c$.}{\qed} \end{proof} \proposition{Theorem.} \begin{proof}% \obs{If four quantities are in proportion, they are in proportion by \textnormal{\textbf{division}}\label{division}; that is, the difference of the first two terms is to the second term as the difference of the last two terms is to the fourth term.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \proveq{$a-b:b$}{$= c-d:d$.} \eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{} \eq[\indent Then]{$\dfrac{a}{b}-1$}{$= \dfrac{c}{d}-1$;}{} \eq[that is,]{$\dfrac{a-b}{b}$}{$= \dfrac{c-d}{d}$.}{} \eq[\indent Or]{$a-b:b$}{$= c-d:d$.}{} \eq[\indent In like manner]{$a-b:a$}{$= c-d:c$.}{\qed} \end{proof} \scanpage{148.png}% \proposition{Theorem.} \begin{proof}% \obs{If four quantities are in proportion, they are in proportion by \textnormal{\textbf{composition and division}}; that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$}{} \eq[\indent Then]{$\dfrac{a+b}{a}$}{$=\dfrac{c+d}{c}$.}{§~332} \eq[\indent And]{$\dfrac{a-b}{a}$}{$=\dfrac{c-d}{c}$.}{§~333} \eq[\indent Divide,]{$\dfrac{a+b}{a-b}$}{$=\dfrac{c+d}{c-d}$.}{} \eq[\indent Or]{$a+b:a-b$}{$ = c+d:c-d$.}{\qed} \end{proof} \proposition{Theorem.} \begin{proof}% \obs{In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.} \step[\indent\textbf{Let}]{$\mathbf{a:b = c:d = e:f = g:h}$.}{} \prove{$a+c+e+g : b+d+f+h = a:b$.} \step[\indent Let]{$r = \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{g}{h}$.}{} \step[\indent Then]{$a=br$, $c=dr$, $e=fr$, $g=hr$.}{} \step[\indent And]{$a+c+e+g = ( b+d+f+h )r$.}{} Divide by $( b+d+f+h )$. \step[\indent Then]{$\dfrac{a+c+e+g}{b+d+f+h} = r = \dfrac{a}{b}$.}{} \step[\indent Or]{$a+c+e+g : b+d+f+h = a:b$.}{\qed} \end{proof} \scanpage{149.png}% \proposition{Theorem.} \begin{proof}% \obs{The products of the corresponding terms of two or more proportions are in proportion.} \eq[\indent\textbf{Let}]{$\mathbf{a:b = c:d, \; e:f}$} {$\mathbf{= g:h, \; k:l = m:n}$.}{} \proveq{$aek:bfl$}{$= cgm:dhn$.} \eq[\indent Now]{$\dfrac{a}{b} = \dfrac{c}{d}$, $\dfrac{e}{f}$} {$= \dfrac{g}{h}$, $\dfrac{k}{l} = \dfrac{m}{n}$.}{} The products of the first members and of the second members of these equations give \eq{$\dfrac{aek}{bfl}$}{$= \dfrac{cgm}{dhn}$.}{} \eq[\indent Or]{$aek:bfl$}{$= cgm:dhn$.}{\qed} \end{proof} \pp{\cor{If three quantities are in continued proportion, the first is to the third as the square of the first is to the square of the second.}} \proposition{Theorem.} \begin{proof}% \obs{Like powers of the terms of a proportion are in proportion.} \eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{} \proveq{$a^n:b^n$}{$= c^n:d^n$.} \eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{} \eq[\indent Raise to the $n$th power,] {$\dfrac{a^n}{b^n}$}{$= \dfrac{c^n}{d^n}$.}{} \eq[\indent Or]{$a^n:b^n$}{$= c^n:d^n$.}{\qed} \end{proof} \pp{\defn{\textbf{Equimultiples}\label{equimultiples} of two quantities are the products obtained by multiplying each of them by the same number. Thus, $ma$ and $mb$ are equimultiples of $a$ and $b$.}} \scanpage{150.png}% \proposition{Theorem.} \begin{proof}% \obs{Equimultiples of two quantities are in the same ratio as the quantities themselves.} \textbf{Let $a$~and~$b$ be any two quantities.} \proveq{$ma : mb$}{$= a:b$.} \eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{a}{b}$.}{} Multiply both terms of the first fraction by $m$. \eq[\indent Then]{$\dfrac{ma}{mb}$}{$= \dfrac{a}{b}$.}{} \eq[\indent Or]{$ma:mb$}{$= a:b$.}{\qed} \end{proof} \begin{point}% \textsc{Scholium.} In the treatment of proportion, it is assumed that the \emph{quantities} involved are expressed by their \emph{numerical measures}. It is evident that the ratio of two quantities of the same kind may be represented by a fraction, if the two quantities are expressed in \emph{integers} in terms of a \emph{common unit}. If there is no unit in terms of which both quantities can be expressed in \emph{integers}, it is still possible by taking the unit of measure sufficiently small to find a fraction that will represent the ratio \emph{to any required degree of accuracy.}~\hfill§~269 If we speak of the product of two quantities, it must be understood that we mean simply \emph{the product of the numbers which represent them when they are expressed in terms of a common unit.} In order that four quantities, $a$, $b$, $c$, $d$, may form a proportion, $a$ and $b$ must be quantities of the same kind; and $c$ and $d$ must be quantities of the same kind; though $c$ and $d$ need not be of the same kind as $a$ and $b$. In alternation, however, \emph{the four quantities must be of the same kind.} \end{point} \scanpage{151.png}% \proposition{Theorem.} \begin{proof}% \obs{If a line is drawn through two sides of a triangle parallel to the third side, it divides those sides proportionally.} \figc{151aa342}{In the triangle $ABC$, let $EF$ be drawn parallel to $BC$.} \proveq{$EB : AE$}{$= FC: AF$.} \textsc{Case 1.} \textit{When $AE$ and $EB$ \textup{(Fig.~1)} are commensurable.} \textbf{Proof.} Find a common measure of $AE$ and $EB$, as $MB$. Let $MB$ be contained $m$ times in $EB$, and $n$ times in $AE$. \eq[\indent Then]{$EB:AE$}{$= m:n$.}{} At the points of division on $BE$ and $AE$ draw lines $\parallel$ to $BC$. These lines will divide $AC$ into $m + n$ equal parts, of which $FC$ will contain $m$, and $AF$ will contain $n$.~\hfill§~187 \eq{$\therefore FC:AF$}{$= m:n$.}{} \eq{$\therefore EB:AE$}{$=FC:AF$.}{Ax.~1} \textsc{Case 2.} \textit{When $AE$ and $EB$ \textup{(Fig.~2)} are incommensurable.} \textbf{Proof.} Divide $AE$ into any number of equal parts, and apply one of these parts to $EB$ as many times as $EB$ will contain it. Since $AE$ and $EB$ are incommensurable, a certain number of these parts will extend from $E$ to some point $K$, leaving a remainder $KB$ less than one of these parts. Draw $KH \parallel BC$. \eq[\indent Then]{$EK:AE$}{$= FH:AF$}{Case~1} \scanpage{152.png}% By increasing the \emph{number} of equal parts into which $AE$ is divided, we can make the \emph{length} of each part less than any assigned value, however small, but not zero. Hence, $KB$, which is less than one of these equal parts, has zero for a limit.~\hfill§~275 And the corresponding segment $HC$ has zero for a limit. Therefore, $EK$ approaches $EB$ as a limit,~\hfill§~271 and $FH$ approaches $FC$ as a limit. \step{$\therefore$ the variable $\dfrac{EK}{AE}$ approaches $\dfrac{EB}{AE}$ as a limit,}{§~280} \step{and the variable $\dfrac{FH}{AF}$ approaches $\dfrac{FC}{AF}$ as a limit.}{} \step[\indent But]{$\dfrac{EK}{AE}$ is constantly equal to $\dfrac{FH}{AF}$}{Case~1} \step{$\therefore \dfrac{EB}{AE} = \dfrac{FC}{AF}$.}{§~284} \hfill\qed \end{proof} \begin{point}% \cor[1]{One side of a triangle is to either part cut off by a straight line parallel to the base as the other side is to the corresponding part.} \eq[\indent For]{$AE:EB$}{$=AF:FC$.}{} \eq[\indent By composition,]{$AE+EB:AE$}{$= AF+FC:AF$.}{§~332} \eq[\indent Or]{$AB:AE$}{$= AC:AF$.}{} \end{point} \begin{point}% \cor[2]{If two lines are cut by any number of parallels the corresponding intercepts are proportional.} \figc{152aa344}{} Draw $AN \parallel$ to $CD$. Then \setlength{\eqalign}{.33\dentwidth} \eq{$AL=CG$, $LM$}{$=GK$, $MN=KD$.}{§~180} \eq[\indent Now]{$AH:AM$}{$=AF:AL=FH:LM$}{} \eq{}{$=HB:MN$.}{§~343} \eq[\indent Or]{$AF:CG$}{$=FH:GK=HB:KD$.}{} \setlength{\eqalign}{.5\dentwidth} \end{point} \scanpage{153.png}% \proposition{Theorem.} \begin{proof}% \obs{If a straight line divides two sides of a triangle proportionally, it is parallel to the third side.} \figc{153aa345}{In the triangle $ABC$, let $EF$ be drawn so that} \eq{$\dfrac{AB}{AE}$}{$= \dfrac{AC}{AF}$.}{} \proveq{$EF$ is}{$\parallel$ to $BC$.} \step[\indent\textbf{Proof.}]{From $E$ draw $EH \parallel$ to $BC$.}{} \eq[\indent Then]{$AB:AE$}{$= AC:AH$,}{§~343} \pnote{(one side of a triangle is to either part cut off by a line parallel to the base as the other side to the corresponding part).} \eq[\indent But]{$AB:AE$}{$= AC:AF$.}{Hyp.} \eq{$\therefore AC:AF$}{$= AC:AH$.}{Ax.~1} \eq{$\therefore AF$}{$= AH$.}{} \step{$\therefore EF$ and $EH$ coincide.}{§~47} \eq[\indent But]{$EH$ is}{$\parallel$ to $BC$.}{Const.} \step{$\therefore EF$, which coincides with $EH$, is $\parallel$ to $BC$.}{\qed} \end{proof} \ex{Find the fourth proportional to $91$,~$65$, and~$133$.} \ex{Find the mean proportional between $39$~and~$351$.} \ex{Find the third proportional to $54$~and~$3$.} \scanpage{154.png}% \begin{point}% If a given line $AB$ is divided at $M$, a point between the extremities $A$ and $B$, it is said to be divided \textbf{internally} into the segments $MA$ and $MB$; and if it is divided at $M'$, a point in the prolongation of $AB$, it is said to be divided \textbf{externally} into the segments $M'A$ and $M'B$. \figc{154aa346}{} In either case the segments are the \emph{distances} from the point of division to the extremities of the line. If the line is divided internally, the \emph{sum} of the segments is equal to the line; and if the line is divided externally, the \emph{difference} of the segments is equal to the line. Suppose it is required to divide the given line $AB$ \textbf{internally and externally in the same ratio}; as, for example, the ratio of the two numbers $3$~and~$5$. \figc{154bb346}{} We divide $AB$ into $5 + 3$, or~$8$, equal parts, and take $3$~parts from $A$; we then have the point $M$, such that \eq{$MA:MB$}{$= 3:5$.}{(1)} Secondly, we divide $AB$ into $5-3$, or~$2$, equal parts, and lay off on the prolongation of $AB$, to the left of $A$, three of these equal parts; we then have the point $M'$, such that \eq{$M'A:M'B$}{$= 3:5$.}{(2)} Comparing (1) and (2), \eq{$MA:MB$}{$= M'A:M'B$.}{} \end{point} \pp{\defn{If a given straight line is divided internally and externally into segments having the same ratio, the line is said to be \indexbf{divided harmonically}.}} \scanpage{155.png}% \proposition{Theorem.} \begin{proof}% \obs{The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.} \figc{155aa348}{Let $CM$ bisect the angle $C$ of the triangle $CAB$.} \proveq{$MA:MB$}{$= CA:CB$.} \textbf{Proof.} Draw $AE \parallel$ to $MC$, meeting $BC$ produced at $E$. \eq[\indent Then]{$MA:MB$}{$= CE:CB$,}{§~342} \pnote{(if a line is drawn through two sides of a $\triangle$ parallel to the third side, it divides those sides proportionally).} \eq[\indent Also,]{$\angle ACM$}{$= \angle CAE$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines);} \eq[and]{$\angle BCM$}{$= \angle CEA$,}{§~112} \pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).} \eq[\indent But]{$\angle ACM$}{$= \angle BCM$.}{Hyp.} \eq{$\therefore \angle CAE$}{$= \angle CEA$.}{Ax.~1} \eq{$\therefore CE$}{$=CA$.}{§~147} Put $CA$ for its equal, $CE$, in the first proportion. \eq[\indent Then]{$MA:MB$}{$=CA:CB$.}{\qed} \end{proof} \ex{In a triangle $ABC$, $AB=12$, $AC=14$, $BC=13$. Find the segments of $BC$ made by the bisector of the angle $A$.} \ex{In a triangle $CAB$, $CA=6$, $CB=12$, $AB=15$. Find the segments of $AB$ made by the bisector of the angle $C$.} \scanpage{156.png}% \proposition{Theorem.} \begin{proof}% \obs{The bisector of an exterior angle of a triangle divides the opposite side externally into segments which are proportional to the adjacent sides.} \figc{156aa349}{Let $CM'$ bisect the exterior angle $ACE$ of the triangle $CAB$, and meet $BA$ produced at $M'$.} \proveq{$M'A:M'B$}{$=CA:CB$.} \step[\indent\textbf{Proof.}]{Draw $AF \parallel$ to $M'C$, meeting $BC$ at $F$.}{} \eq[\indent Then]{$M'A:M'B$}{$CF:CB$.}{§~343} \eq[\indent Now]{$\angle M'CE$}{$=\angle AFC$,}{§~112} \eq[and]{$\angle M'CA$}{$=\angle CAF$,}{§~110} \pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).} \eq[\indent But]{$\angle M'CE$}{$=\angle M'CA$.}{Hyp.} \eq{$\therefore \angle AFC$}{$= \angle CAF$.}{Ax.~1} \eq{$\therefore CA$}{$=CF$.}{§~147} Put $CA$ for its equal, $CF$, in the first proportion. \eq[\indent Then]{$M'A:M'B$}{$=CA:CB$.}{\qed} \end{proof} \textbf{Question.} To what case does this theorem not apply? (See \hyperref[page:69]{Ex.~41, page~\pageref{page:69}}.) \pp{\cor{The bisectors of an interior angle and an exterior angle at one vertex of a triangle meeting the opposite side divide that side harmonically.}~\hfill§~347} \scanpage{157.png}% \section{SIMILAR POLYGONS.} \begin{point}% \defn{\indexbf{Similar polygons} are polygons that have their homologous angles equal, and their homologous sides proportional.} \figc{157aa351}{} Thus, the polygons $ABCDE$ and $A'B'C'D'E'$ are similar, if the $\angle_s A$, $B$, $C$, etc., are equal, respectively, to the $\angle_s A'$, $B'$, $C'$, etc., and if \step{$AB:A'B' = BC:B'C' = CD:C'D'$, etc.}{} \end{point} \pp{\defn{\indexbf{Homologous lines} in similar polygons are lines similarly situated.}} \pp{\defn{The ratio of any two homologous lines in similar polygons, is called the \indexbf{ratio of similitude} of the polygons.}} The primary idea of similarity is \textbf{likeness of form}. The two conditions \emph{necessary} to similarity are: \begin{myenum} \item \emph{For every angle in one of the figures there must be an equal angle in the other.} \item \emph{The homologous sides must be proportional.} \end{myenum} Thus, $Q$ and $Q'$ are not similar; the homologous sides are proportional, but the homologous angles are not equal. Also $R$ and $R'$ are not similar; the homologous angles are equal, but the sides are not proportional. \figc{157bb353}{} In the case of \emph{triangles}, either condition involves the other (see §~354 and §~358). \scanpage{158.png}% \proposition{Theorem.} \label{similar triangles} \begin{proof}% \obs{Two mutually equiangular triangles are similar.} \figc{158aa354}{In the triangles $ABC$ and $A'B'C'$, let the angles $A$, $B$, $C$ be equal to the angles $A'$, $B'$, $C'$, respectively.} \prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.} Since the $\triangle_s$ are mutually equiangular, we have only to prove that \step{$AB:A'B' = AC:A'C' = BC:B'C'$.}{§~351} \textbf{Proof.} Place the $\triangle A'B'C'$ on the $\triangle ABC$ so that $\angle A'$ shall coincide with its equal, the $\angle A$; and $B'C'$ take the position $EH$. \eq[\indent Then]{$\angle AEH$}{$= \angle B$}{Hyp.} \eq{$\therefore EH$ is}{$\parallel$ to $BC$.}{§~114} \eq{$\therefore AB:AE$}{$= AC:AH$.}{§~343} \eq[\indent That is,]{$AB:A'B'$}{$= AC:A'C'$.}{} Similarly, by placing $\triangle A'B'C'$ on $\triangle ABC$, so that $\angle B'$ shall coincide with its equal, the $\angle B$, we may prove that \eq{$AB:A'B'$}{$= BC:B'C'$}{\qed} \end{proof} \pp{\cor[1]{Two triangles are similar if two angles of the one are equal, respectively, to two angles of the other.}} \pp{\cor[2]{Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other.}} \scanpage{159.png}% \proposition{Theorem.} \begin{proof}% \obs{If two triangles have an angle of the one equal to an angle of the other, and the including sides proportional, they are similar}. \figc{159aa357}{In the triangles $ABC$ and $A'B'C'$, let $\angle A$ = $\angle A'$, and let} \eq{$\mathbf{AB : A'B'}$}{$\mathbf{= AC : A'C'}$.}{} \prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.} In this case we prove the $\triangle_s$ similar by proving them mutually equiangular. \textbf{Proof.} Place the $\triangle A'B'C'$ on the $\triangle ABC$, so that the $\angle A'$ shall coincide with its equal, the $\angle A$. Then the $\triangle A'B'C'$ will take the position of $\triangle AEH$. \eq[\indent Now]{$\dfrac{AB}{A'B'}$}{$=\dfrac{AC}{A'C'}$.}{Hyp.} \eq[\indent That is,]{$\dfrac{AB}{AE}$}{$=\dfrac{AC}{AH}$.}{} \step{$\therefore EH$ is $\parallel$ to $BC$,}{§~345} \pnote{(if a line divides two sides of a $\triangle$ proportionally, it is $\parallel$ to the third side).} \step{$\therefore\angle AEH = \angle B$, and $\angle AHE = \angle C$.}{§~112} \step{$\therefore\triangle AEH$ is similar to $\triangle ABC$.}{§~354} \step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed} \end{proof} \scanpage{160.png}% \proposition{Theorem.} \begin{proof}% \obs{If two triangles have their sides respectively proportional, they are similar}. \figc{160aa358}{In the triangles $ABC$ and $A'B'C'$, let} \step{$\mathbf{AB : A'B' = AC : A'C' = BC : B'C'}$.}{} \prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.} \textbf{Proof.} Upon $AB$ take $AE$ equal to $A'B'$, and upon $AC$ take $AH$ equal to $A'C'$; and draw $EH$. \eq[\indent Now]{$AB:A'B'$}{$= AC: A'C'$.}{Hyp.} \eq[\indent Or, since]{$AE$}{$ = A'B'$ and $AH = A'C'$,}{} \eq{$AB:AE$}{$= AC:AH$.}{} \step{$\therefore\triangle_s ABC$ and $AEH$ are similar.}{§~357} \eq{$\therefore AB:AE$}{$= BC:EH$;}{§~351} \eq[that is,]{$AB:A'B'$}{$= BC:EH$.}{} \eq[\indent But]{$AB:A'B'$}{$= BC: B'C'$.}{Hyp.} \eq{$\therefore BC:EH$}{$= BC:B'C'$.}{Ax.~1} \eq{$\therefore EH$}{$= B'C'$.}{} \step{Hence, the $\triangle_s AEH$ and $A'B'C'$ are equal.}{§~150} \step[\indent But]{$\triangle AEH$ is similar to $\triangle ABC$.}{} \step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed} \end{proof} \scanpage{161.png}% \proposition{Theorem.} \begin{proof}% \obs{Two triangles which have their sides respectively parallel, or respectively perpendicular, are similar.} \figc{161aa359}{Let $ABC$ and $A'B'C'$ have their sides respectively parallel; and $DEF$ and $D'E'F'$ have their sides respectively perpendicular.} \prove{the $\triangle_s ABC$ and $A'B'C'$ are similar; and that the $\triangle_s DEF$ and $D'E'F'$ are similar.} \textbf{Proof.} 1.~Prolong $BC$ and $AC$ to $B'A'$, forming $\angle_s x$ and $y$. \step{Then $\angle B' = \angle x$ (§~112), and $\angle B = \angle x$.}{§~110} \eq[\indent Therefore,]{$\angle B'$}{$= \angle B$}{Ax.~1} \eq[\indent In like manner,]{$\angle A'$}{$= \angle A$.}{} \step{Therefore, $\triangle A'B'C'$ is similar to $\triangle ABC$.}{§~355} 2.~Prolong $DE$ and $FD$ to meet $D'E'$ at $H$ and $D'F'$ at $K$. The quadrilateral $EHE'O$ has $\angle_s EHE'$ and $E'OE$ right angles, by hypothesis. \setlength{\eqalign}{.33\dentwidth} \eq[\indent Therefore,]{}{$\angle E'$ and $\angle OEH$ are supplementary.}{§~206} \eq[\indent But]{}{$\angle DEF$ and $\angle OEH$ are supplementary.}{§~86} \setlength{\eqalign}{.5\dentwidth} \eq{Therefore, $\angle DEF$}{$= \angle E'$.}{§~85} \eq[\indent In like manner,]{$\angle EDF$}{$= \angle D'$.}{} \step{Therefore, $\triangle DEF$ is similar to $\triangle D'E'F'$.}{§~355} \hfill\qed \end{proof} \pp{\cor{The parallel sides and the perpendicular sides are homologous sides of the triangles.}} \scanpage{162.png}% \proposition{Theorem.} \begin{proof}% \obs{The homologous altitudes of two similar triangles have the same ratio as any two homologous sides.} \figc{162aa361}{In the two similar triangles $ABC$ and $A'B'C'$, let $CO$ and $C'O'$ be homologous altitudes.} \prove{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.} \textbf{Proof.} In the rt.~$\triangle_s COA$ and $C'O'A'$, \step{$\angle A = \angle A'$,}{§~351} \pnote{(being homologous $\triangle_s$ of the similar $\triangle_s ABC$ and $A'B'C'$).} \step{$\therefore\triangle_s COA$ and $C'O'A'$ are similar,}{§~356} \pnote{(two rt.~$\triangle_s$ having an acute $\angle$ of the one equal to an acute $\angle$ of the other are similar).} \step{$\therefore\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}$.}{§~351} In the similar $\triangle_s ABC$ and $A'B'C'$, \step{$\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{§~351} \step[Therefore,]{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}= \dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{\qed} \end{proof} \ex{The base and altitude of a triangle are $7$~feet $6$~inches and $5$~feet $6$~inches, respectively. If the homologous base of a similar triangle is $5$~feet $6$~inches, find its homologous altitude.} \scanpage{163.png}% \proposition{Theorem.} \begin{proof}% \obs{If two parallels are cut by three or more transversals that pass through the same point, the corresponding segments are proportional}. \figc{163aa362}{Let the two parallels $AE$ and $A'E'$ be cut by the transversals $OA$, $OB$, $OC$, $OD$, $OE$ in $A$, $A'$, $B$, $B'$, etc.} \prove{$\dfrac{AB}{A'B'}= \dfrac{BC}{B'C'}= \dfrac{CD}{C'D'}= \dfrac{DE}{D'E'}$.} \textbf{Proof.} Since $A'E'$ is $\parallel$ to $AE$, the pairs of $\triangle_s OAB$ and $OA'B'$, $OBC$ and $OB'C'$, etc., are similar.~\hfill§~354 \eq{$\therefore\dfrac{AB}{A'B'} = \dfrac{OB}{OB'}$} {and $\dfrac{BC}{B'C'} = \dfrac{OB}{OB'}$.}{§~351} \pnote{(homologous sides of similar $\triangle_s$ are proportional).} \eq{$\therefore \dfrac{AB}{A'B'}$}{$= \dfrac{BC}{B'C'}$.}{Ax.~1} In a similar way it may be shown that \eq{$\dfrac{BC}{B'C'} = \dfrac{CD}{C'D'}$} {and $\dfrac{CD}{C'D'} = \dfrac{DE}{D'E'}$.}{\qed} \end{proof} \note{A condensed form of writing the above is \par \step{\( \dfrac{AB} {A'B'}=\left(\dfrac{OB} {OB'}\right)=\dfrac{BC} {B'C'}=\left(\dfrac{OC} {OC'}\right)=\dfrac{CD} {C'D'}=\left(\dfrac{OD} {OD'}\right)=\dfrac{DE} {D'E'} \). }{} \par A parenthesis about a ratio signifies that this ratio is used to prove the equality of the ratios immediately preceding and following it.} \scanpage{164.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} If three or more non-parallel straight lines intercept proportional segments upon two parallels, they pass through a common point.} \figc{164aa363}{Let $AB$, $CD$, $EF$ cut the parallels $AE$ and $BF$ so that} \eq{$\mathbf{AC : BD}$}{$\mathbf{= CE : DF}$.}{} \prove{$AB$, $CD$, $EF$ prolonged meet in a point.} \textbf{Proof.} Prolong $AB$ and $CD$ until they meet in $O$. \step{Draw $OE$.}{} \step{Designate by $F'$ the point where $OE$ cuts $BF$.}{} \eq[\indent Then]{$AC:BD$}{$=CE:DF'$.}{§~362} \eq[\indent But]{$AC:BD$}{$=CE:DF$.}{Hyp.} \eq{$\therefore CE:DF'$}{$= CE:DF$.}{Ax.~1} \eq{$\therefore DF'$}{$= DF$.}{} \step{$\therefore F'$ coincides with $F$.}{} \step{$\therefore EF$ coincides with $EF'$.}{§~47} \step{$\therefore EF$ prolonged passes through $O$.}{} \step{$\therefore AB$, $CD$, and $EF$ prolonged meet in the point $O$.}{\qed} \end{proof} \scanpage{165.png}% \proposition{Theorem.} \begin{proof}% \obs{The perimeters of two similar polygons have the same ratio as any two homologous sides.} \figc{165aa364}{Let the two similar polygons be $ABCDE$ and $A'B'C'D'E'$, and let $P$ and $P'$ represent their perimeters.} \proveq{$P:P'$}{$= AB: A'B'$.} \step[\indent\textbf{Proof.}]{$AB : A'B' = BC : B'C' = CD : C'D'$, etc.}{§~351} \step{$\therefore AB + BC + \text{etc.}\ : A'B' + B'C' + \text{etc.}\ = AB : A'B'$,}{§~335} \pnote{(in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent).} \eq[\indent That is,]{$P : P'$}{$= AB : A'B'$.}{\qed} \end{proof} \ex{If the line joining the middle points of the bases of a trapezoid is produced, and the two legs are also produced, the three lines will meet in the same point.} \ex{$AB$ and $AC$ are chords drawn from any point $A$ in the circumference of a circle, and $AD$ is a diameter. The tangent to the circle at $D$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Show that the triangles $ABC$ and $AEF$ are similar.} \ex{$AD$ and $BE$ are two altitudes of the triangle $CAB$. Show that the triangles $CED$ and $CAB$ are similar.} \ex{If two circles are tangent to each other, the chords formed by a straight line drawn through the point of contact have the same ratio as the diameters of the circles.} \scanpage{166.png}% \proposition{Theorem.} \begin{proof}% \obs{If two polygons are similar, they are composed of the same number of triangles, similar each to each, and similarly placed.} \figc{166aa365}{Let the polygons $ABCDE$ and $A'B'C'D'E'$ be similar.} From two homologous vertices, as $E$ and $E'$, draw diagonals $EB$, $EC$, and $E'B'$, $E'C'$. \prove{the $\triangle_s EAB$, $EBC$, $ECD$ are similar, respectively, to the $\triangle_s E'A'B'$, $E'B'C'$, $E'C'D'$.} \textbf{Proof.} The $\triangle_s EAB$ and $E'A'B'$ are similar.~\hfill§~357 \eq[\indent For]{$\angle A$}{$= \angle A'$,}{§~351} \eq[and]{$AE:A'E'$}{$=AB:A'B'$.}{§~351} \eq[\indent Now]{$\angle ABC$}{$= \angle A'B'C'$,}{§~351} \eq[and]{$\angle ABE$}{$=\angle A'B'E'$.}{§~351} %proofrule \eq[\indent By subtracting,]{$\angle EBC$}{$=\angle E'B'C'$.}{Ax.~3} \eq[\indent Now]{$EB:E'B'$}{$=AB:A'B'$}{§~351} \eq[and]{$BC:B'C'$}{$=AB:A'B'$}{§~351} \eq{$\therefore EB:E'B'$}{$=BC:B'C'$.}{Ax.~1} \step{$\therefore \triangle_s EBC$ and $E'B'C'$ are similar.}{§~357} In like manner $\triangle_s ECD$ and $E'C'D'$ are similar.~\hfill\qed \end{proof} \scanpage{167.png}% \proposition{Theorem.} \begin{proof}% \obs{\textsc{Conversely:} If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar.} \figc{167aa366}{In the two polygons $ABCDE$ and $A'B'C'D'E'$, let the triangles $AEB$, $BEC$, $CED$ be similar, respectively, to the triangles $A'E'B'$, $B'E'C'$, $C'E'D'$; and similarly placed.} \prove{$ABCDE$ is similar to $A'B'C'D'E'$.} \eq[\indent\textbf{Proof.}]{$\angle A$}{$= \angle A'$}{§~351} \eq[\indent Also,]{$\angle ABE$}{$= \angle A'B'E'$,}{} \eq[and]{$\angle EBC$}{$= \angle E'B'C'$.}{§~351} %proofrule \eq[\indent By adding,]{$\angle ABC$}{$= \angle A'B'C'$.}{Ax.~2} In like manner, $\angle BCD = \angle B'C'D'$, $\angle CDE=\angle C'D'E'$, etc. \step{Hence, the polygons are mutually equiangular.}{} Also, \( \dfrac{AB}{A'B'} = \left(\dfrac{EB}{E'B'}\right) = \dfrac{BC}{B'C'} = \left(\dfrac{EC}{E'C'}\right) = \dfrac{CD}{C'D'} \), etc.~\hfill§~351 Hence, the polygons have their homologous sides proportional. \step{Therefore, the polygons are similar.}{§~351} \hfill\qed \end{proof} \scanpage{168.png}% \pagebreak \section[EXERCISES.]{THEOREMS.} \ex{If two circles are tangent to each other externally, the corresponding segments of two lines drawn through the point of contact and terminated by the circumferences are proportional.} \ex{In a parallelogram $ABCD$, a line $DE$ is drawn, meeting the } diagonal $AC$ in $F$, the side $BC$ in $G$, and the side $AB$ produced in $E$. Prove that $\overline{DF}^2 = FG × FE$. \ex{Two altitudes of a triangle are inversely proportional to the corresponding bases.} \ex{Two circles touch at $P$. Through $P$ three lines are drawn, meeting one circle in $A$, $B$, $C$, and the other in $A'$, $B'$, $C'$, respectively. Prove that the triangles $ABC$, $A'B'C'$ are similar.} \begin{proofex}% Two chords $AB$, $CD$ intersect at $M$, and $A$ is the middle point of the arc $CD$. Prove that the product $AB × AM$ is constant if the chord $AB$ is made to turn about the fixed point $A$. Draw the diameter $AE$, and draw $BE$. \end{proofex} \begin{proofex}% If two circles touch each other, their common external tangent is the mean proportional between their diameters. Let $AB$ be the common tangent. Draw the diameters $AC$, $BD$. Join the point of contact $P$ to $A$, $B$, $C$, and $D$. Show that $APD$ and $BPC$ are straight lines $\perp$ to each other, and that $\triangle_s CAB$, $ABD$ are similar. \end{proofex} \begin{proofex}% If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Draw any two of the chords, and join the points where they meet the circumferences. The $\triangle_s$ thus formed are similar (Ex.~120). \end{proofex} \figc{168aa263}{} \begin{proofex}% In an inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides. Draw $DE$, making $\angle CDE = \angle ADB$. The $\triangle_s ABD$ and $ECD$ are similar; and the $\triangle_s BCD$ and $AED$ are similar. \end{proofex} \ex{Two isosceles triangles with equal vertical angles are similar.} \begin{proofex}% The bisector of the vertical angle $A$ of the triangle $ABC$ intersects the base at $D$ and the circumference of the circumscribed circle at $E$. Show that $ AB × AC = AD × AE$. \end{proofex} \scanpage{169.png}% \clearpage \section{NUMERICAL PROPERTIES OF FIGURES.} \proposition{Theorem.} \label{160} \begin{proof}% \obs{If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse:} \begin{myenum} \item \emph{The triangles thus formed are similar to the given triangle, and to each other.} \item \emph{The perpendicular is the mean proportional between the segments of the hypotenuse.} \item \emph{Each leg of the right triangle is the mean proportional between the hypotenuse and its adjacent segment.} \end{myenum} \figc{169aa367}{In the right triangle $ABC$, let $CF$ be drawn from the vertex of the right angle $C$, perpendicular to $AB$.} \prove[\textup{\textbf{1.~}}To prove that ]{$\triangle$'s $BCA$, $CFA$, $BFC$ are similar.} \textbf{Proof.} The rt.~$\triangle_s CFA$ and $BCA$ are similar,~\hfill§~356 \step{since the $\angle a'$ is common.}{} The rt.~$\triangle_s BFC$ and $BCA$ are similar,~\hfill§~356 \step{since the $\angle b$ is common.}{} Since the $\triangle_s CFA$ and $BFC$ are each similar to $\triangle BCA$, they are similar to each other.~\hfill§~354 \proveq[\indent\textup{\textbf{2.~}}To prove that]{$AF:CF$}{$=CF:FB$.} \textbf{Proof.} In the similar $\triangle_s CFA$ and $BFC$, \eq{$AF:CF$}{$=CF:FB$.}{§~351} \scanpage{170.png}% \filbreak \label{161} \proveq[\indent\textup{\textbf{3.~}}To prove that]{$AB:AC$}{$= AC:AF$,} \eq[\emph{and}]{$AB:BC$}{$= BC:BF$.}{} \textbf{Proof.} In the similar $\triangle_s BCA$ and $CFA$, \eq{$AB:AC$}{$= AC:AF$}{§~351} In the similar $\triangle_s BCA$ and $BFC$, \eq{$AB:BC$}{$=BC:BF$.}{§~351} \hfill\qed \end{proof} \begin{point}% \cor[1]{The squares of the two legs of a right triangle are proportional to the adjacent segments of the hypotenuse.} From the proportions in §~367,3, \step{$\overline{AC}^2=AB × AF$, and $\overline{BC}^2=AB × BF$.}{§~327} \step[\indent Hence,]{\( \dfrac{\overline{AC}^2}{\overline{BC}^2} = \dfrac{AB × AF}{AB × BF} = \dfrac{AF}{BF} \).}{} \end{point} \begin{point}% \cor[2]{The squares of the hypotenuse and either leg are proportional to the hypotenuse and the adjacent segment.} \step[\indent For]{\( \dfrac{\overline{AB}^2}{\overline{AC}^2} = \dfrac{AB × AB}{AB × AF} = \dfrac{AB}{AF} \).}{} \end{point} \figc{170aa370}{} \begin{point}% \cor[3]{The perpendicular from any point in the circumference to the diameter of a circle is the mean proportional between the segments of the diameter. The chord drawn from any point in the circumference to either extremity of the diameter is the mean proportional between the diameter and the adjacent segment.} \step[\indent For]{the $\angle ACB$ is a rt.~$\angle$.}{§~290} \end{point} \scanpage{171.png}% \proposition{Theorem.} \label{162} \begin{proof}% \obs{The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse.} \figc{171aa371}{Let $ABC$ be a right triangle with its right angle at $C$.} \proveq{$\overline{AC}^2 + \overline{CB}^2$}{$= \overline{AB}^2$.} \eq[\indent\textbf{Proof.}]{Draw $CF$}{$\perp$ to $AB$.}{} \eq[\indent Then]{$\overline{AC}^2$}{$= AB × AF$,}{} \eq[and]{$\overline{CB}^2$}{$= AB × BF$.}{§~367} %proofrule \eq[\indent By adding,]{$\overline{AC}^2 + \overline{CB}^2$} {$= AB(AF + BF) = \overline{AB}^2$}{\qed} \end{proof} \pp{\cor[1]{The square of either leg of a right triangle is equal to the difference of the square of the hypotenuse and the square of the other leg.}} \figcc{171bb373}{171cc374} \begin{point}% \cor[2]{The diagonal and a side of a square are incommensurable.} \step[\indent For]{$\overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 = 2 \overline{AB}^2$.}{} \step{$\therefore AC = AB \sqrt{2}$.}{} \end{point} \pp{\defn{The \textbf{projection} of any line upon a second line is the segment of the second line included between the perpendiculars drawn to it from the extremities of the first line. Thus, $PR$ is the projection of $CD$ upon $AB$.}} \scanpage{172.png}% \proposition{Theorem.} \label{163} \begin{proof}% \obs{In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides by the projection of the other upon that side.} \figc{172aa375}{Let $C$ be an acute angle of the triangle $ABC$, and $DC$ the projection of $AC$ upon $BC$.} \prove{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.} \textbf{Proof.} If $D$ falls upon the base (Fig.~1), \eq{$DB$}{$= BC - DC$.}{} If $D$ falls upon the base produced (Fig.~2), \eq{$DB$}{$= DC - BC$.}{} In either case, \step{$\overline{DB}^2 = \overline{BC}^2 + \overline{DC}^2 - 2 BC × DC$.}{} Add $\overline{AD}^2$ to both sides of this equality, and we have \step{\( \overline{AD}^2 + \overline{DB}^2 = \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 - 2 BC × DC \).}{} \eq[\indent But]{$\overline{AD}^2 + \overline{DB}^2$}{$= \overline{AB}^2$}{} \eq[and]{$\overline{AD}^2 + \overline{DC}^2$}{$= \overline{AC}^2$}{§~371} Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality. \step[\indent Then]{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.}{\qed} \end{proof} \scanpage{173.png}% \proposition{Theorem.} \label{164} \begin{proof}% \obs{In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides by the projection\label{projection} of the other upon that side.} \figc{173aa376}{Let $C$ be the obtuse angle of the triangle $ABC$, and $CD$ be the projection of $AC$ upon $BC$ produced.} \prove{\( \overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).} \step[\indent\textbf{Proof.}]{$DB = BC + DC$.}{} \step[\indent Squaring,]{\( \overline{DB}^2 = \overline{BC}^2 + \overline{DC}^2 + 2 BC × DC \).}{} Add $\overline{AD}^2$ to both sides, and we have \step{\( \overline{AD}^2 + \overline{DB}^2 = \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 + 2 BC × DC \).}{} \step[\indent But]{\( \overline{AD}^2 + \overline{DB}^2 = \overline{AB}^2 \text{, and } \overline{AD}^2 + \overline{DC}^2 = \overline{AC}^2 \).}{§~371} Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality. \step[\indent Then]{\( \overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).}{\qed} \end{proof} \note[1]{By the Principle of Continuity the last three theorems may be included in one theorem. Let the student explain.} \note[2]{The last three theorems enable us to compute the lengths of the altitudes of a triangle if the lengths of the three sides are known.} \scanpage{174.png}% \proposition{Theorem.} \begin{proof}% 1.~\obs{The sum of the squares of two sides of a triangle is equal to twice the square of half the third side increased by twice the square of the median upon that side.} 2.~\obs{The difference of the squares of two sides of a triangle is equal to twice the product of the third side by the projection of the median upon that side.} \figc{174aa377}{In the triangle $ABC$, let $AM$ be the median and $MD$ the projection of $AM$ upon the side $BC$. Also, let $AB$ be greater than $AC$.} \proveq{\textup{1.} $\overline{AB}^2+\overline{AC}^2$}{$= 2\overline{BM}^2+2\overline{AM}^2$.} \proveq[]{\textup{2.} $\overline{AB}^2-\overline{AC}^2$}{$= 2BC × MD$.} \textbf{Proof.} Since $AB > AC$, the $\angle AMB$ will be obtuse, and the $\angle AMC$ will be acute.~\hfill§~155 \setlength{\eqalign}{.33\dentwidth} \eq[\indent Then]{$\overline{AB}^2$}{$=\overline{BM}^2+\overline{AM}^2+2BM × MD$,}{§~376} \eq[and]{$\overline{AC}^2$}{$=\overline{MC}^2+\overline{AM}^2-2MC × MD$.}{§~375} \setlength{\eqalign}{.5\dentwidth} Add these two equalities, and observe that $BM=MC$. \eq[\indent Then]{$\overline{AB}^2+\overline{AC}^2$}{$=2 \overline{BM}^2+2 \overline{AM}^2$.}{} Subtract the second equality from the first. \eq[\indent Then]{$\overline{AB}^2-\overline{AC}^2$}{$=2 BC × MD$.}{\qed} \end{proof} \note{This theorem enables us to compute the lengths of the medians of a triangle if the lengths of the three sides are known.} \scanpage{175.png}% \proposition{Theorem.} \begin{proof}% \obs{If two chords intersect in a circle, the product of the segments of one is equal to the product of the segments of the other.} \figc{175aa378}{Let any two chords $MN$ and $PQ$ intersect at $O$.} \proveq{$OM × ON$}{$= OQ × OP$.} \step[\indent\textbf{Proof.}]{Draw $MP$ and $NQ$.}{} \eq{$\angle a$}{$= \angle a'$,}{§~289} \pnote{(each being measured by $\frac{1}{2} \arc PN$).} \eq[\indent And]{}{$\angle c = \angle c'$,}{§~289} \pnote{(each being measured by $\frac{1}{2} \arc MQ$).} \step{$\therefore$ the $\triangle_s NOQ$ and $POM$ are similar.}{§~355} \eq{$\therefore OQ:OM$}{$=ON:OP$.}{§~351} \eq{$\therefore OM × ON$}{$= OQ × OP$.}{§~327} \hfill\qed \end{proof} \begin{point}\textsc{Scholium.} This proportion may be written \step{$\dfrac{OM}{OQ} = \dfrac{OP}{ON}$, or $\dfrac{OM}{OQ} = \dfrac{1}{\dfrac{ON}{OP}}$;}{} that is, the ratio of two corresponding segments is equal to the \emph{reciprocal} of the ratio of the other two segments. Hence, these segments are said to be \emph{reciprocally proportional}. \end{point} \scanpage{176.png}% \pp{\defn{\textbf{A secant from a point to a circle}\label{secant2} is understood to mean the segment of the secant lying between the point and the \emph{second point} of intersection of the secant and circumference.}} \proposition{Theorem.} \begin{proof}% \obs{If from a point without a circle a secant and a tangent are drawn, the tangent is the mean proportional between the whole secant and its external segment.} \figc{176aa381}{Let $AD$ be a tangent and $AC$ a secant drawn from the point $A$ to the circle $BCD$.} \prove{$AC : AD = AD : AB$.} \step[\indent\textbf{Proof.}]{Draw $DC$ and $DB$.}{} \step{The $\triangle_s ADC$ and $ABD$ are similar.}{§~355} \step{For $\angle b$ is common; and $\angle a' = \angle a$,}{§§~289,~295} \pnote{(each being measured by $\frac{1}{2} \arc BD$).} \step{$\therefore AC : AD = AD : AB$.}{§~351} \hfill\qed \end{proof} \begin{point}% \cor{If from a fixed point without a circle a secant is drawn, the product of the secant and its external segment is constant in whatever direction the secant is drawn.} \step[\indent For]{$AC × AB = \overline{AD}^2$.}{§~327} \end{point} \scanpage{177.png}% \proposition{Theorem.} \begin{proof}% \obs{The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments made by the bisector upon the third side of the triangle.} \figc{177aa383}{Let $NO$ bisect the angle $MNP$ of the triangle $MNP$.} \prove{$\overline{NO}^2 = NM × NP - OM × OP$.} \step[\indent\textbf{Proof.}]{Circumscribe the $\odot MNP$ about the $\triangle MNP$.}{§~314} Produce $NO$ to meet the circumference in $Q$, and draw $PQ$. \step{The $\triangle_s NQP$ and $NMO$ are similar.}{§~355} \eq[\indent For]{$\angle b$}{$= \angle b'$}{Hyp.} \eq[and]{$\angle a$}{$= \angle a'$}{§~289} \eq[\indent Whence]{$NQ:NM$}{$= NP:NO$.}{§~351} \eq{$\therefore NM × NP$}{$= NQ × NO$}{} \eq{}{$= (NO+OQ)NO$}{} \eq{}{$= \overline{NO}^2 + NO × OQ$.}{} \eq[\indent But]{$NO × OQ$}{$= MO × OP$.}{§~378} \eq{$\therefore MN × NP$}{$= \overline{NO}^2 + MO × OP$.}{} \step[\indent Whence]{$\overline{NO}^2 = NM × NP = MO × OP$.}{Ax.~3} \hfill\qed \end{proof} \note{This theorem enables us to compute the lengths of the bisectors of the angles of a triangle if the lengths of the sides are known.} \scanpage{178.png}% \proposition{Theorem.} \begin{proof}% \obs{In any triangle the product of two sides is equal to the product of the diameter of the circumscribed circle by the altitude upon the third side.} \figc{178aa384}{Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed about the triangle NMQ.} \step{Draw the diameter $NP$, and draw $PQ$.}{} \prove{$NM × NQ = NP × NO$.} \textbf{Proof.} In the $\triangle_s NOM$ and $NQP$, \step{$\angle NOM$ is a rt.~$\angle$,}{Hyp.} \step{$\angle NQP$ is a rt.~$\angle$,}{§~290} \eq{and $\angle a$}{$= \angle a'$,}{§~289} \pnote{(each being measured by $\frac{1}{2} \arc NQ$).} \step{$\therefore \triangle_s NOM$ and $NQP$ are similar.}{§~356} \eq[\indent Whence]{$NM:NP$}{$= NO:NQ$.}{§~351} \eq{$\therefore NM × NQ$}{$= NP × NO$.}{§~327} \hfill\qed \end{proof} \note{This theorem enables us to compute the length of the radius of a circle circumscribed about a triangle, if the lengths of the three sides of the triangle are known. } \ex{If $OE$, $OF$, $OG$ are the perpendiculars from any point $O$ within the triangle $ABC$ upon the sides $AB$, $BC$, $CA$, respectively, show that \( \overline{AE}^2 + \overline{BF}^2 + \overline{CG}^2 = \overline{EB}^2 + \overline{FC}^2 + \overline{GA}^2 \).} \scanpage{179.png}% \section[EXERCISES.]{THEOREMS.} \begin{proofex}% The sum of the squares of the segments of two perpendicular chords is equal to the square of the diameter of the circle. If $AB$, $CD$ are the chords, draw the diameter $BE$, draw $AC$, $ED$, $BD$. Prove that $AC = ED$, and apply §~371. \end{proofex} \ex{The tangents to two intersecting circles drawn from any point in their common chord produced, are equal. (§~381.)} \ex{The common chord of two intersecting circles, if produced, will bisect their common tangents. (§~381.)} \figc{179aa270}{} \begin{proofex}% If three circles intersect one another, the common chords all pass through the same point. Let two of the chords $AB$ and $CD$ meet at $O$. Join the point of intersection $E$ to $O$, and suppose that $EO$ produced meets the same two circles at two different points $P$ and $Q$. Then prove that $OP = OQ$ (§~378); hence, that the points $P$ and $Q$ coincide. \end{proofex} \ex{If two circles are tangent to each other, the common internal tangent bisects the two common external tangents.} \begin{proofex}% If the perpendiculars from the vertices of the triangle $ABC$ upon the opposite sides intersect at $D$, show that \step{\( \overline{AB}^2-\overline{AC}^2 = \overline{BD}^2-\overline{CD}^2 \).}{} \end{proofex} \ex{In an isosceles triangle, the square of a leg is equal to the square of any line drawn from the vertex to the base, increased by the product of the segments of the base.} \ex{The squares of two chords drawn from the same point in a circumference have the same ratio as the projections of the chords on the diameter drawn from the same point.} \ex{The difference of the squares of two sides of a triangle is equal to the difference of the squares of the segments of the third side, made by the perpendicular on the third side from the opposite vertex.} \begin{proofex}% $E$ is the middle point of $BC$, one of the parallel sides of the trapezoid $ABCD$; $AE$ and $DE$ produced meet $DC$ and $AB$ produced at $F$ and $G$, respectively. Show that $FG$ is parallel to $DA$. $\triangle_s AGD$ and $BGE$ are similar; and $\triangle_s AFD$ and $EFC$ are similar. \end{proofex} \scanpage{180.png}% \ex{If two tangents are drawn to a circle at the extremities of a diameter, the portion of a third tangent intercepted between them is divided at its point of contact into segments whose product is equal to the square of the radius.} \ex{If two exterior angles of a triangle are bisected, the line drawn from the point of intersection of the bisectors to the opposite angle of the triangle bisects that angle.} \ex{The sum of the squares of the diagonals of a quadrilateral is equal to twice the sum of the squares of the lines that join the middle points of the opposite sides.} \figcc{180aa280}{180bb281} \begin{proofex}% The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of the diagonals, increased by four times the square of the line joining the middle points of the diagonals. Apply §~377 to the $\triangle_s$ $ABC$ and $ADC$, add the results, and eliminate $\overline{BE}^2 + \overline{DE}^2$ by applying §~377 to the $\triangle BDE$. \end{proofex} \begin{proofex}% The square of the bisector of an exterior angle of a triangle is equal to the product of the external segments determined by the bisector upon one of the sides, diminished by the product of the other two sides. Let $CD$ bisect the exterior $\angle BCH$ of the $\triangle ABC$. $\triangle_s$ $ACD$ and $FCB$ are similar (§~355). Apply §~382. \end{proofex} \ex{If a point $O$ is joined to the vertices of a triangle $ABC$; through any point $A'$ in $OA$ a line parallel to $AB$ is drawn, meeting $OB$ at $B'$; through $B'$ a line parallel to $BC$, meeting $OC$ at $C'$; and $C'$ is joined to $A'$; the triangle $A'B'C'$ is similar to the triangle $ABC$.} \ex{If the line of centres of two circles meets the circumferences at the consecutive points $A$, $B$, $C$, $D$, and meets the common external tangent at $P$, then $PA × PD = PB × PC$.} \begin{proofex}% The line of centres of two circles meets the common external tangent at~$P$, and a secant is drawn from~$P$, cutting the circles at the consecutive points $E$, $F$, $G$,~$H$. Prove that $PE × PH = PF × PG$. Draw radii to the points of contact, and to $E$, $F$, $G$, $H$. Let fall $\perp_s$ on $PH$ from the centres of the $\odot_s$. The various pairs of $\triangle_s$ are similar. \end{proofex} \ex{If a line drawn from a vertex of a triangle divides the opposite side into segments proportional to the adjacent sides, the line bisects the angle at the vertex.} \scanpage{181.png}% \clearpage \section{PROBLEMS OF CONSTRUCTION.} \proposition{Problem.} \begin{proof}% \obs{To divide a given straight line into parts proportional to any number of given lines.} \figc{181aa385}{Let $AB$, $m$, $n$, and $p$ be given straight lines.} \prove[To divide ]{$AB$ into parts proportional to $m$, $n$, and $p$.} \step{Draw $AX$, making any convenient $\angle$ with $AB$.}{} \step{On $AX$ take $AC$ equal to $m$, $CE$ to $n$, $EF$ to $p$.}{} \step{Draw $BF$.}{} \step{From $E$ and $C$ draw $EK$ and $CH \parallel$ to $FB$.}{} \step{Through $A$ draw a line $\parallel$ to $BF$.}{} \step{$K$ and $H$ are the division points required.}{} \step[\indent\textbf{Proof.}]{$\dfrac{AH}{AC} = \dfrac{HK}{CE} = \dfrac{KB}{EF}$,}{§~344} \pnote{(if two lines are cut by any number of parallels, the corresponding intercepts are proportional).} Substitute $m$, $n$, and $p$ for their equals $AC$, $CE$, and $EF$. \step[\indent Then]{$\dfrac{AH}{m} = \dfrac{HK}{n} = \dfrac{KB}{p}$.}{\qef} \end{proof} \ex{Divide a line $12$~inches long into three parts proportional to the numbers $3$,~$5$,~$7$.} \scanpage{182.png}% \proposition{Problem.} \begin{proof}% \obs{To find the fourth proportional to three given straight lines.} \figc{182aa386}{Let the three given lines be $m$, $n$, and $p$.} \prove[To find ]{the fourth proportional to $m$, $n$, and $p$.} \step {Draw $Ax$ and $Ay$ containing any convenient angle.} {} \step {On $Ax$ take $AB$ equal to $m$, $BC$ to $n$.} {} \step {On $Ay$ take $AD$ equal to $p$.} {} \step {Draw $BD$.} {} \step {From $C$ draw $CF \parallel$ to $BD$, meeting $Ay$ at $F$.} {} \step {$DF$ is the fourth proportional required.} {} \eq [\indent\textbf{Proof.}] {$AB:BC$ } {$ = AD:DF$,} {§~342} \pnote{(a line drawn through two sides of a $\triangle \parallel$ to the third side divides those sides proportionally).} \step{Substitute $m$, $n$, and $p$ for their equals $AB$, $BC$, and $AD$.} {} \eq [\indent Then] {$m:n$} {$ = p:DF$} {\qef} \end{proof} \ex {The square of the altitude of an equilateral triangle is equal to three fourths of the square of one side of the triangle.} \scanpage{183.png}% \proposition{Problem.} \begin{proof}% \obs{To find the third proportional to two given straight lines.} \figc{183aa387}{Let $m$ and $n$ be the two given straight lines.} \prove[To find ]{the third proportional to $m$ and $n$.} \step{Construct any convenient angle $A$,}{} \step{and take $AB$ equal to $m$, $AC$ equal to $n$.}{} \step{Produce $AB$ to $D$, making $BD$ equal to $AC$.}{} \step{Draw $BC$.}{} \step{Through $D$ draw $DE \parallel$ to $BC$, meeting $AC$ produced at $E$.}{} \step{$CE$ is the third proportional required.}{} \eq[\indent\textbf{Proof.}]{$AB:BD$}{$= AC:CE$,}{§~342} \pnote{(a line drawn through two sides of a $\triangle$ parallel to the third side divides those sides proportionally).} Substitute, in the above proportion, $AC$ for its equal $BD$. \eq[\indent Then]{$AB:AC$}{$= AC:CE$,}{} \eq[that is,]{$m:n$}{$=n:CE$.}{\qef} \end{proof} \begin{proofex}% Construct $x$, if (1) $x=\dfrac{ab}{c}$, (2) $x = \dfrac{a^2}{c}$. Special cases: (1)~$a = 2$, $b = 8$, $c = 4$; (2)~$a = 3$, $b = 7$, $c = 11$; (3)~$a = 2$, $c = 3$; (4)~$a = 3$, $c = 5$; (5)~$a = 2c$. \end{proofex} \scanpage{184.png}% \proposition{Problem.} \begin{proof}% \obs{To find the mean proportional between two given straight lines.} \figc{184aa388}{Let the two given lines be $m$ and $n$.} \prove[To find ]{the mean proportional between $m$ and $n$.} \step{On the straight line $AE$}{} \step{take $AC$ equal to $m$, and $CB$ equal to $n$.}{} \step{On $AB$ as a diameter describe a semicircumference.}{} \step{At $C$ erect the $\perp$ $CH$ meeting the circumference at $H$.}{} \step{$CH$ is the mean proportional between $m$ and $n$.}{} \eq[\indent\textbf{Proof.}]{$AC:CH$}{$=CH:CB$}{§~370} \pnote{(the $\perp$ let fall from a point in a circumference to the diameter of a circle is the mean proportional between the segments of the diameter).} Substitute for $AC$ and $CB$ their equals $m$ and $n$. \eq[\indent Then]{$m:CH$}{$=CH:n$.}{\qef} \end{proof} \pp{\defn{A straight line is divided \textbf{in extreme and mean ratio}\label{extrememean}, when one of the segments is the mean proportional between the whole line and the other segment.}} \begin{proofex}% Construct $x$, if $ x=\sqrt{ab} $. Special cases: (1)~$a = 2$, $b = 3$; (2)~$a = 1$, $b = 6$; (3)~$a = 3$, $b = 7$. \end{proofex} \scanpage{185.png}% \proposition{Problem.} \begin{proof}% \obs{To divide a given line in extreme and mean ratio.} \figc{185aa390}{Let $AB$ be the given line.} \prove[To divide ]{$AB$ in extreme and mean ratio.} \step{At $B$ erect a $\perp BE$ equal to half of $AB$.}{} \step{From $E$ as a centre, with a radius equal to $EB$, describe a $\odot$.}{} \step{Draw $AE$, meeting the circumference in $F$ and $G$.}{} \step{On $AB$ take $AC$ equal to $AF$.}{} \step{On $BA$ produced take $AC'$ equal to $AG$.}{} Then $AB$ is divided internally at $C$ and externally at $C'$ in extreme and mean ratio. \step{$AG:AB = AB:AF$.}{§~381} \begin{center} \begin{tabular}{r@{}l@{}l | r@{}l@{}l} $\overline{AB}^2$& $= AF × AG$ && $\overline{AB}^2$& $= AG × AF$ \\ & $= AC(AF+AG)$ && & $= C'A(AG-AF)$ \\ & $= AC(AC+AB)$ && & $= C'A(C'A-AB)$ \\ & $= \overline{AC}^2 + AB × AC$. && & $= \overline{C'A}^2 - AB × C'A$. \\ & $\therefore \overline{AB}^2-AB × AC$ & $=\overline{AC}^2$. & & $\therefore \overline{AB}^2+AB × C'A$ & $=\overline{C'A}^2$. \\ & $\therefore AB(AB-AC)$ & $=\overline{AC}^2$. & & $\therefore AB(AB+C'A)$ & $=\overline{C'A}^2$. \\ & $\therefore AB × CB$ & $=\overline{AC}^2$. & & $\therefore AB × C'B$ & $=\overline{C'A}^2$. \end{tabular} \end{center} \hfill\qef \end{proof} \scanpage{186.png}% \proposition{Problem.} \begin{proof}% \obs{Upon a given line homologous to a given side of a given polygon, to construct a polygon similar to the given polygon.} \figc{186aa391}{Let $A'E'$ be the given line homologous to $AE$ of the given polygon $ABCDE$.} \prove[To construct ]{on $A'E'$ a polygon similar to the given polygon.} \step{From $E$ draw the diagonals $EB$ and $EC$.}{} \step{From $E'$ draw $E'B'$, $E'C'$, and $E'D'$,}{} \step{making $ \triangle$'s $A'E'B'$, $B'E'C'$, and $C'E'D'$ equal, respectively, to}{} \step{$ \triangle_s AEB$, $BEC$, and $CED$.}{} \step{From $A'$ draw $A'B'$, making $\angle E'A'B'$ equal to $\angle EAB$,}{} \step{and meeting $E'B'$ at $B'$.}{} \step{From $B'$ draw $B'C'$, making $\angle E'B'C'$ equal to $\angle EBC$,}{} \step{and meeting $E'C'$ at $C'$.}{} \step{From $C'$ draw $C'D'$, making $\angle E'C'D'$ equal to $\angle ECD$,}{} \step{and meeting $E'D'$ at $D'$.}{} \step{Then $A'B'C'D'E'$ is the required polygon.}{} \step[\indent\textbf{Proof.}]{The $\triangle_s ABE$, $A'B'E'$, etc., are similar.}{§~354} \step{Therefore, the two polygons are similar.}{§~366} \hfill\qef \end{proof} \scanpage{187.png}% \section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.} \ex{To divide one side of a given triangle into segments proportional to the adjacent sides (§~348).} \figcccc{187aa291}{187bb292}{187cc293}{187dd294} \begin{proofex}% To find in one side of a given triangle a point whose distances from the other sides shall be to each other in the given ratio $m : n$. Take $AG = m \perp$ to $AC$, $GH=n \perp$ to $BC$. Draw $CD \parallel$ to $OG$. \end{proofex} \ex{Given an obtuse triangle; to draw a line from the vertex of the obtuse angle to the opposite side which shall be the mean proportional between the segments of that side.} \begin{proofex}% Through a given point $P$ within a given circle to draw a chord $AB$ so that the ratio $AP: BP$ shall equal the given ratio $m : n$. Draw $OPC$ so that $OP:PC = n:m$. Draw $CA$ equal to the fourth proportional to $n$, $m$, and the radius of the circle. \end{proofex} \begin{proofex}% To draw through a given point $P$ in the arc subtended by a chord $AB$ a chord which shall be bisected by $AB$. On radius $OP$ take $CD$ equal to $CP$. Draw $DE \parallel$ to $BA$. \end{proofex} \figcccc{187ee295}{187ff296}{187gg297}{187hh298} \begin{proofex}% To draw through a given external point $P$ a secant $PAB$ to a given circle so that the ratio $PA:AB$ shall equal the given ratio $m : n$. \[ PD:DC = m:n. \quad PD:PA = PA:PC. \] \end{proofex} \begin{proofex}% To draw through a given external point $P$ a secant $PAB$ to a given circle so that $\overline{AB}^2 = PA × PB$. \[ PC:CD = CD:PD. \quad PA = CD. \] \end{proofex} \scanpage{188.png}% \ex{To find a point $P$ in the arc subtended by a given chord $AB$ so that the ratio $PA:PB$ shall equal the given ratio $m : n$.} \ex{To draw through one of the points of intersection of two circles a secant so that the two chords that are formed shall be in the given ratio $m:n$.} \ex{Having given the greater segment of a line divided in extreme and mean ratio, to construct the line.} \ex{To construct a circle which shall pass through two given points and touch a given straight line.} \ex{To construct a circle which shall pass through a given point and touch two given straight lines.} \ex{To inscribe a square in a semicircle.} \figc{188aa303}{} \begin{proofex}% To inscribe a square in a given triangle. Let $DEFG$ be the required inscribed square. Draw $CM \parallel$ to $AB$, meeting $AF$ produced in $M$. Draw $CH$ and $MN \perp$ to $AB$, and produce $AB$ to meet $MN$ at $N$. The $\triangle_s ACM$, $AGF$ are similar; also, the $\triangle_s AMN$, $AFE$ are similar. By these triangles show that the figure $CMNH$ is a square. By constructing this square, the point $F$ can be found. \end{proofex} \ex{To inscribe in a given triangle a rectangle similar to a given rectangle.} \ex{To inscribe in a circle a triangle similar to a given triangle.} \ex{To inscribe in a given semicircle a rectangle similar to a given rectangle.} \ex{To circumscribe about a circle a triangle similar to a given triangle.} \ex{To construct the expression, $x = \dfrac{2abc}{de}$; that is, $\dfrac{2ab}{d} × \dfrac{c}{e}$.} \ex{To construct two straight lines, having given their sum and their ratio.} \ex{To construct two straight lines, having given their difference and their ratio.} \ex{Given two circles, with centres $O$ and $O'$, and a point $A$ in their plane, to draw through the point $A$ a straight line, meeting the circumferences at $B$ and $C$, so that $AB:AC=m:n$.} \scanpage{189.png}% \subsection{PROBLEMS OF COMPUTATION.} \begin{proofex}% To compute the altitudes of a triangle in terms of its sides. \figc{189aa312}{} At least one of the angles $A$ or $B$ is acute. Suppose $B$ is acute. \eq[\indent In the $\triangle$ $CDB$,]{$h^2$}{$=a^2- \overline{BD}^2$,}{§~372} \eq[\indent In the $\triangle$ $ABC$,]{$b^2$}{$=a^2+c^2 - 2c ×\ BD$.}{§~376} \eq[\indent Whence]{$BD$}{$=\dfrac{a^2+c^2-b^2}{2c}$.}{} \setlength{\eqalign}{0.25\dentwidth} \eq[\indent Hence,]{$h^2$}{\(=a^2-\dfrac{(a^2+c^2-b^2)^2} {4c^2}= \dfrac{4a^2c^2-(a^2+c^2-b^2)^2} {4c^2} \)}{} \eq{}{\( =\dfrac{(2ac+a^2+c^2-b^2)(2ac-a^2-c^2+b^2)} {4c^2} \)}{} \eq{}{\( =\dfrac{\{(a+c)^2-b^2\}\{b^2-(a-c)^2\}} {4c^2} \)}{} \eq{}{\( =\dfrac{(a+b+c)(a+c-b)(b+a-c)(b-a+c)} {4c^2}. \)}{} \setlength{\eqalign}{0.5\dentwidth} \eq[\indent Let]{$a+b+c$}{$=2s$.}{} \eq[\indent Then]{$a+c-b$}{$=2(s-b)$,}{} \eq{$b+a-c$}{$= 2(s-c)$,}{} \eq{$b-a+c$}{$=2(s-a)$.}{} \setlength{\eqalign}{0.25\dentwidth} \eq[\indent Hence,]{$h^2$}{\( =\dfrac{2s × 2(s-a) × 2(s-b) × 2(2-c)} {4c^2} \).}{} By simplifying, and extracting the square root, \label{formtrialtitude}% \eq{$h$}{\( =\dfrac{2}{c} \sqrt{s(s-a)(s-b)(s-c)} \).}{} \setlength{\eqalign}{0.5\dentwidth} \end{proofex} \figc{189bb313}{} \begin{proofex}% To compute the medians of a triangle in terms of its sides. \setlength{\eqalign}{0.33\dentwidth} \eq[\indent By §~377,]{}{\( a^2+b^2 = 2m^2+2\left(\dfrac{c}{2}\right)^2 \).}{} \eq[\indent Whence]{$4m^2$}{$=2(a^2+b^2)-c^2$.}{} \label{formtrimedian}% \eq{$\therefore m$}{$=\dfrac{1}{2} \sqrt{2(a^2+b^2)-c^2}$.}{} \end{proofex} \setlength{\eqalign}{0.5\dentwidth} \scanpage{190.png}% \figc{190aa314}{} \begin{proofex}% To compute the bisectors of a triangle in terms of the sides. \setlength{\eqalign}{.33\dentwidth} \eq[\indent By §~383,]{$t^2$}{$=ab-AD × BD$.}{} \eq[\indent By §348,]{$\dfrac{AD}{B}$}{$=\dfrac{BD}{a}=\dfrac{AD+BD}{a+b}=\dfrac{c}{a+b}$.}{} \eq{$\therefore AD$}{$=\dfrac{bc}{a+b}$, and $BD=\dfrac{ac}{a+b}$.}{} \eq[\indent Whence]{$t^2$}{$= ab - \dfrac{abc^2}{(a+b)^2}$}{} \eq{}{$=ab\left[1-\dfrac{c^2}{(a+b)^2}\right]$}{} \eq{}{$=\dfrac{ab\{(a+b)^2-c^2\}}{(a+b)^2}$}{} \eq{}{$=\dfrac{ab(a+b+c)(a+b-c)}{(a+b)^2}$}{} \eq{}{$=\dfrac{ab × 2s × 2(s-c)}{(a+b^2)}$.}{} \label{formtribisector}% \eq[\indent Whence]{$t$}{$= \dfrac{2}{a+b} \sqrt{abs(s-c)}$.}{} \setlength{\eqalign}{.5\dentwidth} \end{proofex} \figc{190bb315}{} \begin{proofex}% To compute the radius of the circle circumscribed about a triangle in terms of the sides of the triangle. \eq[\indent By §384,]{$AC × AB$}{$= AE × AD$,}{} \eq[or,]{$bc$}{$= 2 B × AD$.}{} \setlength{\eqalign}{0.33\dentwidth} \eq[\indent But]{$AD$}{$=\dfrac{2}{a}\sqrt{s(s-a)(s-b)(s-c)}$.}{Ex.~312} \label{formradcircum}% \eq{$\therefore R$}{$=\dfrac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$.}{} \setlength{\eqalign}{0.5\dentwidth} \end{proofex} \ex{If the sides of a triangle are $3$,~$4$, and~$5$, is the angle opposite~$5$ right, acute, or obtuse?} \ex{If the sides of a triangle are $7$,~$9$, and~$12$, is the angle opposite~$12$ right, acute, or obtuse?} \ex{If the sides of a triangle are $7$,~$9$, and~$11$, is the angle opposite~$11$ right, acute, or obtuse?} \ex{The legs of a right triangle are $8$~inches and $12$~inches; find the lengths of the projections of these legs upon the hypotenuse, and the distance of the vertex of the right angle from the hypotenuse.} \ex{If the sides of a triangle are $6$~inches, $9$~inches, and $12$~inches, find the lengths (1)~of the altitudes; (2)~of the medians; (3)~of the bisectors; (4)~of the radius of the circumscribed circle. } \scanpage{191.png}% \ex{A line is drawn parallel to a side $AB$ of a triangle $ABC$, cutting $AC$ in $D$, $BC$ in $E$. If $AD:DC = 2:3$, and $AB = 20$ inches, find $DE$.} \ex{The sides of a triangle are $9$,~$12$,~$15$. Find the segments of the sides made by bisecting the angles.} \ex{A tree casts a shadow $90$~feet long, when a post $6$~feet high casts a shadow~$4$ feet long. How high is the tree?} \ex{The lower and upper bases of a trapezoid are $a$, $b$, respectively; and the altitude is~$h$. Find the altitudes of the two triangles formed by producing the legs until they meet.} \ex{The sides of a triangle are $6$,~$7$,~$8$, respectively. In a similar triangle the side homologous to~$8$ is~$40$. Find the other two sides.} \ex{The perimeters of two similar polygons are $200$~feet and $300$~feet. If a side of the first is $24$~feet, find the homologous side of the second.} \ex{How long a ladder is required to reach a window $24$~feet high, if the lower end of the ladder is $10$~feet from the side of the house?} \ex{If the side of an equilateral triangle is~$a$, find the altitude.} \ex{If the altitude of an equilateral triangle is~$h$, find the side.} \ex{Find the length of the longest chord and of the shortest chord that can be drawn through a point $6$~inches from the centre of a circle whose radius is $10$~inches.} \ex{The distance from the centre of a circle to a chord $10$~feet long is $12$~feet. Find the distance from the centre to a chord $24$~feet long.} \ex{The radius of a circle is $5$~inches. Through a point $3$~inches from the centre a diameter is drawn, and also a chord perpendicular to the diameter. Find the length of this chord, and the distance from one end of the chord to the ends of the diameter.} \ex{The radius of a circle is $6$~inches. Find the lengths of the tangents drawn from a point $10$~inches from the centre, and also the length of the chord joining the points of contact.} \ex{The sides of a triangle are $407$~feet, $368$~feet, and $351$~feet. Find the three bisectors and the three altitudes. } \scanpage{192.png}% \ex{If a chord $8$~inches long is $8$~inches distant from the centre of the circle, find the radius, and the chords drawn from the end of the chord to the ends of the diameter which bisects the chord.} \ex{From the end of a tangent $20$~inches long a secant is drawn through the centre of the circle. If the external segment of this secant is $8$~inches, find the radius of the circle.} \ex{The radius of a circle is $13$~inches. Through a point $5$~inches from the centre any chord is drawn. What is the product of the two segments of the chord? What is the length of the shortest chord that can be drawn through the point?} \ex{The radius of a circle is $9$~inches and the length of a tangent $12$~inches. Find the length of a line drawn from the extremity of the tangent to the centre of the circle.} \ex{Two circles have radii of $8$~inches and $3$~inches, respectively, and the distance between their centres is $15$~inches. Find the lengths of their common tangents.} \ex{Find the segments of a line $10$~inches long divided in extreme and mean ratio.} \ex{The sides of a triangle are $4$,~$5$,~$5$. Is the largest angle acute, right, or obtuse?} \ex{Find the third proportional to two lines whose lengths are $28$~feet and $42$~feet.} \ex{If the sides of a triangle are $a$,~$b$,~$c$, respectively, find the lengths of the three altitudes.} \ex{The diameter of a circle is $30$~feet and is divided into five equal parts. Find the lengths of the chords drawn through the points of division perpendicular to the diameter.} \ex{The radius of a circle is $2$~inches. From a point $4$~inches from the centre a secant is drawn so that the internal segment is $1$~inch. Find the length of the secant.} \ex{The sides of a triangular pasture are $1551$ yards, $2068$ yards, $2585$ yards. Find the median to the longest side.} \ex{The diagonal of a rectangle is $d$, and the perimeter is $p$. Find the sides.} \ex{The radius of a circle is $r$. Find the length of a chord whose distance from the centre is $\frac{1}{2} r$.} \scanpage{193.png}% \chapter{BOOK IV\@. AREAS OF POLYGONS.} \markboth{\Headings{BOOK IV\@. PLANE GEOMETRY.}} {\Headings{AREAS OF POLYGONS.}}% \pp{\defn{The \textbf{unit of surface} is a square whose side is a \emph{unit of length}.}} \pp{\defn{The \textbf{area of a surface}\label{area} is the \emph{number of units of surface} it contains.}} \pp{\defn{Plane figures that \emph{have equal areas but cannot be made to coincide} are called \textbf{equivalent}\label{equivalent2}.}} \note{In propositions relating to \emph{areas}, the words ``rectangle,'' ``triangle,'' etc., are often used for ``area of rectangle,'' ``area of triangle,'' etc.} \proposition{Theorem.} \begin{proof}% \obs{Two rectangles having equal altitudes are to each other as their bases.} \figc{193aa395}{Let the rectangles $AC$ and $AF$ have the same altitude $AD$.} \prove{$\rect AC: \rect AF = \base AB: \base AE$.} \textsc{Case 1.~} \emph{When $AB$ and $AE$ are commensurable.} \textbf{Proof.} Suppose $AB$ and $AE$ have a common measure, as $AO$, which is contained $m$ times in $AB$ and $n$ times in $AE$. \eq[\indent Then]{$AB:AE$}{$ = m:n$.}{} \scanpage{194.png}% \filbreak Apply $AO$ as a unit of measure to $AB$ and $AE$, and at the several points of division erect $\perp_s$. \step[\indent The]{$\rect AC$ is divided into $m$~rectangles,}{} \step[and the]{$\rect AF$ is divided into $n$~rectangles.}{§~107} \step{These rectangles are all equal.}{§~186} \eq[\indent Hence,]{}{$\rect AC: \rect AF = m:n$.}{} \eq[\indent Therefore,]{}{$\rect AC: \rect AF = AB:AE$.}{Ax.~1} \textsc{Case 2.} \emph{When $AB$ and $AE$ are incommensurable.} \figc{194aa395}{} \textbf{Proof.} Divide $AB$ into any number of equal parts, and apply one of them to $AE$ as many times as $AE$ will contain it. Since $AB$ and $AE$ are incommensurable, a certain number of these parts will extend from $A$ to some point $K$, leaving a remainder $KE$ less than one of the equal parts of $AB$. \step{Draw $KH \parallel$ to $EF$.}{} Then $AB$ and $AK$ are commensurable by construction. \step[\indent Therefore,]{\( \dfrac{\rect AH}{\rect AC} = \dfrac{AK}{AB}. \)}{Case~1} If the number of equal parts into which $AB$ is divided is indefinitely increased, the varying values of these ratios will continue equal, and approach for their respective limits the ratios \step{\( \dfrac{\rect AF}{\rect AC} \) and \( \dfrac{AE}{AB} \). (See §~287.)}{} \step{\( \therefore \dfrac{\rect AF}{\rect AC} = \dfrac{AE}{AB}. \)}{§~284} \hfill\qed \end{proof} \pp{\cor{Two rectangles having equal bases are to each other as their altitudes.}} \scanpage{195.png}% \proposition{Theorem.} \begin{proof}% \obs{Two rectangles are to each other as the products of their bases by their altitudes.} \vspace{1ex} \figc{195aa397}{Let $R$ and $R'$ be two rectangles, having for their bases $b$ and $b'$, and for their altitudes $a$ and $a'$, respectively.} \proveq{$\dfrac{R}{R'}$}{$= \dfrac{a × b}{a' × b'}$.} \textbf{Proof.} Construct the rectangle $S$, with its base equal to that of $R$, and its altitude equal to that of $R'$. \eq[\indent Then]{$\dfrac{R}{S}$}{$=\dfrac{a}{a'}$,}{§~396} \eq[and]{$\dfrac{S}{R'}$}{$=\dfrac{b}{b'}$.}{§~395} The products of the corresponding members of these equations give \eq{$\dfrac{R}{R'}$}{$=\dfrac{a × b}{a' × b'}$.}{\qed} \end{proof} \ex{Find the ratio of a rectangular lawn $72$~yards by $49$~yards to a grass turf $18$~inches by $14$~inches.} \ex{Find the ratio of a rectangular courtyard $18\frac{1}{2}$~yards by $15\frac{1}{2}$~yards to a flagstone $31$~inches by $18$~inches.} \ex{A square and a rectangle have the same perimeter, $100$~yards. The length of the rectangle is $4$~times its breadth. Compare their areas.} \ex{On a certain map the linear scale is $1$~inch to $5$~miles. How many acres are represented on this map by a square the perimeter of which is $1$~inch?} \scanpage{196.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a rectangle is equal to the product of its base by its altitude.} \figc{196aa398}{Let $R$ be a rectangle, $b$ its base, and $a$ its altitude.} \proveq{the area of $R$}{$= a × b$.} \textbf{Proof.} Let $U$ be the unit of surface. \eq{\( \dfrac{R}{U} = \dfrac{a × b}{1 × 1} \)}{$= a × b$,}{} \pnote{(two rectangles are to each other as the products of their bases and altitudes).} \step[\indent But]{$\dfrac{R}{U} =$ the \emph{number} of units of surface in $R$.}{§~393} \label{formarearect}% \eq{$\therefore$ the area of $R$}{$= a × b$.}{\qed} \end{proof} \figc{196bb399}{} \begin{point}\textsc{Scholium.} When the base and altitude each contain the linear unit an integral number of times, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of surface. Thus, if the base contains seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of surface. \end{point} \scanpage{197.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a parallelogram is equal to the product of its base by its altitude.} \figc{197aa400}{Let $AEFD$ be a parallelogram, $b$ its base, and $a$ its altitude.} \prove{the area of the \textnormal{$\Par AEFD = a × b$.}} \textbf{Proof.} From $A$ draw $AB$ $\parallel$ to $DC$ to meet $FE$ produced. Then the figure $ABCD$ is a rectangle, with the same base and the same altitude as the $\Par AEFD$. \step{The rt.~$\triangle_s ABE$ and $DCF$ are equal.}{§~151} \step{For $AB = CD$, and $AE = DF$.}{§~178} From $ABFD$ take the $\triangle DCF$; the $\rect ABCD$ is left. From $ABFD$ take the $\triangle ABE$; the $\Par AEFD$ is left. \step{$\therefore \rect ABCD \Bumpeq \Par AEFD$}{Ax.~3} \step{But the area of the $\rect ABCD = a × b$.}{§~398} \label{formareapar}% \step{$\therefore$ the area of the $\Par AEFD = a × b$.}{Ax.~1} \hfill\qed \end{proof} \pp{\cor[1]{Parallelograms having equal bases and equal altitudes are equivalent.}} \pp{\cor[2]{Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases; any two parallelograms are to each other as the products of their bases by their altitudes.}} \scanpage{198.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a triangle is equal to half the product of its base by its altitude.} \figc{198aa403}{Let $a$ be the altitude and $b$ the base of the triangle $ABC$.} \prove{the area of the $\triangle{}ABC=\frac{1}{2}a × b$.} \textbf{Proof.} Construct on $AB$ and $BC$ the parallelogram $ABCH$. \step[\indent Then]{$\triangle ABC=\frac{1}{2}\Par ABCH$.}{§~179} \step{The area of the $\Par ABCH=a × b$.}{§~400} \label{formareatri}% \step{Therefore, the area of $\triangle ABC=\frac{1}{2}a × b$.}{Ax.~7} \hfill\qed \end{proof} \pp{\cor[1]{Triangles having equal bases and equal altitudes are equivalent.}} \pp{\cor[2]{Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the products of their bases by their altitudes.}} \pp{\cor[3]{The product of the legs of a right triangle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle.}} \ex{The lines which join the middle point of either diagonal of a quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts.} \scanpage{199.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a trapezoid is equal to half the sum of its bases multiplied by the altitude.} \figc{199aa407}{Let $b$ and $b'$ be the bases and $a$ the altitude of the trapezoid $ABCH$.} \prove{the area of the $ABCH=\frac{1}{2}a(b+b')$.} \step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{} \eq[\indent Then]{the area of the $\triangle ABC$}{$=\frac{1}{2}a × b$,}{} \eq[and]{the area of the $\triangle AHC$}{$=\frac{1}{2}a × b'$.}{§~403} \label{formareatrap}% \eq{$\therefore$ the area of $ABCH$}{$=\frac{1}{2}a(b+b')$.}{Ax.~2} \hfill\qed \end{proof} \pp{\cor{The area of a trapezoid is equal to the product of the median by the altitude.}~\hfill§~190} \figc{199bb409}{} \begin{point}% \textsc{Scholium.} The area of an irregular polygon may be found by dividing the polygon into triangles, and by finding the area of each of these triangles separately. Or, we may draw the longest diagonal, and let fall perpendiculars upon this diagonal from the other vertices of the polygon. The sum of the areas of the right triangles, rectangles, and trapezoids thus formed is the area of the polygon. \end{point} \scanpage{200.png}% \proposition{Theorem.} \begin{proof}% \obs{The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.} \figc{200aa410}{Let the triangles $ABC$ and $ADE$ have the common angle $A$.} \prove {$\dfrac{\triangle ABC} {\triangle ADE} = \dfrac{AB × AC} {AD × AE}$.} \step [\indent Proof.] {Draw $BE$.}{} \step [\indent Now] {$\dfrac{\triangle ABC}{\triangle ABE}= \dfrac{AC}{AE}$,} {} \step [and] {$\dfrac{\triangle ABE}{\triangle ADE}= \dfrac{AB}{AD}$.} {§~405} The products of the first members and of the second members of these equalities give \step {$\dfrac{\triangle ABC}{\triangle ADE}=\dfrac{AB × AC}{AD × AE}$.} {\qed} \end {proof} \ex{The areas of two triangles which have an angle of the one supplementary to an angle of the other are to each other as the products of the sides including the supplementary angles.} \scanpage{201.png}% \clearpage \section{COMPARISON OF POLYGONS.} \proposition{Theorem.} \begin{proof}% \obs{The areas of two similar triangles are to each as the squares of any two homologous sides.} \figc{201ab411}{Let the two similar triangles be $ACB$ and $A'C'B'$.} \proveq{$\dfrac{\triangle ACB}{\triangle A'C'B'}$} {$= \dfrac{\overline{AB}^2}{\overline{A'B'}^2}$.} \textbf{Proof.} Draw the altitudes $CO$ and $C'O'$. \step[\indent Then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} = \dfrac{AB × CO}{A'B' × C'O'} = \dfrac{AB}{A'B'} × \dfrac{CO}{C'O'} \),}{§~405} \pnote{(two $\triangle_s$ are to each other as the products of their bases by their altitudes).} \eq[\indent But]{$\dfrac{AB}{A'B'}$}{$= \dfrac{CO}{C'O'}$.}{§~361} \pnote{(the homologous altitudes of two similar $\triangle_s$ have the same ratio as any two homologous sides).} Substitute, in the above equality, for $\dfrac{CO}{C'O'}$ its equal $\dfrac{AB}{A'B'}$; \step[then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} = \dfrac{AB }{A'B'} × \dfrac{AB }{A'B'} = \dfrac{\overline{AB}^2 }{\overline{A'B'}^2} \).}{\qed} \end{proof} \ex{Prove this proposition by §~410.} \scanpage{202.png}% \proposition{Theorem.} \begin{proof}% \obs{The areas of two similar polygons are to each other as the squares of any two homologous sides.} \figc{202ab412}{Let $S$ and $S'$ denote the areas of the two similar polygons $ABC$ etc.\ and $A'B'C'$ etc.} \proveq{$S:S'$}{$=\overline{AB}^2:\overline{A'B'^2}$.} \textbf{Proof.} By drawing all the diagonals from any homologous vertices $E$ and $E'$, the two similar polygons are divided into similar triangles.~\hfill§~365 \step{\( \displaystyle \therefore \frac{\overline{AB}^2}{\overline{A'B'^2}}= \frac{\triangle ABE}{\triangle A'B'E'}= \left(\frac{\overline{BE}^2}{\overline{B'E'^2}}\right)= \frac{\triangle BCE}{\triangle B'C'E'}=\text{etc.} \)}{§~411} \step[\indent That is,]{\( \displaystyle \frac{\triangle ABE}{\triangle A'B'E'}= \frac{\triangle BCE}{\triangle B'C'E'}= \frac{\triangle CDE}{\triangle C'D'E'} \).}{} \( \displaystyle \therefore \frac{\triangle ABE + \triangle BCE + \triangle CDE} {\triangle A'B'E' + \triangle B'C'E' + \triangle C'D'E'}= \frac{\triangle ABE}{\triangle A'B'E'}= \frac{\overline{AB}^2}{\overline{A'B'^2}} \).\hsp§~335 \step{\( \displaystyle \therefore S:S'=\overline{AB}^2:\overline{A'B'^2} \)}{\qed} \end{proof} \pp{\cor[1]{The areas of two similar polygons are to each other as the squares of any two homologous lines.}} \pp{\cor[2]{The homologous sides of two similar polygons have the same ratio as the square roots of their areas.}} \scanpage{203.png}% \proposition{Theorem.} \begin{proof}% \obs{The square on the hypotenuse of a right triangle is equivalent to the sum of the squares on the two legs.} \figc{203aa415}{Let $BE$, $CH$, $AF$, be squares on the three sides of the right triangle $ABC$.} \prove{$BE \Bumpeq CH + AF$.} \textbf{Proof.} Through $A$ draw $AL \parallel$ to $CE$, and draw $AD$ and $CF$. Since $\angle_s{} BAC$, $BAG$, and $CAH$ are rt.\ $\angle_s$, $CAG$ and $BAH$ are straight lines.~\hfill§~90 \eq[\indent The]{$\triangle ABD$}{$=\triangle FBC$.}{§~143} \eq[\indent For]{$BD$}{$=BC$,}{} \eq{$BA$}{$=BF$,}{§~168} \eq[and]{$\angle ABD$}{$=\angle FBC$,}{Ax.~2} \pnote{(each being the sum of a rt.\ $\angle$ and the $\angle ABC$).} \step{Now the rectangle $BL$ is double the $\triangle ABD$,}{} \pnote{(having the same base $BD$, and the same altitude, the distance between the $\parallel_s AL$ and $BD$),} \step{and the square $AF$ is double the $\triangle FBC$,}{} \pnote{(having the same base $FB$, and the same altitude $AB$).} $\therefore$ the rectangle $BL$ is equivalent to the square $AF$.\hfill~Ax.~6 In like manner, by drawing $AE$ and $BK$, it may be proved that the rectangle $CL$ is equivalent to the square $CH$. Hence, the square $BE$, the sum of the rectangles $BL$ and $CL$, is equivalent to the sum of the squares $CH$ and $AF$.~\hfill\qed \end{proof} \pp{\cor{The square on either leg of a right triangle is equivalent to the difference of the square on the hypotenuse and the square on the other leg.}} \scanpage{204.png}% \section[EXERCISES.]{THEOREMS.} \figccc{204aa356}{204bb357}{204cc358} \begin{proofex}% The square constructed upon the sum of two straight lines is equivalent to the sum of the squares constructed upon these two lines, increased by twice the rectangle of these lines: Let $AB$ and $BC$ be the two straight lines, and $AC$ their sum. Construct the squares $ACGK$ and $ABED$ upon $AC$ and $AB$, respectively. Prolong $BE$ and $DE$ until they meet $KG$ and $CG$, respectively. Then we have the square $EFGH$, with sides each equal to $BC$. Hence, the square $ACGK$ is the sum of the squares $ABED$ and $EFGH$, and the rectangles $DEHK$ and $BCFE$, the dimensions of which are equal to $AB$ and $BC$. \end{proofex} \begin{proofex}% The square constructed upon the difference of two straight lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines. Let $AB$ and $AC$ be the two straight lines, and $BC$ their difference. Construct the square $ABFG$ upon $AB$, the square $ACKH$ upon $AC$, and the square $BEDC$ upon $BC$ (as shown in the figure). Prolong $ED$ to meet $AG$ in $L$. The dimensions of the rectangles $LEFG$ and $HKDL$ are $AB$ and $AC$, and the square $BCDE$ is evidently the difference between the whole figure and the sum of these rectangles; that is, the square constructed upon $BC$ is equivalent to the sum of the squares constructed upon $AB$ and $AC$, diminished by twice the rectangle of $AB$ and $AC$. \end{proofex} \begin{proofex}% The difference between the squares constructed upon two straight lines is equivalent to the rectangle of the sum and difference of these lines. Let $ABDE$ and $BCFG$ be the squares constructed upon the two straight lines $AB$ and $BC$. The difference between these squares is the polygon $ACGFDE$, which is composed of the rectangles $ACHE$ and $GFDH$. Prolong $AE$ and $CH$ to $I$ and $K$, respectively, making $EI$ and $HK$ each equal to $BC$, and draw $IK$. The rectangles $GFDH$ and $EHKI$ are equal. The difference between the squares $ABDE$ and $BCGF$ is then equivalent to the rectangle $ACKI$, which has for dimensions $AI$, equal to $AB + BC$, and $EH$, equal to $AB - BC$. \end{proofex} \scanpage{205.png}% \ex{The area of a rhombus is equal to half the product of its diagonals.} \ex{Two isosceles triangles are equivalent if their legs are equal each to each, and the altitude of one is equal to half the base of the other.} \ex{The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle.} \ex{Two parallelograms are equal if two adjacent sides of the one are equal, respectively, to two adjacent sides of the other, and the included angles are supplementary.} \ex{If $ABC$ is a right triangle, $C$ the vertex of the right angle, $BD$ a line cutting $AC$ in $D$, then \( \overline{BD}^2 + \overline{AC}^2 = \overline{AB}^2 + \overline{DC}^2 \).} \ex{Upon the sides of a right triangle as homologous sides three similar polygons are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the legs.} \ex{If the middle points of two adjacent sides of a parallelogram are joined, a triangle is formed which is equivalent to one eighth of the parallelogram.} \ex{If any point within a parallelogram is joined to the four vertices, the sum of either pair of triangles having parallel bases is equivalent to half the parallelogram.} \ex{Every straight line drawn through the intersection of the diagonals of a parallelogram divides the parallelogram into two equal parts.} \ex{The line which joins the middle points of the bases of a trapezoid divides the trapezoid into two equivalent parts.} \ex{Every straight line drawn through the middle point of the median of a trapezoid cutting both bases divides the trapezoid into two equivalent parts.} \ex{If two straight lines are drawn from the middle point of either leg of a trapezoid to the opposite vertices, the triangle thus formed is equivalent to half the trapezoid.} \ex{The area of a trapezoid is equal to the product of one of the legs by the distance from this leg to the middle point of the other leg.} \ex{The figure whose vertices are the middle points of the sides of any quadrilateral is equivalent to half the quadrilateral.} \scanpage{206.png}% \clearpage \section{PROBLEMS OF CONSTRUCTION.} \proposition{Problem.} \begin{proof}% \obs{To construct a square equivalent to the sum of two given squares.} \figc{206aa417}{Let $R$ and $R'$ be two given squares.} \prove[To construct ]{a square equivalent to $R'+R$.} \step {Construct the rt.\ $\angle A$.} {} \step {Take $AC$ equal to a side of $R'$,} {} \step {and $AB$ equal to a side of $R$; and draw $BC$.} {} \step {Construct the square $S$, having each of its sides equal to $BC$.} {} \step [\indent Then] {$S$ is the square required.} {} \step [\indent Proof.] {$\overline{BC}^2 \Bumpeq \overline{AC}^2 + \overline{AB}^2$,} {§~415} \pnote {(the square on the hypotenuse of a rt.\ $\triangle$ is equivalent to the sum of the squares on the two legs).} \step {$\therefore S \Bumpeq R'+R$.} {} \hfill\qef \end{proof} \ex{If the perimeter of a rectangle is $72$~feet, and the length is equal to twice the width, find the area.} \ex{How many tiles $9$~inches long and $4$~inches wide will be required to pave a path $8$~feet wide surrounding a rectangular court $120$~feet long and $36$~feet wide?} \ex{The bases of a trapezoid are $16$~feet and $10$~feet; each leg is equal to $5$~feet. Find the area of the trapezoid.} \scanpage{207.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a square equivalent to the difference of two given squares.} \figc{207aa418}{Let $R$ be the smaller square and $R'$ the larger.} \prove[To construct ]{a square equivalent to $R'-R$.} \step{Construct the rt.\ $\angle A$.}{} \step{Take $AB$ equal to a side of $R$.}{} \step{From $B$ as a centre, with a radius equal to a side of $R'$,}{} \step{describe an arc cutting the line $AX$ at $C$.}{} \step{Construct the square $S$, having each of its sides equal to $AC$.}{} \step[\indent Then]{$S$ is the square required.}{} \step[\indent\textbf{Proof.}]{\( \overline{AC}^2 \Bumpeq \overline{BC}^2-\overline{AB}^2 \),}{§~416} \pnote{(the square on either leg of a rt.\ $\triangle$ is equivalent to the difference of the square on the hypotenuse and the square on the other leg).} \step{$\therefore S \Bumpeq R'-R$.}{} \hfill\qef \end{proof} \ex{Construct a square equivalent to the sum of two squares whose sides are $3$~inches and $4$~inches.} \ex{Construct a square equivalent to the difference of two squares whose sides are $2\frac{1}{2}$~inches and $2$~inches.} \ex{Find the side of a square equivalent to the sum of two squares whose sides are $24$~feet and $32$~feet.} \ex{Find the side of a square equivalent to the difference of two squares whose sides are $24$~feet and $40$~feet.} \ex{A rhombus contains $100$~square feet, and the length of one diagonal is $10$~feet. Find the length of the other diagonal. } \scanpage{208.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a polygon similar to two given similar polygons and equivalent to their sum.} \figc{208aa419}{Let $R$ and $R'$ be two similar polygons, and $AB$ and $A'B'$ two homologous sides.} \prove[To construct ]{a similar polygon equivalent to $R+R'$.} \step{Construct the rt.\ $\angle P$.}{} \step{Take $PH$ equal to $A'B'$, and $PO$ equal to $AB$.}{} \step{Draw $OH$, and take $A''B''$ equal to $OH$.}{} \step{Upon $A''B''$, homologous to $AB$, construct $R''$ similar to $R$.}{} \step{Then $R''$ is the polygon required.}{} \eq[\indent\textbf{Proof.}]{$\overline{PO}^2 + \overline{PH}^2$}{$= \overline{OH}^2$.}{§~415} Put for $PO$, $PH$, and $OH$ their equals $AB$, $A'B'$, and $A''B''$. \eq[\indent Then]{$\overline{AB}^2 + \overline{A'B'^2}$}{$= \overline{A''B''^2}$.}{} \step[\indent Now]{$\dfrac{R}{R''} = \dfrac{\overline{AB}^2}{\overline{A''B''^2}}$, and $\dfrac{R'}{R''} = \dfrac{\overline{A'B'^2}}{\overline{A''B''^2}}$.} {§~412} \step[\indent By addition,]{$\dfrac{R+R'}{R''} = \dfrac{\overline{AB}^2 + \overline{A'B'^2}}{\overline{A''B''^2}} = 1$.} {Ax.~2} \step{$\therefore R'' \Bumpeq R+R'$.}{\qef} \end{proof} \scanpage{209.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a triangle equivalent to a given polygon.} \figc{209aa420}{Let $ABCDHE$ be the given polygon.} \prove[To construct ]{a triangle equivalent to the given polygon.} Let $D$, $H$, and $E$ be any three consecutive vertices of the polygon. Draw the diagonal $DE$. \step{From $H$ draw $HF \parallel$ to $DE$.}{} \step{Produce $AE$ to meet $HF$ at $F$, and draw $DF$.}{} Again, draw $CF$, and draw $DK \parallel$ to $CF$ to meet $AF$ produced at $K$, and draw $CK$. In like manner continue to reduce the number of sides of the polygon until we obtain the $\triangle CIK$. \step{Then $\triangle CIK$ is the triangle required.}{} \textbf{Proof.} The polygon $ABCDF$ has one side less than the polygon \newline$ACBDHE$, but the two polygons are equivalent. \step{For the part $ACBDE$ is common,}{} \step{and the $\triangle DEF \Bumpeq \triangle DEH$,}{§~404} \pnote{(for the base $DE$ is common, and their vertices $F$ and $H$ are in the line $FH \parallel$ to the base).} In like manner it may be proved that \step{$ABCK \Bumpeq ABCDF$, and $CIK \Bumpeq ABCK$.}{\qef} \end{proof} \scanpage{210.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a square equivalent to a given parallelogram.} \figc{210aa421}{Let $ABCD$ be the parallelogram, $b$ its base, and $a$ its altitude.} \prove[To construct ]{a square equivalent to the $\Par ABCD$.} \step{Upon a line $MX$ take $MN$ equal to $a$, $NO$ equal to $b$.}{} \step{Upon $MO$ as a diameter, describe a semicircle.}{} \step{At $N$ erect $NP \perp$ to $MO$, meeting the circumference at $P$.}{} Then the square $R$, constructed upon a line equal to $NP$, is equivalent to the $\Par ACBD$. \eq[\indent\textbf{Proof.}]{$MN:NP$}{$= NP:NO$,}{§~370} \pnote{(a $\perp$ let fall from any point of a circumference to the diameter is the mean proportional between the segments of the diameter).} \step{$\therefore \overline{NP}^2 = MN × NO = a × b$.}{§~327} \step[\indent Therefore,]{$R \Bumpeq \Par ABCD$.}{\qef} \end{proof} \pp{\cor[1]{A square may be constructed equivalent to a given triangle, by taking for its side the mean proportional between the base and half the altitude of the triangle.}} \pp{\cor[2]{A square may be constructed equivalent to a given polygon, by first reducing the polygon to an equivalent triangle, and then constructing a square equivalent to the triangle.}} \scanpage{211.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line.} \figc{211aa424}{Let $R$ be the given square, and let the sum of the base and altitude of the required parallelogram be equal to the given line $MN$.} \prove[To construct ]{a $\Par$ equivalent to $R$, with the sum of its base and altitude equal to $MN$.} \step{Upon $MN$ as a diameter, describe a semicircle.}{} At $M$ erect $MP$, a $\perp$ to $MN$, equal to a side of the given square $R$. \step{Draw $PQ \parallel$ to $MN$, cutting the circumference at $S$.}{} \step{Draw $SC \perp$ to $MN$.}{} Any $\Par$ having $CM$ for its altitude and $CN$ for its base is equivalent to~$R$. \eq[\indent\textbf{Proof.}]{$SC$}{$=PM$.}{§§~104, 180} \eq{$\therefore \overline{SC}^2$}{$= \overline{PM}^2 = R$.}{} \eq{$MC:SC$}{$= SC:CN$,}{§~370} \pnote{(a $\perp$ let fall from any point of a circumference to the diameter is the mean proportional between the segments of the diameter).} \step[\indent Then]{$\overline{SC}^2 \Bumpeq MC × CN$.}{§~327} \hfill\qef \end{proof} \note{This problem may be stated as follows:} \emph{To construct two straight lines the sum and product of which are known.} \scanpage{212.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line.} \figc{212aa425}{Let $R$ be the given square, and let the difference of the base and altitude of the required parallelogram be equal to the given line $MN$.} \textit{To construct a $\Par$ equivalent to $R$, with the difference of its base and altitude equal to $MN$.} Upon the given line $MN$ as a diameter, describe a circle. From $M$ draw $MS$, tangent to the $\odot$, and equal to a side of the given square $R$. Through the centre of the $\odot$ draw $SB$ intersecting the circumference at $C$ and $B$. Then any $\Par$, as $R'$, having $SB$ for its base and $SC$ for its altitude, is equivalent to $R$. \step[\indent\textbf{Proof.}]{$SB:SM=SM:SC$,}{§~381} \pnote{(if from a point without a $\odot$ a secant and a tangent are drawn, the tangent is the mean proportional between the whole secant and the external segment).} \step[\indent Then]{$\overline{SM}^2 \Bumpeq SB × SC$,}{§~327} \noindent and the difference between $SB$ and $SC$ is the diameter of the $\odot$, that is,~$MN$. \hfill\qef \end{proof} \note{This problem may be stated: \textit{To construct two straight lines the difference and product of which are known.}} \scanpage{213.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a polygon similar to a given polygon $P$ and equivalent to a given polygon $Q$.} \figc{213aa426}{Let $P$ and $Q$ be the two given polygons, and $AB$ a side of $P$.} \prove[To construct ]{a polygon similar to $P$ and equivalent to $Q$.} \step{Find squares equivalent to $P$ and $Q$,}{§~423} \step{and let $m$ and $n$ respectively denote their sides.}{} Find $A'B'$, the fourth proportional to $m$, $n$, and $AB$.~\hfill§~386 Upon $A'B'$, homologous to $AB$, construct $P'$ similar to $P$. \eq[\indent Then]{$P'$}{$\Bumpeq Q$.}{} \eq[\indent\textbf{Proof.}]{$m:n$}{$= AB:A'B'$.}{Const.} \eq{$\therefore m^2:n^2$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~338} \eq[\indent But]{$P \Bumpeq m^2$,}{and $Q \Bumpeq n^2$. }{Const.} \eq{$\therefore P:Q = m^2$}{$:n^2 = \overline{AB}^2:\overline{A'B'}^2$.}{} \eq[\indent But]{$P:P'$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~412} \eq{$\therefore P:Q$}{$= P:P'$.}{Ax.~1} \eq{$\therefore P'$}{$\Bumpeq Q$.}{\qef} \end{proof} \ex{To construct a square equivalent to the sum of any number of given squares.} \ex{To construct a polygon similar to two given similar polygons and equivalent to their difference. } \scanpage{214.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a square which shall have a given ratio to a given square.} \figc{214aa427}{Let R be the given square, and $\dfrac{n}{m}$ the given ratio.} \textit{To construct a square which shall be to $R$ as $n$ is to $m$.} Take $AB$ equal to a side of $R$, and draw $Ay$, making any convenient angle with $AB$. On $Ay$ take $AE$ equal to $m$, $EF$ equal to $n$, and draw $EB$. \step{Draw $FC \parallel$ to $EB$ meeting $AB$ produced at $C$.}{} \step{On $AC$ as a diameter, describe a semicircle.}{} \step{At $B$ erect the $\perp BD$, meeting the semicircumference at $D$.}{} \step{Then $BD$ is a side of the square required.}{} \step[\indent\textbf{Proof.}]{Denote $AB$ by $a$, $BC$ by $b$, and $BD$ by $x$.}{} \eq[\indent Now]{$a:x$}{$= x:b$.}{§~370} \eq[\indent Therefore,]{$a^2:x^2$}{$= a:b$.}{§~337} \eq[\indent But]{$a:b$}{$= m:n$.}{§~342} \eq[\indent Therefore,]{$a^2:x^2$}{$= m:n$.}{Ax.~1} \eq[\indent By inversion,]{$x^2:a^2$}{$= n:m$.}{§~331} Hence, the square on $BD$ will have the same ratio to $R$ as $n$ has to $m$. \hfill\qef \end{proof} \scanpage{215.png}% \proposition{Problem.} \begin{proof}% \obs{To construct a polygon similar to a given polygon and having a given ratio to it.} \figc{215aa428}{Let $R$ be the given polygon, and $\dfrac{n}{m}$ the given ratio.} \prove[To construct ]{a polygon similar to $R$, which shall be to $R$ as $n$ is to $m$.} Construct a line $A'B'$, such that the square on $A'B'$ shall be to the square on $AB$ as $n$ is to $m$.~\hfill§~427 Upon $A'B'$, as a side homologous to $AB$, construct the polygon $S$ similar to $R$.~\hfill§~391 \step{Then $S$ is the polygon required.}{} \eq[\indent\textbf{Proof.}]{$S:R$}{$= \overline{A'B'}^2 : \overline{AB}^2$.}{§~412} \eq[\indent But]{$\overline{A'B'}^2 : \overline{AB}^2$}{$= n:m$.}{Const.} \eq[\indent Therefore,]{$S:R$}{$= n:m$.}{Ax.~1} \hfill\qef \end{proof} \ex{To construct a triangle equivalent to a given triangle, and having one side equal to a given length $l$.} \ex{To transform a triangle into an equivalent right triangle.} \ex{To transform a given triangle into an equivalent right triangle, having one leg equal to a given length.} \ex{To transform a given triangle into an equivalent right triangle, having the hypotenuse equal to a given length.} \scanpage{216.png}% \section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.} \begin{proofex}% To transform a triangle $ABC$ into an equivalent triangle, having a side equal to a given length $l$, and an angle equal to angle $BAC$. Upon $AB$ (produced if necessary), take $AD$ equal to $l$, draw $BE \parallel$ to $CD$, meeting $AC$ (produced if necessary) at $E$. $\triangle BED \Bumpeq \triangle BEC$. \end{proofex} \ex{To transform a given triangle into an equivalent isosceles triangle, having the base equal to a given length.} \exheader{To construct a triangle equivalent to:} \ex{The sum of two given triangles.} \ex{The difference of two given triangles.} \ex{To transform a given triangle into an equivalent equilateral triangle.} \exheader{To transform a parallelogram into an equivalent:} \ex{Parallelogram having one side equal to a given length.} \ex{Parallelogram having one angle equal to a given angle.} \ex{Rectangle having a given altitude.} \exheader{To transform a square into an equivalent:} \ex{Equilateral triangle.} \ex{Right triangle having one leg equal to a given length.} \ex{Rectangle having one side equal to a given length.} \exheader{To construct a square equivalent to:} \ex{Five eighths of a given square.} \ex{Three fifths of a given pentagon.} \ex{To divide a given triangle into two equivalent parts by a line through a given point $P$ in one of the sides.} \ex{To find a point within a triangle, such that the lines joining this point to the vertices shall divide the triangle into three equivalent parts.} \ex{To divide a given triangle into two equivalent parts by a line parallel to one of the sides.} \ex{To divide a given triangle into two equivalent parts by a line perpendicular to one of the sides.} \scanpage{217.png}% \subsection{PROBLEMS OF COMPUTATION.} \figccc{217aa404}{217bb405}{217cc406} \begin{proofex}% To find the area of an equilateral triangle in terms of its side. Denote the side by $a$, the altitude by $h$, and the area by $S$. \setlength{\eqalign}{0.33\dentwidth} \eq[\indent Then]{$h^2$}{$a^2 - \dfrac{a^2}{4} = \dfrac{3a^2}{4} = \dfrac{a^2}{4} × 3$.}{§~372} \eq{$\therefore h$}{$= \dfrac{a}{2}\sqrt{3}$.}{} \eq[\indent But]{$S$}{$= \dfrac{a × h}{2}$.}{§~403} \label{formareaequitri}% \eq{$\therefore S$} {$=\dfrac{a}{2} × \dfrac{a\sqrt{3}}{2} = \dfrac{a^2\sqrt{3}}{4}$.}{} \setlength{\eqalign}{0.5\dentwidth} \end{proofex} \begin{proofex}% To find the area of a triangle in terms of its sides. \label{formareatri2}% \setlength{\eqalign}{0.33\dentwidth} \eq[\indent By Ex.~312,]{$h$}{$= \dfrac{2}{b} \sqrt{s(s - a)(s - b)(s - c)}$.}{} \eq[\indent Hence,]{$S$}{$= \dfrac{b}{2} × \dfrac{2}{b} \sqrt{s(s - a)(s - b)(s - c)}$}{§~403} \eq{}{$= \sqrt{s(s - a)(s - b)(s - c)}$.}{} \setlength{\eqalign}{0.5\dentwidth} \end{proofex} \begin{proofex}% To find the area of a triangle in terms of the radius of the circumscribed circle. If $R$ denotes the radius of the circumscribed circle, and $h$ the altitude of the triangle, we have, by §~384, \eq{$b × c$}{$= 2 R × h$.}{} Multiply by $a$, and we have, \eq{$a × b × c$}{$= 2 R × a × h$.}{} \eq[\indent But]{$a × h$}{$= 2 S$.}{§~403} \eq{$\therefore a × b × c$}{$= 4 R × S$.}{} \label{formareatri3}% \eq{$\therefore S$}{$= \dfrac{abc}{4R}$.}{} Show that the radius of the circumscribed circle is equal to $\dfrac{abc}{4S}$. \end{proofex} \scanpage{218.png}% \ex{Find the area of a right triangle, if the length of the hypotenuse is $17$~feet and the length of one leg is $8$~feet.} \ex{Find the ratio of the altitudes of two equivalent triangles, if the base of one is three times that of the other.} \ex{The bases of a trapezoid are $8$~feet and $10$~feet, and the altitude is $6$~feet. Find the base of the equivalent rectangle that has an equal altitude.} \ex{Find the area of a rhombus, if the sum of its diagonals is $12$~feet, and their ratio is~$3:5$.} \ex{Find the area of an isosceles right triangle, if the hypotenuse is $20$~feet.} \ex{In a right triangle the hypotenuse is $13$~feet, one leg is $5$~feet. Find the area.} \ex{Find the area of an isosceles triangle, if base $=b$, and leg $=c$.} \ex{Find the area of an equilateral triangle, if one side $=8$~feet.} \ex{Find the area of an equilateral triangle, if the altitude $=h$.} \ex{A house is $40$~feet long, $30$~feet wide, $25$~feet high to the roof, and $35$~feet high to the ridge-pole. Find the number of square feet in its entire exterior surface.} \ex{The sides of a right triangle are as $3:4:5$. The altitude upon the hypotenuse is $12$~feet. Find the area.} \ex{Find the area of a right triangle, if one leg $=a$, and the altitude upon the hypotenuse $=h$.} \ex{Find the area of a triangle, if the lengths of the sides are $104$~feet, $111$~feet, and $175$~feet.} \ex{The area of a trapezoid is $700$~square feet. The bases are $30$~feet and $40$~feet, respectively. Find the altitude.} \ex{$ABCD$ is a trapezium; $AB = 87$ feet, $BC = 119$ feet, $CD = 41$ feet, $DA = 169$ feet, $AC = 200$ feet. Find the area.} \ex{What is the area of a quadrilateral circumscribed about a circle whose radius is $25$~feet, if the perimeter of the quadrilateral is $400$ feet? What is the area of a hexagon that has a perimeter of $400$ feet and is circumscribed about the same circle of $25$~feet radius (Ex.~361)?} \ex{The base of a triangle is $15$~feet, and its altitude is $8$~feet. Find the perimeter of an equivalent rhombus, if the altitude is $6$~feet.} \scanpage{219.png}% \ex{Upon the diagonal of a rectangle $24$~feet by $10$~feet a triangle equivalent to the rectangle is constructed. What is its altitude?} \ex{Find the side of a square equivalent to a trapezoid whose bases are $56$~feet and $44$~feet, and each leg is $10$~feet.} \ex{Through a point $P$ in the side $AB$ of a triangle $ABC$, a line is drawn parallel to $BC$ so as to divide the triangle into two equivalent parts. Find the value of $AP$ in terms of $AB$.} \ex{What part of a parallelogram is the triangle cut off by a line from one vertex to the middle point of one of the opposite sides?} \ex{In two similar polygons, two homologous sides are $15$~feet and $25$~feet. The area of the first polygon is $450$~square feet. Find the area of the second polygon.} \ex{The base of a triangle is $32$~feet, its altitude $20$~feet. What is the area of the triangle cut off by a line parallel to the base at a distance of $15$~feet from the base?} \ex{The sides of two equilateral triangles are $3$~feet and $4$~feet. Find the side of an equilateral triangle equivalent to their sum.} \ex{If the side of one equilateral triangle is equal to the altitude of another, what is the ratio of their areas?} \ex{The sides of a triangle are $10$~feet, $17$~feet, and $21$~feet. Find the areas of the parts into which the triangle is divided by the bisector of the angle formed by the first two sides.} \ex{In a trapezoid, one base is $10$~feet, the altitude is $4$~feet, the area is $32$~square feet. Find the length of a line drawn between the legs parallel to the bases and distant $1$~foot from the lower base.} \ex{The diagonals of a rhombus are $90$~yards and $120$~yards, respectively. Find the area, the length of one side, and the perpendicular distance between two parallel sides.} \ex{Find the number of square feet of carpet that are required to cover a triangular floor whose sides are, respectively, $26$~feet, $35$~feet, and $51$~feet.} \ex{If the altitude $h$ of a triangle is increased by a length $m$, how much must be taken from the base $a$ that the area may remain the same?} \ex{Find the area of a right triangle, having given the segments $p$, $q$, into which the hypotenuse is divided by a perpendicular drawn to the hypotenuse from the vertex of the right angle.} \scanpage{220.png}% \chapter{BOOK V\@. REGULAR POLYGONS AND CIRCLES.} \markboth{\Headings{BOOK V\@. PLANE GEOMETRY.}} {\Headings{REGULAR POLYGONS AND CIRCLES.}}% \pp{\defn{A \indexbf{regular polygon} is a polygon which is both equilateral and equiangular. The equilateral triangle and the square are examples.}} \proposition{Theorem.} \begin{proof}% \obs{An equilateral polygon inscribed in a circle is a regular polygon.} \figc{220aa430}{Let $ABC$ etc.\ be an equilateral polygon inscribed in a circle.} \prove{the polygon $ABC$ etc.\ is a regular polygon.} \step[\indent\textbf{Proof.}]{The arcs $AB$, $BC$, $CD$, etc., are equal.}{§~243} \step{Hence, arcs $ABC$, $BCD$, etc., are equal.}{Ax.~2} \step{Therefore, arcs $CFA$, $DFB$, etc., are equal.}{Ax.~3} \step{Therefore, $\angle_s A$, $B$, $C$, etc., are equal.}{§~289} Therefore, the polygon $ABC$ etc.\ is a regular polygon, being equilateral and equiangular.~\hfill§~429 \hfill\qed \end{proof} \scanpage{221.png}% \proposition{Theorem.} \begin{proof}% \obs{A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon.} \figc{221aa431}{Let $ABCDE$ be a regular polygon.} \prove[\textup{1.} To prove that ]{a circle may be circumscribed about $ABCDE$.} \textbf{Proof.} Let $O$ be the centre of the circle which may be passed through $A$, $B$, and $C$.~\hfill§~258 \step{Draw $OA$, $OB$, $OC$, and $OD$.}{} \eq[\indent Then]{$\angle ABC$}{$= \angle BCD$,}{§~429} \eq[and]{$\angle OBC$}{$= \angle OCB$.}{§~145} %proofrule \eq[\indent By subtraction,]{$\angle OBA$}{$= \angle OCD$.}{Ax.~3} \step{The $\triangle_s OBA$ and $OCD$ are equal.}{§~143} \eq[\indent For]{$\angle OBA$}{$= \angle OCD$,}{} \eq{$OB$}{$= OC$,}{§~217} \eq[and]{$AB$}{$= CD$.}{§~429} \eq{$\therefore OA$}{$= OD$.}{§~128} $\therefore$ the circle passing through $A$, $B$, $C$, passes through $D$. In like manner it may be proved that the circle passing through $B$, $C$, and $D$ also passes through $E$; and so on. \scanpage{222.png}% Therefore, the circle described from $O$ as a centre, with a radius $OA$, will be circumscribed about the polygon.~\hfill§~231 \prove[\textup{2.} To prove that ]{a circle may be inscribed in $ABCDE$.} \textbf{Proof.} Since the sides of the regular polygon are equal chords of the circumscribed circle, they are equally distant from the centre.~\hfill§~249 Therefore, the circle described from $O$ as a centre, with the distance from $O$ to a side of the polygon as a radius, will be inscribed in the polygon (§~232).~\hfill\qed \end{proof} \pp{\defn{The radius of the circumscribed circle, $OA$, is called the \textbf{radius} of the polygon\label{polyradius}.}} \pp{\defn{The radius of the inscribed circle, $OF$, is called the \textbf{apothem}\label{apothem} of the polygon.}} \pp{\defn{The common centre, $O$, of the circumscribed and inscribed circles is called the \textbf{centre} of the polygon\label{centrepoly}.}} \pp{\defn{The angle between radii drawn to the extremities of any side is called the \textbf{angle at the centre} of the polygon.}} By joining the centre to the vertices of a regular polygon, the polygon can be decomposed into as many equal isosceles triangles as it has sides. \pp{\cor[1]{The angle at the centre of a regular polygon\label{anglecentreregpoly} is equal to four right angles divided by the number of sides of the polygon. Hence, the angles at the centre of any regular polygon are all equal.}} \pp{\cor[2]{The radius drawn to any vertex of a regular polygon bisects the angle at the vertex.}} \pp{\cor[3]{The angle at the centre of a regular polygon and an interior angle of the polygon are supplementary.}} \step[\indent For]{$\angle_s FOB$ and $FBO$ are complementary.}{§~135} $\therefore$ their doubles $AOB$ and $FBC$ are supplementary.\hfill~Ax.~6 \scanpage{223.png}% \proposition{Theorem.} \begin{proof}% \obs{If the circumference of a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regular circumscribed polygon.} \figc{223aa439}{Suppose the circumference divided into equal arcs $AB$, $BC$, etc. Let $AB$, $BC$, etc., be the chords, $FBG$, $GCH$, etc., the tangents.} 1.~\prove{$ABCDE$ is a regular polygon.} \step[\indent\textbf{Proof.}]{The sides $AB$, $BC$, $CD$, etc., are equal.}{§~241} \step{Therefore, the polygon is regular.}{§~430} 2.~\prove{To prove that $FGHIK$ is a regular polygon.} \textbf{Proof.} The $\triangle_s AFB$, $BGC$, $CHD$, etc., are all equal isosceles triangles.\hfill\allowbreak\null\nobreak\hfill\nobreak§§~295,139 \step{$\therefore \angle_s F$, $G$, $H$, etc., are equal, and $FB$, $BG$, $GC$, etc., are equal.}{} \step{$\therefore FG=GH=HI$, etc.}{Ax.~6} \step{$\therefore FGHIK$ is a regular polygon.}{§~429} \hfill\qed \end{proof} \pp{\cor[1]{Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides as the inscribed polygon.}} \scanpage{224.png}% \figccc{224aa441}{224bb442}{224cc443} \begin{point}% \cor[2]{Tangents to a circle at the middle points of the arcs subtended by the sides of a regular inscribed polygon form a circumscribed regular polygon, whose sides are parallel to the sides of the inscribed polygon and whose vertices lie on the radii (prolonged) of the inscribed polygon.} For two corresponding sides, $AB$ and $A'B'$, are perpendicular to $OM$ (§§~248, 254), and are parallel (§~104); and the tangents $MB'$ and $NB'$, intersecting at a point equidistant from $OM$ and $ON$ (§~261), intersect upon the bisector of the $\angle MON$ (§~162); that is, upon the radius $OB$. \end{point} \pp{\cor[3]{If the vertices of a regular inscribed polygon are joined to the middle points of the arcs subtended by the sides of the polygon, the joining lines form a regular inscribed polygon of double the number of sides.}} \pp{\cor[4]{Tangents at the middle points the arcs between adjacent points of contact of the sides of a regular circumscribed polygon form a regular circumscribed polygon of double the number of sides.}} \begin{point}% \cor[5]{The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double the number of sides; and the perimeter of a circumscribed polygon is greater than the perimeter of a circumscribed polygon of double the number of sides.} For two sides of a triangle are together greater than the third side.~\hfill§~138 \end{point} \scanpage{225.png}% \proposition{Theorem.} \begin{proof}% \obs{Two regular polygons of the same number of sides are similar.} \figc{225aa445}{Let $Q$ and $Q'$ be two regular polygons, each having $n$ sides.} \prove{$Q$ and $Q'$ are similar.} \textbf{Proof.} The sum of the interior $\angle_s$ of each polygon is equal to \step{$(n-2)2$ rt.~$\angle_s$,}{§~205} \pnote{(the sum of the interior $\angle_s$ of a polygon is equal to 2 rt.\ $\angle_s$ taken as many times less two as the polygon has sides).} \step{Each angle of either polygon $= \dfrac{(n-2) 2 \text{ rt.\ } \angle_s}{n}$,}{§~206} \pnote{(for the $\angle_s$ of a regular polygon are all equal, and hence each $\angle$ is equal to the sum of the $\angle_s$ divided by their number).} Hence, the two polygons $Q$ and $Q'$ are mutually equiangular. \step{Since $AB = BC$, etc., and $A'B' = B'C'$, etc.,}{§~429} \step{\( AB:A'B' = BC:B'C' \), etc.}{} Hence, the two polygons have their homologous sides proportional. \step{Therefore the two polygons are similar.}{§~351} \hfill\qed \end{proof} \pp{\cor{The areas of two regular polygons of the same number of sides are to each other as the squares of any two homologous sides.}~\hfill§~412} \scanpage{226.png}% \proposition{Theorem.} \begin{proof}% \obs{The perimeters of two regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, and also as the radii of their inscribed circles.} \figc{226aa447}{Let $P$ and $P'$ denote the perimeters, $O$ and $O'$ the centres, of the two regular polygons.} From $O$, $O'$ draw $OA$, $O'A'$, $OB$, $O'B'$, and the $\perp_s OM$, $O'M'$. \prove{$P:P' = OA:O'A' = OM:O'M'$.} \step[\indent\textbf{Proof.}]{Since the polygons are similar,}{§~445} \eq{$P:P'$}{$= AB:A'B'$.}{§~364} The $\triangle_s OAB$ and $O'A'B'$ are isosceles.~\hfill§~431 \eq[\indent Now]{$\angle O$}{$= \angle O'$,}{§~436} \eq[and]{$OA:OB$}{$= O'A':O'B'$.}{} \step{$\therefore$ the $\triangle_s OAB$ and $O'A'B'$ are similar.}{§~357} \eq{$\therefore AB:A'B'$}{$= OA:O'A'$.}{§~351} \eq[\indent Also,]{$AB:A'B'$}{$= OM:O'M'$.}{§~361} \step{$\therefore P:P' = OA:O'A' = OM:O'M'$.}{Ax.~1} \hfill\qed \end{proof} \pp{\cor{The areas of two regular polygons of the same number of sides are to each other as the squares of the radii of the circumscribed circles, and of the inscribed circles.}~\hfill§~413} \scanpage{227.png}% \proposition{Theorem.} \begin{proof}% \obs{If the number of sides of a regular inscribed polygon is indefinitely increased, the apothem of the polygon approaches the radius of the circle as its limit.} \figc{227aa449}{Let $AB$ be a side and $OP$ the apothem of a regular polygon of $n$ sides inscribed in the circle whose radius is $OA$.} \prove{$OP$ approaches $OA$ as a limit, when $n$ increases indefinitely.} \eq[\indent\textbf{Proof.}]{$OP$}{$ K$ always, and $S < K$ always (Ax.~8), the difference between $K$ and either $S'$ or $S$ is less than the difference $S'-S$, and consequently can be made less than any assigned quantity, but cannot be made zero. Therefore, $K$ is the common limit of $S'$ and $S$.~\hfill§~275 \hfill\qed \end{proof} \scanpage{231.png}% \proposition{Theorem.} \begin{proof}% \obs{Two circumferences have the same ratio as their radii.} \figc{231aa455}{Let $C$ and $C'$ be the circumferences, $R$ and $R'$ the radii, of the two circles $Q$ and $Q'$.} \proveq{$C:C'$}{$= R:R'$.} \textbf{Proof.} Inscribe in the $\odot_s$ two similar regular polygons, and denote their perimeters by $P$ and $P'$. \eq[\indent Then]{$P:P'$}{$= R:R'$.}{§~447} Conceive the number of sides of these regular polygons to be indefinitely increased, the polygons continuing similar. Then $P$ and $P'$ will have $C$ and $C'$ as limits.~\hfill§~454 But $P:P'$ will always be equal to $R:R'$.~\hfill§~447 \eq{$\therefore C:C'$}{$= R:R'$.}{§~285} \hfill\qed \end{proof} \begin{point}% \cor{The ratio of the circumference of a circle to its diameter is constant.} \eq[\indent For]{$C:C'$}{$= R: R'$.}{§~455} \eq{$\therefore C:C'$}{$= 2R:2R'$.}{§~340} \eq[\indent By alternation,]{$C: 2R$}{$= C' : 2R'$.}{§~330} \end{point} \pp{\defn{The constant ratio of the circumference of a circle to its diameter is represented by the Greek letter $\pi$\label{pi}.}} \label{formcircum}% \pp{\cor{$\pi = \dfrac{C}{2R}$. $\therefore C=2\pi R$.}} \scanpage{232.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a regular polygon is equal to half the product of its apothem by its perimeter.} \figc{232aa459}{Let $P$ represent the perimeter, $R$ the apothem, and $S$ the area of the regular polygon $ABC$ etc.} \prove{$S = \frac{1}{2}R × P$.} \step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OC$, etc.}{} \step{The polygon is divided into as many $\triangle_s$ as it has sides.}{} \step{The apothem is the common altitude of these $\triangle_s$,}{} \step{and the area of each $\triangle=\frac{1}{2}R$ multiplied by the base.}{§~403} Hence, the area of all the $\triangle_s$ is equal to $\frac{1}{2}R$ multiplied by the sum of all the bases. But the sum of the areas of all the $\triangle_s$ is equal to the area of the polygon.\hfill~Ax.~9 And the sum of all the bases of the $\triangle_s$ is equal to the perimeter of the polygon.\hfill~Ax.~9 \label{formareapoly}% \step{$\therefore S = \frac{1}{2}R × P$.}{\qed} \end{proof} \pp{\defn{In different circles \indexbf{similar arcs}, \indexbf{similar sectors}, and \indexbf{similar segments} are such as correspond to \emph{equal angles at the centre}.}} \scanpage{233.png}% \proposition{Theorem.} \begin{proof}% \obs{The area of a circle is equal to half the product of its radius by its circumference.} \figc{233aa461}{Let $R$ represent the radius, $C$ the circumference, and $S$ the area, of the circle whose centre is $O$.} \proveq{$S$}{$= \frac{1}{2}R × C$.} \textbf{Proof.} Circumscribe any regular polygon about the circle, and denote its perimeter by $P$, and its area by $S'$. \eq[\indent Then]{$S'$}{$= \frac{1}{2} R × P$.}{§~459} Conceive the number of sides of the polygon to be indefinitely increased. \step{Then $P$ approaches $C$ as its limit,}{§~454} \step{$\frac{1}{2}R × P$ approaches $\frac{1}{2}R× C$ as its limit,}{§~279} \step{and $S'$ approaches $S$ as its limit.}{§~454} \step[\indent But]{$S' = \frac{1}{2} R × P$, always.}{§~459} \step{$\therefore S = \frac{1}{2}R× C$.}{§~284} \hfill\qed \end{proof} \label{formareasector}% \begin{point}% \cor[1]{The area of a sector is equal to half the product of its radius by its arc.} For the sector and its arc are like parts of the circle and its circumference, respectively. \end{point} \begin{point}% \cor[2]{The area of a circle is equal to $\pi$ times the square of its radius.} \label{formareacircle}% For the area of the \( \odot = \frac{1}{2} R × C = \frac{1}{2} R × 2\pi R = \pi R^2 \). \end{point} \scanpage{234.png}% \begin{point}% \cor[3]{The areas of two circles are to each other as the squares of their radii.} For, if $S$ and $S'$ denote the areas, and $R$ and $R'$ the radii, \[ S:S' = \pi R^2:\pi R'^2 = R^2:R'^2. \] \end{point} \pp{\cor[4]{Similar arcs are to each other as their radii; similar sectors are to each other as the squares of their radii.}} \proposition{Theorem.} \begin{proof}% \obs{The areas of two similar segments are to each other as the squares of their radii.} \figc{234aa466}{Let $AC$ and $A'C'$ be the radii of the two similar sectors $ACB$ and $A'C'B'$, and let $ABP$ and $A'B'P'$ be the corresponding segments.} \proveq{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.} \step[\indent\textbf{Proof.}]{Sector $ACB :$ Sector $A'C'B' = \overline{AC}^2:\overline{A'C'}^2$.}{§~465} \step{The $\triangle_s ACB$ and $A'C'B'$ are similar.}{§~357} \eq{$\therefore \triangle ACB:\triangle A'C'B'$} {$=\overline{AC}^2:\overline{A'C'}^2$.}{§~411} \eq{$\therefore$ sector $ACB :$ sector $A'C'B'$} {$=\triangle ACB : \triangle A'C'B'$.}{Ax.~1} \eq{$\therefore$ sector $ACB : \triangle ACB$} {$=$ sector $A'C'B' : \triangle A'C'B'$.}{§~330} \step{\( \therefore \dfrac{\text{sector } ACB-\triangle ACB} {\text{sector } A'C'B'-\triangle A'C'B'} = \dfrac{\triangle ACB}{\triangle A'C'B'} = \dfrac{\overline{AC}^2}{\overline{A'C'}^2} \).}{§~333} \eq[\indent That is,]{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.}{\qed} \end{proof} \scanpage{235.png}% \clearpage \section{PROBLEMS OF CONSTRUCTION.} \proposition{Problem.} \begin{proof}% \obs{To inscribe a square in a given circle.} \figc{235aa467}{Let $O$ be the centre of the given circle.} \prove[]{To inscribe a square in the given circle.} \step{Draw two diameters $AC$ and $BD \perp$ to each other.}{} \step{Draw $AB$, $BC$, $CD$, and $DA$.}{} \step{Then $ABCD$ is the square required.}{} \step[\indent\textbf{Proof.}]{The $\angle_s ABC$, $BCD$, etc., are rt.\ $\angle_s$,}{§~290} \pnote{(each being inscribed in a semicircle),} \step{and the sides $AB$, $BC$, etc., are equal,}{§~241} \pnote{(in the same $\odot$ equal arcs are subtended by equal chords).} \step{Hence the quadrilateral $ABCD$ is a square.}{§~168} \hfill\qef \end{proof} \pp{\cor{By bisecting the arcs $AB$, $BC$, etc., a regular polygon of eight sides may be inscribed in the circle; and, by continuing the process, regular polygons of sixteen, thirty-two, sixty-four, etc., sides may be inscribed.}} \ex{The area of a circumscribed square is equal to twice the area of the inscribed square.} \ex{The area of a circular ring is equal to that of a circle whose diameter is a chord of the outer circle tangent to the inner circle.} \scanpage{236.png}% \proposition{Problem.} \begin{proof}% \obs{To inscribe a regular hexagon in a given circle.} \figc{236aa469}{Let $O$ be the centre of the given circle.} \prove[To inscribe ]{a regular hexagon in the given circle.} \step{From $O$ draw any radius, as $OC$.}{} \step{From $C$ as a centre, with a radius equal to $OC$,}{} \step{describe an arc intersecting the circumference at $F$.}{} \step{Draw $OF$ and $CF$.}{} \step{Then $CF$ is a side of the regular hexagon required.}{} \step[\indent\textbf{Proof.}]{The $\triangle OFC$ is equiangular,}{§~146} \pnote{(since it is equilateral by construction).} Hence, the $\angle FOC$ is $\frac{1}{3}$ of $2$~rt.~$\angle_s$, or $\frac{1}{6}$ of $4$~rt.~$\angle_s$.~\hfill§~136 \step{$\therefore$ the arc $FC$ is $\frac{1}{6}$ of the circumference,}{} and the chord $FC$ is a side of a regular inscribed hexagon. Hence, to inscribe a regular hexagon apply the radius six times as a chord.~\hfill\qef \end{proof} \pp{\cor[1]{By joining the alternate vertices $A$, $C$, $D$, an equilateral triangle is inscribed in the circle.}} \pp{\cor[2]{By bisecting the arcs $AB$, $BC$, etc., a regular polygon of twelve sides may be inscribed in the circle; and, by continuing the process, regular polygons of twenty-four, forty-eight, etc., sides may be inscribed.}} \scanpage{237.png}% \proposition{Problem.} \begin{proof}% \obs{To inscribe a regular decagon in a given circle.} \figc{237aa472}{Let $O$ be the centre of the given circle.} \prove[To inscribe ]{a regular decagon in the given circle.} \step{Draw any radius $OC$,}{} \step{and divide it in extreme and mean ratio, so that $OC$ shall}{} \step{be to $OS$ as $OS$ is to $SC$.}{§~389} \step{From $C$ as a centre, with a radius equal to $OS$,}{} \step{describe an arc intersecting the circumference at $B$.}{} \step{Draw $BC$.}{} \step{Then $BC$ is a side of the regular decagon required.}{} \step[\indent\textbf{Proof.}]{Draw $BS$ and $BO$.}{} \eq[\indent Now]{$OC:OS$}{$= OS:SC$,}{Const.} \eq[and]{$BC$}{$= OS$.}{Const.} \eq{$\therefore OC:BC$}{$= BC:SC$.}{} \eq[\indent Moreover,]{$\angle OCB$}{$= \angle SCB$.}{Iden.} \step{Hence, the $\triangle_s OCB$ and $BCS$ are similar.}{§~357} \step{But the $\triangle OCB$ is isosceles.}{§~217} \step{$\therefore \triangle BCS$, which is similar to the $\triangle OCB$, is isosceles,}{} \step{and $CB = BS = SO$.}{§~120} \scanpage{238.png}% \step{$\therefore$ the $\triangle SOB$ is isosceles, and the $\angle O = \angle SBO$.}{§~145} \step{But the ext.~$\angle CSB = \angle O + \angle SBO = 2 \angle O$.}{§~137} \eq[\indent Hence,]{$\angle SCB$}{$= 2\angle O$,}{} \eq[and]{$\angle OBC$}{$= 2\angle O$.}{} \step{$\therefore$ the sum of the $\angle_s$ of the $\triangle OCB = 5 \angle O = 2 \text{rt.\ } \angle_s$,}{} \step[and]{$\angle O = \frac{1}{5}$ of $2$~rt.~$\angle_s$, or $\frac{1}{10}$ of $4$~rt.~$\angle_s$.}{} \step{Therefore, the arc $BC$ is $\frac{1}{10}$ of the circumference,}{} and the chord $BC$ is a side of a regular inscribed decagon. Therefore, to inscribe a regular decagon, divide the radius internally in extreme and mean ratio, and apply the greater segment ten times as a chord.~\hfill\qef \end{proof} \pp{\cor[1]{By joining the alternate vertices of a regular inscribed decagon, a regular pentagon is inscribed.}} \pp{\cor[2]{By bisecting the arcs $BC$, $CF$, etc., a regular polygon of twenty sides may be inscribed in the circle; and, by continuing the process, regular polygons of forty, eighty, etc., sides may be inscribed.}} If $R$ denotes the radius of a regular inscribed polygon, $r$ the apothem, $a$ one side, $A$ an interior angle, and $C$ the angle at the centre, show that \ex{In a regular inscribed triangle $a = R\sqrt{3}$, $r =\frac{1}{2}R$, $A = 60°$, $C = 120°$.} \ex{In an inscribed square $a = R\sqrt{2}$, $r = \frac{1}{2}R\sqrt{2}$, $A=90°$, $C=90°$.} \ex{In a regular inscribed hexagon $a = R$, $r = \frac{1}{2}R\sqrt{3}$, $A=120°$, $C=60°$.} \ex{In a regular inscribed decagon \[ a = \frac{R(\sqrt{5}-1)}{2},\ r = \frac{1}{4} R \sqrt{10+2\sqrt{5}},\ A = 144°,\ C = 36°. \]} \scanpage{239.png}% \proposition{Problem.} \begin{proof}% \obs{To inscribe in a given circle a regular pentedecagon, or polygon of fifteen sides.} \figc{239aa475}{Let $Q$ be the given circle.} \prove[To inscribe ]{in $Q$ a regular pentedecagon.} \step{Draw $EH$ equal to the radius of the circle,}{} \step{and $EF$ equal to a side of the regular inscribed decagon.}{§~472} \step{Draw $FH$.}{} Then $FH$ is a side of the regular pentedecagon required. \step[\indent\textbf{Proof.}]{The arc $EH$ is $\frac{1}{6}$ of the circumference,}{§~469} \step{and the arc $EF$ is $\frac{1}{10}$ of the circumference.}{Const.} Hence, the arc $FH$ is $\frac{1}{6} - \frac{1}{10}$, or $\frac{1}{15}$, of the circumference. And the chord $FH$ is a side of a regular inscribed pentedecagon. By applying $FH$ fifteen times as a chord, we have the polygon required.~\hfill\qef \end{proof} \pp{\cor{By bisecting the arcs $FH$, $HA$, etc., a regular polygon of thirty sides may be inscribed; and, by continuing the process, regular polygons of sixty, one hundred twenty, etc., sides may be inscribed.}} \scanpage{240.png}% \proposition{Problem.} \begin{proof}% \obs{To inscribe in a given circle a regular polygon similar to a given regular polygon.} \figc{240aa477}{Let $ABC$ etc.\ be the given regular polygon, and $O'$ the centre of the given circle.} \prove[To inscribe ]{in the circle a regular polygon similar to $ABC$ etc.} \step{From $O$, the centre of the given polygon,}{} \step{draw $OD$ and $OC$.}{} \step{From $O'$, the centre of the given circle,}{} \step{draw $O'C'$ and $O'D'$,}{} \step{making the $\angle O'$ equal to the $\angle O$.}{} \step{Draw $C'D'$.}{} \step{Then $C'D'$ is a side of the regular polygon required.}{} \textbf{Proof.} Each polygon has as many sides as the $\angle O$, or $\angle O'$, is contained times in $4$~rt.~$\angle_s$. Therefore, the polygon $C'D'E'$ etc.\ is similar to the polygon $CDE$ etc.,~\hfill§~445 \pnote{(two regular polygons of the same number of sides are similar).} \hfill\qef \end{proof} \ex{The area of an inscribed regular octagon is equal to that of the rectangle whose sides are equal to the sides of the inscribed and the circumscribed squares. } \scanpage{241.png}% \proposition{Problem.} \begin{proof}% \obs{Given the side and the radius of a regular inscribed polygon, to find the side of the regular inscribed polygon of double the number of sides.} \figc{241aa478}{Let $AB$ be a side of the regular inscribed polygon.} \prove[To find ]{$AD$, a side of the regular inscribed polygon of double the number of sides.} \step{Denote the radius by $R$, and $AB$ by $a$.}{} \step{From $D$ draw $DH$ through the centre $O$, and draw $OA$, $AH$.}{} \step{$DH$ is $\perp$ to $AB$ at its middle point $C$.}{§~161} \eq[\indent In the rt.~$\triangle OCA$,]{$\overline{OC}^2$} {$= R^2 - \frac{1}{4}a^2$.}{§~372} \eq[\indent Therefore,]{$OC$}{$= \sqrt{R^2 - \frac{1}{4}a^2}$,}{} \eq[and]{$DC$}{$= R - \sqrt{R^2 - \frac{1}{4}a^2}$.}{} \step{The $\angle DAH$ is a rt.~$\angle$.}{§~290} In the rt.~$\triangle DAH$, \( \overline{AD}^2 = DH × DC \).~\hfill§~367 But $DH = 2R$, and \( DC = R - \sqrt{R^2 - \frac{1}{4}a^2} \). \eq{$\therefore AD$}{$= \sqrt{2R(R - \sqrt{R^2 - \frac{1}{4}a^2})}$}{} \eq{}{$= \sqrt{R(2R - \sqrt{4R^2 - a^2})}$.}{\qef} \end{proof} \pp{\cor{If $R=1$, $AD = \sqrt{2-\sqrt{4-a^2}}$.}} \scanpage{242.png}% \proposition{Problem.} \begin{proof}% \obs{To find the numerical value of the ratio of the circumference of a circle to its diameter.} \figc{242aa480}{Let $C$ be the circumference, when the radius is unity.} \prove[To find ]{the numerical value of $\pi$.} By §~458, $2\pi R = C$. \qquad $\therefore \pi = \frac{1}{2}C$ when $R = 1$. Let $S_6$ be the length of a side of a regular polygon of $6$~sides, $S_{12}$ of $12$~sides, and so on. If $R=1$, by §~469, $S_6=1$ and by §~479 we have \[ \begin{array}{lcc} \multicolumn{1}{c}{\text{\footnotesize Form of Computation.}} & \text{\footnotesize Length of Side.} & \text{\footnotesize Length of Perimeter.} \\ % S_{12} = \sqrt{2 - \sqrt{4 - 1^2}} & 0.51763809 & 6.21165708 \\ % S_{24} = \sqrt{2 - \sqrt{4 - (0.51763809)^2}} & 0.26105238 & 6.26525722 \\ % S_{48} = \sqrt{2 - \sqrt{4 - (0.26105238)^2}} & 0.13080626 & 6.27870041 \\ % S_{96} = \sqrt{2 - \sqrt{4 - (0.13080626)^2}} & 0.06543817 & 6.28206396 \\ % S_{192} = \sqrt{2 - \sqrt{4 - (0.06543817)^2}} & 0.03272346 & 6.28290510 \\ % S_{384} = \sqrt{2 - \sqrt{4 - (0.03272346)^2}} & 0.01636228 & 6.28311544 \\ % S_{768} = \sqrt{2 - \sqrt{4 - (0.01636228)^2}} & 0.00818121 & 6.28316941 \\ \end{array} \] $\therefore C = 6.28317$ approximately; that is, $\pi = 3.14159$ nearly. \hfill\qef \end{proof} \begin{point}% \textsc{Scholium.~}$\pi$ is incommensurable. We generally take \[ \pi = 3.1416, \text{ and } \frac{1}{\pi} = 0.31831. \] \end{point} \scanpage{243.png}% \section{MAXIMA AND MINIMA.} \begin{point}% \defn{Among geometrical magnitudes which satisfy given conditions, the \emph{greatest} is called the \indexbf{maximum}; and the \emph{smallest} is called the \indexbf{minimum}.} Thus, the diameter of a circle is the maximum among all chords; and the perpendicular is the minimum among all lines drawn to a given line from a given external point. \end{point} \pp{\defn{\indexbf{Isoperimetric} polygons are polygons which have equal \newline perimeters.}} \proposition{Theorem.} \begin{proof}% \obs{Of all triangles having two given sides, that in which these sides include a right angle is the maximum.} \figc{243aa484}{Let the triangles $ABC$ and $EBC$ have the sides $AB$ and $BC$ equal to $EB$ and $BC$, respectively; and let the angle $ABC$ be a right angle.} \proveq{$\triangle ABC$}{$> \triangle EBC$.} \step[\indent\textbf{Proof.}]{From $E$ draw the altitude $ED$.}{} The $\triangle_s ABC$ and $EBC$, having the same base, $BC$, are to each other as their altitudes $AB$ and $ED$.~\hfill§~405 \eq[\indent Now]{$EB$}{$> ED$.}{§~97} \eq[\indent But]{$EB$}{$= AB$.}{Hyp.} \eq{$\therefore AB$}{$> ED$.}{} \eq{$\therefore \triangle ABC$}{$> \triangle EBC$.}{§~405} \hfill\qed \end{proof} \scanpage{244.png}% \proposition{Theorem.} \begin{proof}% \obs{Of all isoperimetric triangles having the same base the isosceles triangle is the maximum.} \figc{244aa485}{Let the $\triangle_s ACB$ and $ADB$ have equal perimeters, and let $AC$ and $CB$ be equal, and $AD$ and $DB$ be unequal.} \prove {$\triangle ACB > \triangle ADB$.} \step [\indent Proof.] {Produce $AC$ to $H$, making $CH = AC$; and draw $HB$.} {} \step {Produce $HB$, take $DP$ equal to $DB$, and draw $AP$.} {} \step {Draw $CE$ and $DF \perp$ to $AB$, and $CK$ and $DM \parallel$ to $AB$.} {} \step {The $\angle ABH$ is a right $\angle$, for it may be inscribed in the semicircle whose centre is $C$ and radius $CA$.} {§~290} $ADP$ is not a straight line, for then the $\angle_s DBA$ and $DAB$ would be equal, being complements of the equal $\angle_s DBM$ and $DPM$, respectively; and $DA$ and $DB$ would be equal (§~147), which is contrary to the hypothesis. Hence, \step {$AP < AD + DP$, $\therefore < AD+DB$, $\therefore < AC+CB$, $\therefore < AH$.} {} \eq {$\therefore BH $} {$> BP$.} {§~102} \eq {$\therefore CE(=\frac{1}{2} BH) $} {$> DF (=\frac{1}{2}BP)$.} {Ax.~7} \eq [\indent Therefore,] {$\triangle ACB $} {$> \triangle ADB$.} {§~405} \step {} {\qed} \end {proof} \scanpage{245.png}% \proposition{Theorem.} \begin{proof}% \obs{Of all polygons with sides all given but one, the maximum can be inscribed in a semicircle which has the undetermined side for its diameter.} \figc{245aa486}{Let $ABCDE$ be the maximum of polygons with sides $AB$, $BC$, $CD$, $DE$, and the extremities $A$ and $E$ on the straight line $MN$.} \prove{$ABCDE$ can be inscribed in a semicircle.} \textbf{Proof.} From \emph{any} vertex, as $C$, draw $CA$ and $CE$. The $\triangle ACE$ must be the maximum of all $\triangle_s$ having the sides $CA$ and $CE$, and the third side on $MN$; otherwise by increasing or diminishing the $\angle ACE$, keeping the lengths of the sides $CA$ and $CE$ unchanged, but sliding the extremities $A$ and $E$ along the line $MN$, we could increase the $\triangle ACE$, while the rest of the polygon would remain unchanged; and therefore increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence, the $\triangle ACE$ is the maximum of $\triangle_s$ that have the sides $CA$ and $CE$. \step{Therefore, the $\angle ACE$ is a right angle.}{§~484} \step{Therefore, $C$ lies on the semicircumference.}{§~290} Hence, \emph{every} vertex lies on the circumference; that is, the maximum polygon can be inscribed in a semicircle having the undetermined side for a diameter.~\hfill\qed \end{proof} \scanpage{246.png}% \proposition{Theorem.} \begin{proof}% \obs{Of all polygons with given sides, that which can be inscribed in a circle is the maximum.} \figc{246aa487}{Let $ABCDE$ be a polygon inscribed in a circle, and $A'B'C'D'E'$ be a polygon, equilateral with respect to $ABCDE$, which cannot be inscribed in a circle.} \prove{that $ABCDE > A'B'C'D'E'$.} \step [\indent Proof.] {Draw the diameter $AH$, and draw $CH$ and $DH$.} {} \step {Upon $C'D'$ construct the $\triangle C'H'D' = \triangle CHD$, and draw $A'H'$.} {} Since, by hypothesis, a $\odot$ cannot pass through \emph{all} the vertices of $A'B'C'D'E'$, \emph{one} or \emph{both} of the parts $ABCH$, $AEDH$ must be greater than the corresponding part of $A'B'C'H'D'E'$.\hfill§~486 If either of these parts is \emph{not greater than} its corresponding part, it is equal to it,\hfill§~486 \pnote{(for $ABCH$ and $AEDH$ are the maxima of polygons that have sides equal to $AB$, $BC$, $CH$, and $AE$, $ED$, $DH$, respectively, and the remaining side undetermined).} \eq {$\therefore ABCHDE $} {$> A'B'C'H'D'E'$.} {Ax.~4} \step {Take away from the two figures the equal $\triangle_s CHD$ and $C'H'D'$.}{} \eq [\indent Then] {$ABCDE $} {$> A'B'C'D'E'$.} {Ax.~5} \step {} {\qed} \end {proof} \scanpage{247.png}% \proposition{Theorem.} \begin{proof}% \obs{Of isoperimetric polygons of the same number of sides, the maximum is equilateral.} \figc{247aa488}{Let $ABCD$ etc.\ be the maximum of isoperimetric polygons of any given number of sides.} \prove{$AB$, $BC$, $CD$, etc., are equal.} \step[\indent\textbf{Proof.}]{Draw $AC$.}{} The $\triangle ABC$ must be the maximum of all the $\triangle_s$ which are formed upon $AC$ with a perimeter equal to that of $\triangle ABC$. Otherwise a greater $\triangle AKC$ could be substituted for $\triangle ABC$, without changing the perimeter of the polygon. But this is inconsistent with the hypothesis that the polygon $ABCD$ etc.\ is the maximum polygon. \step{$\therefore$ the $\triangle ABC$ is isosceles.}{§~485} \step{$\therefore AB = BC$.}{} In like manner it may be proved that $BC=CD$, etc.~\hfill\qed \end{proof} \begin{point}% \cor{The maximum of isoperimetric polygons of the same number of sides is a regular polygon.} For the maximum polygon is equilateral (§~488), and can be inscribed in a circle (§~487), and is, therefore, regular.~\hfill§~430 \end{point} \scanpage{248.png}% \proposition{Theorem.} \begin{proof}% \obs{Of isoperimetric regular polygons, that which has the greatest number of sides is the maximum.} \figc{248aa490}{Let $Q$ be a regular polygon of three sides, and $Q'$ a regular polygon of four sides, and let the two polygons have equal perimeters.} \prove{$Q'$ is greater than $Q$.} \textbf{Proof.} Draw $CD$ from $C$ to any point in $AB$. Invert the $\triangle CDA$ and place it in the position $DCE$, letting $D$ fall at $C$, $C$ at $D$, and $A$ at $E$. The polygon $DBCE$ is an irregular polygon of four sides, which by construction has the same perimeter as $Q'$, and the same area as $Q$. Then the irregular polygon $DBCE$ of four sides is less than the isoperimetric regular polygon $Q'$ of four sides.~\hfill§~489 In like manner it may be shown that $Q'$ is less than an isoperimetric regular polygon of five sides, and so on.~\hfill\qed \end{proof} \ex{Of all equivalent parallelograms that have equal bases, the rectangle has the minimum perimeter.} \ex{Of all equivalent rectangles, the square has the minimum perimeter.} \ex{Of all triangles that have the same base and the same altitude, the isosceles has the minimum perimeter.} \ex{Of all triangles that can be inscribed in a given circle, the equilateral is the maximum and has the maximum perimeter.} \scanpage{249.png}% \proposition{Theorem.} \begin{proof}% \obs{Of regular polygons having a given area, that which has the greatest number of sides has the least perimeter.} \figc{249aa491}{Let $Q$ and $Q'$ be regular polygons having the same area, and let $Q'$ have the greater number of sides.} \proveq[\indent To prove ]{the perimeter of $Q$}{$>$ the perimeter of $Q'$.} \textbf{Proof.} Let $Q''$ be a regular polygon having the same perimeter as $Q'$, and the same number of sides as $Q$. \eq[\indent Then]{$Q'$}{$> Q''$}{§~490} \pnote{(of isoperimetric regular polygons, that which has the greatest number of sides is the maximum).} \eq[\indent But]{$Q$}{$\Bumpeq Q'$.}{Hyp.} \eq{$\therefore Q$}{$> Q''$.}{} \eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q''$.}{} \eq{But the perimeter of $Q'$}{$=$ the perimeter of $Q''$.}{Hyp.} \eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q'$.}{\qed} \end{proof} \ex{To inscribe in a semicircle the maximum rectangle.} \ex{Of all polygons of a given number of sides which may be inscribed in a given circle, that which is regular has the maximum area and the maximum perimeter.} \ex{Of all polygons of a given number of sides which may be circumscribed about a given circle, that which is regular has the minimum area and the minimum perimeter. } \scanpage{250.png}% \section[EXERCISES.]{THEOREMS.} \ex{Every equilateral polygon circumscribed about a circle is regular if it has an \emph{odd} number of sides.} \ex{Every equiangular polygon inscribed in a circle is regular if it has an \emph{odd} number of sides.} \ex{Every equiangular polygon circumscribed about a circle is regular.} \ex{The side of a circumscribed equilateral triangle is equal to twice the side of the similar inscribed triangle.} \ex{The apothem of an inscribed regular hexagon is equal to half the side of the inscribed equilateral triangle.} \ex{The area of an inscribed regular hexagon is three fourths of the area of the circumscribed regular hexagon.} \ex{The area of an inscribed regular hexagon is the mean proportional between the areas of the inscribed and the circumscribed equilateral triangles.} \ex{The square of the side of an inscribed equilateral triangle is equal to three times the square of a side of the inscribed regular hexagon.} \ex{The area of an inscribed equilateral triangle is equal to half the area of the inscribed regular hexagon.} \ex{The square of the side of an inscribed equilateral triangle is equal to the sum of the squares of the sides of the inscribed square and of the inscribed regular hexagon.} \ex{The square of the side of an inscribed regular pentagon is equal to the sum of the squares of the radius of the circle and the side of the inscribed regular decagon.} \exheader{If $R$ denotes the radius of a circle, and $a$ one side of an inscribed regular polygon, show that:} \ex{In a regular pentagon, $a = \frac{1}{2} R \sqrt{10 - 2 \sqrt{5}}$.} \ex{In a regular octagon, $a = R \sqrt{2 - \sqrt{2}}$.} \ex{In a regular dodecagon, $a = R \sqrt{2 - \sqrt{3}}$.} \ex{If two diagonals of a regular pentagon intersect, the longer segment of each is equal to a side of the pentagon.} \scanpage{251.png}% \ex{The apothem of an inscribed regular pentagon is equal to half the sum of the radius of the circle and the side of the inscribed regular decagon.} \ex{The side of an inscribed regular pentagon is equal to the hypotenuse of the right triangle which has for legs the radius of the circle and the side of the inscribed regular decagon.} \ex{The radius of an inscribed regular polygon is the mean proportional between its apothem and the radius of the similar circumscribed regular polygon.} \ex{If squares are constructed outwardly upon the six sides of a regular hexagon, the exterior vertices of these squares are the vertices of a regular dodecagon.} \ex{If the alternate vertices of a regular hexagon are joined by straight lines, show that another regular hexagon is thereby formed. Find the ratio of the areas of these two hexagons.} \ex{If on the legs of a right triangle as diameters semicircles are described external to the triangle, and from the whole figure a semicircle on the hypotenuse is subtracted, the remaining figure is equivalent to the given right triangle.} \ex{The star-shaped polygon, formed by producing the sides of a regular hexagon, is equivalent to twice the given hexagon.} \ex{The sum of the perpendiculars drawn to the sides of a regular polygon from any point within the polygon is equal to the apothem multiplied by the number of sides.} \ex{If two chords of a circle are perpendicular to each other, the sum of the four circles described on the four segments as diameters is equivalent to the given circle.} \ex{If the diameter of a circle is divided into any two segments, and upon these segments as diameters semicircumferences are described upon opposite sides of the diameter, these semicircumferences divide the circle into two parts which have the same ratio as the two segments of the diameter.} \ex{The diagonals that join any vertex of a regular polygon to all the vertices not adjacent divide the angle at that vertex into as many equal parts less two as the polygon has sides.} \scanpage{252.png}% \subsection{PROBLEMS OF CONSTRUCTION.} \ex{To circumscribe an equilateral triangle about a given circle.} \ex{To circumscribe a square about a given circle.} \ex{To circumscribe a regular hexagon about a given circle.} \ex{To circumscribe a regular octagon about a given circle.} \ex{To circumscribe a regular pentagon about a given circle.} \ex{To draw through a given point a line so as to divide a given circumference into two parts having the ratio $3:7$.} \ex{To construct a circumference equal to the sum of two given circumferences.} \ex{To construct a circumference equal to the difference of two given circumferences.} \ex{To construct a circle equivalent to the sum of two given circles.} \ex{To construct a circle equivalent to the difference of two given circles.} \ex{To construct a circle equivalent to three times a given circle.} \ex{To construct a circle equivalent to three fourths of a given circle.} \ex{To construct a circle whose ratio to a given circle shall be equal to the given ratio $m:n$.} \ex{To divide a given circle by a concentric circumference into two equivalent parts.} \ex{To divide a given circle by concentric circumferences into five equivalent parts.} \ex{To construct an angle of~$18°$; of~$36°$; of~$9°$.} \ex{To construct an angle of$12°$; of~$24°$; of~$6°$.} \exheader{To construct with a side of a given length:} \ex{An equilateral triangle.} \ex{A square.} \ex{A regular hexagon.} \ex{A regular octagon.} \ex{A regular pentagon.} \ex{A regular decagon.} \ex{A regular dodecagon.} \ex{A regular pentedecagon.} \scanpage{253.png}% \subsection{PROBLEMS OF COMPUTATION.} \ex{Find the area of a circle whose radius is $12$ inches.} \ex{Find the circumference and the area of a circle whose diameter is $8$ feet.} \ex{A regular pentagon is inscribed in a circle whose radius is $R$. If the length of a side is $a$, find the apothem.} \ex{A regular polygon is inscribed in a circle whose radius is $R$. If the length of a side is $a$, show that the apothem is $\frac{1}{2} \sqrt{R^2 - a^2}$.} \ex{Find the area of a regular decagon inscribed in a circle whose radius is $16$ inches.} \ex{Find the side of a regular dodecagon inscribed in a circle whose radius is $20$ inches.} \ex{Find the perimeter of a regular pentagon inscribed in a circle whose radius is $25$ feet.} \ex{The length of each side of a park in the shape of a regular decagon is $100$ yards. Find the area of the park.} \ex{Find the cost, at $\$2$ per yard, of building a wall around a cemetery in the shape of a regular hexagon, that contains $16,627.84$ square yards.} \ex{The side of an inscribed regular polygon of $n$ sides is $16$ feet. Find the side of an inscribed regular polygon of $2n$ sides.} \ex{If the radius of a circle is $R$, and the side of an inscribed regular polygon is $a$, show that the side of the similar circumscribed regular polygon is $\dfrac{2aR}{\sqrt{4R^2-a^2}}$.} \ex{What is the width of the circular ring between two concentric circumferences whose lengths are $650$ feet and $425$ feet?} \ex{Find the angle subtended at the centre by an arc $5$ feet $10$ inches long, if the radius of the circle is $9$ feet $4$ inches.} \ex{The chord of a segment is $10$ feet, and the radius of the circle is $16$ feet. Find the area of the segment.} \ex{Find the area of a sector, if the angle at the centre is $20°$, and the radius of the circle is $20$ inches.} \scanpage{254.png}% \ex{The chord of half an arc is $12$~feet, and the radius of the circle is $18$~feet. Find the height of the segment subtended by the whole arc.} \ex{Find the side of a square which is equivalent to a circle whose diameter is $35$~feet.} \ex{The diameter of a circle is $15$~feet. Find the diameter of a circle twice as large. Three times as large.} \ex{Find the radii of the concentric circumferences that divide a circle $11$~inches in diameter into five equivalent parts.} \ex{The perimeter of a regular hexagon is $840$ feet, and that of a regular octagon is the same. By how many square feet is the octagon larger than the hexagon?} \ex{The diameter of a bicycle wheel is $28$~inches. How many revolutions does the wheel make in going $10$~miles?} \ex{Find the diameter of a carriage wheel that makes $264$~revolutions in going half a mile.} \ex{The sides of three regular octagons are $6$~feet, $7$~feet, $8$~feet, respectively. Find the side of a regular octagon equivalent to the sum of the three given octagons.} \ex{A circular pond $100$ yards in diameter is surrounded by a walk $10$~feet wide. Find the area of the walk.} \ex{The span (chord) of a bridge in the form of a circular arc is $120$~feet, and the highest point of the arch is $15$~feet above the piers. Find the radius of the arc.} \ex{Three equal circles are described each tangent to the other two. If the common radius is~$R$, find the area contained between the circles.} \ex{Given $p$, $P$, the perimeters of regular polygons of $n$~sides inscribed in and circumscribed about a given circle. Find $p'$, $P'$, the perimeters of regular polygons of $2n$~sides inscribed in and circumscribed about the given circle.} \ex{Given the radius $R$, and the apothem $r$ of an inscribed regular polygon of $n$ sides. Find the radius $R'$ and the apothem $r'$ of an isoperimetrical regular polygon of $2n$ sides.} \scanpage{255.png}% \subsection{MISCELLANEOUS EXERCISES.} \subsection{THEOREMS.} \ex{If two adjacent angles of a quadrilateral are right angles, the bisectors of the other two angles are perpendicular.} \ex{If two opposite angles of a quadrilateral are right angles, the bisectors of the other two angles are parallel.} \ex{The two lines that join the middle points of the opposite sides of a quadrilateral bisect each other.} \ex{The line that joins the feet of the perpendiculars dropped from the extremities of the base of an isosceles triangle to the opposite sides is parallel to the base.} \ex{If $AD$ bisects the angle $A$ of a triangle $ABC$, and $BD$ bisects the exterior angle $CBF$, then angle $ADB$ equals one half angle $ACB$.} \ex{The sum of the acute angles at the vertices of a pentagram\label{pentagram} (five-pointed star) is equal to two right angles.} \begin{proofex}% The altitudes $AD$, $BE$, $CF$ of the triangle $ABC$ bisect the angles of the triangle $DEF$. Circles with $AB$, $BC$, $AC$ as diameters will pass through $E$ and $D$, $E$ and $F$, $D$ and $F$, respectively. \end{proofex} \ex{The segments of any straight line intercepted between the circumferences of two concentric circles are equal.} \ex{If a circle is circumscribed about any triangle, the feet of the perpendiculars dropped from any point in the circumference to the sides of the triangle lie in one straight line.} \ex{Two circles are tangent internally at $P$, and a chord $AB$ of the larger circle touches the smaller circle at $C$. Prove that $PC$ bisects the angle $APB$.} \ex{The diagonals of a trapezoid divide each other into segments which are proportional.} \ex{If through a point $P$ in the circumference of a circle two chords are drawn, the chords and the segments between $P$ and a chord parallel to the tangent at $P$ are reciprocally proportional.} \scanpage{256.png}% \ex{The perpendiculars from two vertices of a triangle upon the opposite sides divide each other into segments reciprocally proportional.} \ex{The perpendicular from any point of a circumference upon a chord is the mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord.} \ex{In an isosceles right triangle either leg is the mean proportional between the hypotenuse and the perpendicular upon it from the vertex of the right angle.} \ex{If two circles intersect in the points $A$ and $B$, and through $A$ any secant $CAD$ is drawn limited by the circumferences at $C$ and $D$, the straight lines $BC$, $BD$ are to each other as the diameters of the circles.} \ex{The area of a triangle is equal to half the product of its perimeter by the radius of the inscribed circle.} \ex{The perimeter of a triangle is to one side as the perpendicular from the opposite vertex is to the radius of the inscribed circle.} \begin{proofex}% If three straight lines $AA'$, $BB'$, $CC'$, drawn from the vertices of a triangle $ABC$ to the opposite sides, pass through a common point $O$ within the triangle, then \step{\( \dfrac{OA'}{AA'} + \dfrac{OB'}{BB'} + \dfrac{OC'}{CC'} = 1 \).}{} \end{proofex} \ex{$ABC$ is a triangle, $M$ the middle point of $AB$, $P$ any point in $AB$ between $A$ and $M$. If $MD$ is drawn parallel to $PC$, meeting $BC$ at $D$, the triangle $BPD$ is equivalent to half the triangle $ABC$.} \ex{Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio.} \ex{If all the diagonals of a regular pentagon are drawn, another regular pentagon is thereby formed.} \ex{The area of an inscribed regular dodecagon is equal to three times the square of the radius.} \ex{The area of a square inscribed in a semicircle is equal to two fifths the area of the square inscribed in the circle.} \ex{The area of a circle is greater than the area of any polygon of equal perimeter.} \ex{The circumference of a circle is less than the perimeter of any polygon of equal area.} \scanpage{257.png}% \subsection{PROBLEMS OF LOCI.} \ex{Find the locus of the centre of the circle inscribed in a triangle that has a given base and a given angle at the vertex.} \ex{Find the locus of the intersection of the altitudes of a triangle that has a given base and a given angle at the vertex.} \ex{Find the locus of the extremity of a tangent to a given circle, if the length of the tangent is equal to a given line.} \ex{Find the locus of a point, tangents drawn from which to a given circle form a given angle.} \ex{Find the locus of the middle point of a line drawn from a given point to a given straight line.} \ex{Find the locus of the vertex of a triangle that has a given base and a given altitude.} \ex{Find the locus of a point the sum of whose distances from two given parallel lines is equal to a given length.} \ex{Find the locus of a point the difference of whose distances from two given parallel lines is equal to a given length.} \ex{Find the locus of a point the sum of whose distances from two given intersecting lines is equal to a given length.} \ex{Find the locus of a point the difference of whose distances from two given intersecting lines is equal to a given length.} \ex{Find the locus of a point whose distances from two given points are in the given ratio $m:n$.} \ex{Find the locus of a point whose distances from two given parallel lines are in the given ratio $m:n$.} \ex{Find the locus of a point whose distances from two given intersecting lines are in the given ratio $m:n$.} \ex{Find the locus of a point the sum of the squares of whose distances from two given points is constant.} \ex{Find the locus of a point the difference of the squares of whose distances from two given points is constant.} \ex{Find the locus of the vertex of a triangle that has a given base and the other two sides in the given ratio $m:n$. } \scanpage{258.png}% \subsection{PROBLEMS OF CONSTRUCTION.} \ex{To divide a given trapezoid into two equivalent parts by a line parallel to the bases.} \ex{To divide a given trapezoid into two equivalent parts by a line through a given point in one of the bases.} \ex{To construct a regular pentagon, given one of the diagonals.} \ex{To divide a given straight line into two segments such that their product shall be the maximum.} \ex{To find a point in a semicircumference such that the sum of its distances from the extremities of the diameter shall be the maximum.} \ex{To draw a common secant to two given circles exterior to each other such that the intercepted chords shall have the given lengths $a$, $b$.} \ex{To draw through one of the points of intersection of two intersecting circles a common secant which shall have a given length.} \ex{To construct an isosceles triangle, given the altitude and one of the equal base angles.} \ex{To construct an equilateral triangle, given the altitude.} \ex{To construct a right triangle, given the radius of the inscribed circle and the difference of the acute angles.} \ex{To construct an equilateral triangle so that its vertices shall lie in three given parallel lines.} \ex{To draw a line from a given point to a given straight line which shall be to the perpendicular from the given point as $m : n$.} \ex{To find a point within a given triangle such that the perpendiculars from the point to the three sides shall be as the numbers $m$, $n$, $p$.} \ex{To draw a straight line equidistant from three given points.} \ex{To draw a tangent to a given circle such that the segment intercepted between the point of contact and a given straight line shall have a given length.} \ex{To inscribe a straight line of a given length between two given circumferences and parallel to a given straight line. } \scanpage{259.png}% \ex{To draw through a given point a straight line so that its distances from two other given points shall be in a given ratio.} \ex{To construct a square equivalent to the sum of a given triangle and a given parallelogram.} \ex{To construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon.} \ex{To construct a pentagon similar to a given pentagon and equivalent to a given trapezoid.} \ex{To find a point whose distances from three given straight lines shall be as the numbers $m$, $n$, $p$.} \ex{Given an angle and two points $P$ and $P'$ between the sides of the angle. To find the shortest path from $P$ to $P'$ that shall touch both sides of the angle.} \ex{To construct a triangle, given its angles and its area.} \ex{To transform a given triangle into a triangle similar to another given triangle.} \ex{Given three points $A$, $B$, $C$. To find a fourth point $P$ such that the areas of the triangles $APB$, $APC$, $BPC$ shall be equal.} \ex{To construct a triangle, given its base, the ratio of the other sides, and the angle included by them.} \ex{To divide a given circle into $n$ equivalent parts by concentric circumferences.} \ex{In a given equilateral triangle to inscribe three equal circles tangent to each other, each circle tangent to two sides of the triangle.} \ex{Given an angle and a point $P$ between the sides of the angle. To draw through $P$ a straight line that shall form with the sides of the angle a triangle with the perimeter equal to a given length $a$.} \ex{In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to two sides of the square.} \ex{In a given square to inscribe four equal circles, so that each circle shall be tangent to two of the others and also tangent to one side of the square.} \scanpage{260.png}% \chapter{TABLE OF FORMULAS.} \markboth{\Headings{PLANE GEOMETRY.}}{\Headings{TABLE OF FORMULAS.}}% \subsection{PLANE FIGURES.} \subsection{NOTATION.} \begin{tabular}{r@{~}c@{~}l} $P$ &=& perimeter. \\ $h$ &=& altitude. \\ $b$ &=& lower base. \\ $b'$ &=& upper base. \\ $R$ &=& radius of circle. \\ $D$ &=& diameter of circle. \\ $C$ &=& circumference of circle. \\ $r$ &=& apothem of regular polygon. \\ $a$, $b$, $c$ &=& sides of triangle. \\ $s$ &=& \( \frac{1}{2}(a+b+c) \). \\ $p$ &=& perpendicular of triangle. \\ $m,n$ &=& segments of third side of triangle adjacent to \\ && sides $b$ and $a$, respectively. \\ $S$ &=& area. \\ $\pi$ &=& 3.1416. \end{tabular} \newpage \subsection{FORMULAS.} \noindent\begin{supertabular}{lr@{~}c@{~}l@{\qquad}r} \multicolumn{5}{l}{\hspace{-2ex}\textbf{Line Values.}} \\ \multicolumn{5}{r}{\tiny PAGE}\\ Right triangle, & $b^2$ &=& $c × m$; $a^2 = c × n$ & \pageref{160} \\ & $p^2$ &=& $m × n$ & \pageref{160} \\ & $b^2:a^2$ &=& $m:n$ & \pageref{161} \\ & $b^2:c^2$ &::& $m:c$ & \pageref{161} \\ & $a^2+b^2$ &=& $c^2$ & \pageref{162} \\ \scanpage{261.png}% Any triangle, & $a^2$ &=& $b^2+c^2 \pm 2c × m$ & \llap{\pageref{163},}\pageref{164} \\ \multicolumn{4}{l}{Altitude of triangle on side $a$,} \\ & $h$ &=& \( \displaystyle \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} \) & \pageref{formtrialtitude} \\ \multicolumn{4}{l}{Median of triangle on side $a$,} \\ & $m$ &=& \( \frac{1}{2} \sqrt{2(b^2+c^2) - a^2} \) & \pageref{formtrimedian} \\ \multicolumn{4}{l}{Bisector of triangle on side $a$,} \\ & $t$ &=& \( \displaystyle \frac{2}{b+c} \sqrt{bcs(s-a)} \) & \pageref{formtribisector} \\ \multicolumn{4}{l}{Radius of circumscribed circle,} \\ & $R$ &=& \( \displaystyle \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}} \) & \pageref{formradcircum} \\ Circumference of circle, & $C$ &=& $2\pi R$ & \pageref{formcircum} \\ \qquad\DittoMark\qquad\qquad\DittoMark & $C$ &=& $\pi D$ & \pageref{formcircum} \\ \multicolumn{5}{l}{\hspace{-2ex}\textbf{Areas.}} \\ Rectangle, & $S$ &=& $b × h$ & \pageref{formarearect} \\ Square, & $S$ &=& $b^2$ & \pageref{formarearect} \\ Parallelogram, & $S$ &=& $b × h$ & \pageref{formareapar} \\ Triangle, & $S$ &=& $\frac{1}{2}b × h$ & \pageref{formareatri} \\ \qquad\DittoMark & $S$ &=& $\sqrt{s(s-a)(s-b)(s-c)}$ & \pageref{formareatri2} \\ \qquad\DittoMark & $S$ &=& \( \displaystyle \frac{abc}{4R} \) & \pageref{formareatri3} \\ Equilateral triangle, & $S$ &=& \( \displaystyle \frac{a^2}{4}\sqrt{3} \) & \pageref{formareaequitri} \\ Trapezoid, & $S$ &=& $\frac{1}{2}h(b+b')$ & \pageref{formareatrap} \\ Regular polygon, & $S$ &=& $\frac{1}{2}r × P$ & \pageref{formareapoly} \\ Circle, & $S$ &=& $\frac{1}{2}R × C$ & \pageref{formareacircle} \\ \qquad\DittoMark & $S$ &=& $\pi R^2$ & \pageref{formareacircle} \\ Sector, & $S$ &=& $\frac{1}{2}R × \arc$ & \pageref{formareasector} \\ \end{supertabular} \scanpage{262.png}% \twocolumn \chapter{INDEX.} \markboth{INDEX.}{INDEX.} \small \noindent\begin{supertabular}{lr} \multicolumn{2}{r}{\tiny PAGE} \\ \tablehead{\multicolumn{2}{r}{\tiny PAGE} \\}% Abbreviations & \pageref{abbr} \\ Alternation & \pageref{alternation} \\ Altitude of parallelogram & \pageref{altpar} \\ \Ditto of trapezoid & \pageref{alttrap} \\ \Ditto of triangle & \pageref{alttri} \\ Analysis & \pageref{analysis} \\ Angle & \pageref{angle} \\ \Ditto acute & \pageref{acute} \\ \multicolumn{2}{l}{\Ditto at centre of}\\ \quad\quad\quad regular polygon & \pageref{anglecentreregpoly} \\ \Ditto central & \pageref{central} \\ \Ditto exterior of triangle & \pageref{exteriortri} \\ \Ditto inscribed in circle & \pageref{inscribedcirc} \\ \Ditto inscribed in segment & \pageref{inscribedseg} \\ \Ditto oblique & \pageref{oblique} \\ \Ditto obtuse & \pageref{obtuse} \\ \Ditto reflex & \pageref{reflex} \\ \Ditto right & \pageref{right} \\ \Ditto salient & \pageref{salient} \\ \Ditto straight & \pageref{straight} \\ \Ditto vertical & \pageref{vertical angle} \\ Angles, adjacent & \pageref{adjacent1},\pageref{adjacent2} \\ \Ditto alternate-exterior & \pageref{altext} \\ \Ditto alternate-interior & \pageref{altint} \\ \Ditto complementary & \pageref{complementary} \\ \Ditto conjugate & \pageref{conjugate angles} \\ \Ditto exterior & \pageref{exterior} \\ \Ditto exterior-interior & \pageref{extint} \\ \Ditto interior & \pageref{interior} \\ \Ditto supplementary & \pageref{supplementary} \\ \Ditto supplementary-adjacent & \pageref{suppladj} \\ \Ditto vertical & \pageref{vertical angles} \\ Antecedents & \pageref{antecedents} \\ Apothem & \pageref{apothem} \\ Arc & \pageref{arc} \\ Area & \pageref{area} \\ Axiom & \pageref{axiom} \\ \Ditto of parallel lines & \pageref{axiomparallel} \\ Axioms of straight lines & \pageref{axiomstraight} \\ \Ditto general & \pageref{generalaxioms} \\ Axis of symmetry & \pageref{axissym} \\ \\ \textbf{B}ase of isosceles triangle & \pageref{baseiso} \\ \Ditto of parallelogram & \pageref{basepar} \\ \Ditto of triangle & \pageref{basetri} \\ Bases of trapezoid & \pageref{basetrap} \\ Bisector & \pageref{bisector} \\ \\ \textbf{C}entre of circle & \pageref{centrecirc} \\ \Ditto of regular polygon & \pageref{centrepoly} \\ \Ditto of symmetry & \pageref{centresym} \\ Chord & \pageref{chord} \\ Circle & \pageref{circle} \\ \Ditto circumscribed & \pageref{circcircumscribed} \\ \Ditto inscribed & \pageref{circinscribed} \\ Circles, concentric & \pageref{concentric} \\ \Ditto escribed & \pageref{escribed} \\ Circum-centre of triangle & \pageref{circum-centre} \\ Circumference & \pageref{circumference} \\ Commensurable & \pageref{commensurable} \\ Complement & \pageref{complement} \\ Composition & \pageref{composition} \\ Conclusion & \pageref{conclusion} \\ Concurrent lines & \pageref{concurrent} \\ Congruent figures & \pageref{congruent} \\ Consequents & \pageref{consequents} \\ Constant & \pageref{constant} \\ Construction & \pageref{construction} \\ Continued proportion & \pageref{continuedprop} \\ Continuity, Principle of & \pageref{princcont} \\ Contradictory of a theorem & \pageref{contradictory} \\ Converse of a theorem & \pageref{converse1},\pageref{converse2} \\ Convex curve & \pageref{convexcurve} \\ Curved surface & \pageref{curvedsurf} \\ \\ \textbf{D}ecagon & \pageref{decagon} \\ Diagonal & \pageref{diagonal1},\pageref{diagonal2} \\ Diameter & \pageref{diameter} \\ Dimensions & \pageref{dimensions} \\ Distance & \pageref{distance1},\pageref{distance2} \\ Division & \pageref{division} \\ Dodecagon & \pageref{dodecagon} \\ Duality, Principle of & \pageref{princduality} \\ \\ \textbf{E}qual figures & \pageref{equal} \\ Equimultiples & \pageref{equimultiples} \\ Equivalent figures & \pageref{equivalent1},\pageref{equivalent2} \\ Ex-centres of triangle & \pageref{ex-centres} \\ Extreme and mean ratio & \pageref{extrememean} \\ Extremes & \pageref{extremes} \\ \\ \textbf{F}igure, curvilinear & \pageref{curvilinear} \\ \Ditto geometrical & \pageref{geometrical figure} \\ \Ditto plane & \pageref{plane figure} \\ \Ditto rectilinear & \pageref{rectilinear} \\ Foot of perpendicular & \pageref{foot} \\ Fourth proportional & \pageref{fourth} \\ \\ \textbf{G}eometrical solid & \pageref{geometrical1},\pageref{geometrical2} \\ Geometry & \pageref{Geometry} \\ Geometry, Plane & \pageref{Plane Geometry} \\ \Ditto Solid & \pageref{Solid Geometry} \\ \\ \textbf{H}armonic division & \pageref{divided harmonically} \\ Heptagon & \pageref{heptagon} \\ Hexagon & \pageref{hexagon} \\ Homologous angles & \pageref{homologous angles},\pageref{homangles} \\ \Ditto lines & \pageref{Homologous lines} \\ \Ditto sides & \pageref{homologous sides},\pageref{homsides} \\ Hypotenuse & \pageref{hypotenuse} \\ Hypothesis & \pageref{hypothesis} \\ \\ \textbf{I}n-centre of triangle & \pageref{in-centre} \\ Incommensurable ratio & \pageref{incommensurable ratio} \\ Intersection & \pageref{intersection} \\ Inversion & \pageref{inversion} \\ Isoperimetric figures & \pageref{Isoperimetric} \\ \\ \textbf{L}egs of right triangle & \pageref{legs} \\ \Ditto of trapezoid & \pageref{legstrap} \\ Limit & \pageref{limit} \\ Line & \pageref{line},\pageref{line2},\pageref{line3} \\ \Ditto curved & \pageref{curved line} \\ \Ditto of centres & \pageref{line of centres} \\ \Ditto straight & \pageref{straight line} \\ Lines, oblique & \pageref{oblique lines} \\ \Ditto parallel & \pageref{parallel lines} \\ \Ditto perpendicular & \pageref{perpendicular} \\ \\ \textbf{M}ajor arc & \pageref{major} \\ Maximum & \pageref{maximum} \\ Mean proportional & \pageref{mean proportional} \\ Means & \pageref{means} \\ Median of trapezoid & \pageref{mediantrap} \\ Minimum & \pageref{minimum} \\ Minor arc & \pageref{minor} \\ \\ \textbf{N}egative quantities & \pageref{negative} \\ Numerical measure & \pageref{numerical measure} \\ \\ \textbf{O}ctagon & \pageref{octagon} \\ Opposite of a theorem & \pageref{opposite} \\ Origin & \pageref{origin} \\ \\ \textbf{P}arallel lines & \pageref{parallel lines} \\ Parallelogram & \pageref{parallelogram} \\ Pentagon & \pageref{pentagon} \\ Pentagram & \pageref{pentagram} \\ Perigon & \pageref{perigon} \\ Perimeter & \pageref{perimeter},\pageref{perimeter2} \\ Perpendicular bisector & \pageref{perpbisector} \\ Perpendicular lines & \pageref{perpendicular} \\ Pi ($\pi$) & \pageref{pi} \\ Plane & \pageref{planes},\pageref{plane} \\ Point & \pageref{point},\pageref{point2} \\ \Ditto of contact & \pageref{point of contact} \\ \Ditto of tangency & \pageref{point of tangency} \\ Polygon & \pageref{polygon} \\ \Ditto angles of & \pageref{polyangles} \\ \Ditto circumscribed & \pageref{polycircumscribed} \\ \Ditto concave & \pageref{concave polygon} \\ \Ditto convex & \pageref{convex polygon} \\ \Ditto equiangular & \pageref{equiangular polygon} \\ \Ditto equilateral & \pageref{equilateral polygon} \\ \Ditto inscribed & \pageref{polyinscribed} \\ \Ditto regular & \pageref{regular polygon} \\ Polygons mut.\ equiangular & \pageref{mutually equiangular} \\ \Ditto mutually equilateral & \pageref{mutually equilateral} \\ Positive quantities & \pageref{positive} \\ Postulate & \pageref{postulate} \\ Projection & \pageref{projection} \\ Proof & \pageref{proof} \\ Proportion & \pageref{proportion} \\ Proposition & \pageref{proposition} \\ \\ \textbf{Q}uadrant & \pageref{quadrant} \\ Quadrilateral & \pageref{quadrilateral},\pageref{quadrilateral2} \\ Radius of regular polygon & \pageref{polyradius} \\ Ratio & \pageref{ratio} \\ Ratio of similitude & \pageref{ratio of similitude} \\ Reciprocity, Principle of & \pageref{princreciprocity} \\ Rectangle & \pageref{rectangle} \\ Rhomboid & \pageref{rhomboid} \\ Rhombus & \pageref{rhombus} \\ \\ \textbf{S}cholium & \pageref{scholium} \\ Secant & \pageref{secant},\pageref{secant2} \\ Sector & \pageref{sector} \\ Segment of circle & \pageref{segment} \\ \Ditto of line & \pageref{lineseg} \\ Semicircle & \pageref{semicircle} \\ Semicircumference & \pageref{semicircumference} \\ Sides of an angle & \pageref{anglesides} \\ \Ditto of polygon & \pageref{polysides} \\ \Ditto of triangle & \pageref{trisides} \\ Similar arcs & \pageref{similar arcs} \\ \Ditto figures & \pageref{similar} \\ \Ditto polygons & \pageref{Similar polygons} \\ \Ditto sectors & \pageref{similar sectors} \\ \Ditto segments & \pageref{similar segments} \\ \Ditto triangles & \pageref{similar triangles} \\% not sure about this ref Square & \pageref{square} \\ Superposition & \pageref{superposition} \\ Supplement & \pageref{supplement} \\ Surface & \pageref{surface},\pageref{surface2},\pageref{surface3} \\ Symbols & \pageref{symbols} \\ Symmetry & \pageref{symmetry} \\ \\ \textbf{T}angent & \pageref{tangent},\pageref{tangent2} \\ \Ditto common external & \pageref{common external tangent} \\ \Ditto common internal & \pageref{common internal tangent} \\ Terms of a proportion & \pageref{terms} \\ Theorem & \pageref{theorem} \\ Third proportional & \pageref{third} \\ Transversal & \pageref{transversal} \\ Trapezium & \pageref{trapezium} \\ Trapezoid & \pageref{trapezoid} \\ \Ditto isosceles & \pageref{isosceles trapezoid} \\ Triangle & \pageref{triangle},\pageref{triangle2} \\ \Ditto equiangular & \pageref{equiangular triangle} \\ \Ditto equilateral & \pageref{equilateral triangle} \\ \Ditto isosceles & \pageref{isosceles triangle} \\ \Ditto obtuse & \pageref{obtuse triangle} \\ \Ditto right & \pageref{right triangle} \\ \Ditto scalene & \pageref{scalene triangle} \\ \Ditto altitudes of & \pageref{alttri} \\ \Ditto angles of & \pageref{anglestri} \\ \Ditto bisectors of & \pageref{tribisectors} \\ \Ditto medians of & \pageref{trimedians} \\ \Ditto vertices of & \pageref{trivertices} \\ \\ \textbf{V}ariable & \pageref{variable} \\ Vertex of angle & \pageref{vertex} \\ \Ditto of triangle & \pageref{trivertex} \\ Vertices of polygon & \pageref{polyvertices} \\ \end{supertabular} \onecolumn %%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% \cleardoublepage \backmatter \phantomsection \pdfbookmark[-1]{BACK MATTER.}{BACK MATTER} \phantomsection \pdfbookmark[0]{PG LICENSE.}{LICENSE} \fancyhead[C]{\Headings{LICENSE.}} \begin{PGtext} End of Project Gutenberg's Plane Geometry, by George Albert Wentworth *** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY *** ***** This file should be named 33063-pdf.pdf or 33063-pdf.zip ***** This and all associated files of various formats will be found in: http://book.klll.cc/3/3/0/6/33063/ Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy and the Online Distributed Proofreading Team at http://www.pgdp.net Updated editions will replace the previous one--the old editions will be renamed. Creating the works from public domain print editions means that no one owns a United States copyright in these works, so the Foundation (and you!) can copy and distribute it in the United States without permission and without paying copyright royalties. Special rules, set forth in the General Terms of Use part of this license, apply to copying and distributing Project Gutenberg-tm electronic works to protect the PROJECT GUTENBERG-tm concept and trademark. Project Gutenberg is a registered trademark, and may not be used if you charge for the eBooks, unless you receive specific permission. If you do not charge anything for copies of this eBook, complying with the rules is very easy. You may use this eBook for nearly any purpose such as creation of derivative works, reports, performances and research. They may be modified and printed and given away--you may do practically ANYTHING with public domain eBooks. Redistribution is subject to the trademark license, especially commercial redistribution. *** START: FULL LICENSE *** THE FULL PROJECT GUTENBERG LICENSE PLEASE READ THIS BEFORE YOU DISTRIBUTE OR USE THIS WORK To protect the Project Gutenberg-tm mission of promoting the free distribution of electronic works, by using or distributing this work (or any other work associated in any way with the phrase "Project Gutenberg"), you agree to comply with all the terms of the Full Project Gutenberg-tm License (available with this file or online at http://gutenberg.org/license). Section 1. General Terms of Use and Redistributing Project Gutenberg-tm electronic works 1.A. By reading or using any part of this Project Gutenberg-tm electronic work, you indicate that you have read, understand, agree to and accept all the terms of this license and intellectual property (trademark/copyright) agreement. If you do not agree to abide by all the terms of this agreement, you must cease using and return or destroy all copies of Project Gutenberg-tm electronic works in your possession. If you paid a fee for obtaining a copy of or access to a Project Gutenberg-tm electronic work and you do not agree to be bound by the terms of this agreement, you may obtain a refund from the person or entity to whom you paid the fee as set forth in paragraph 1.E.8. 1.B. "Project Gutenberg" is a registered trademark. It may only be used on or associated in any way with an electronic work by people who agree to be bound by the terms of this agreement. There are a few things that you can do with most Project Gutenberg-tm electronic works even without complying with the full terms of this agreement. See paragraph 1.C below. There are a lot of things you can do with Project Gutenberg-tm electronic works if you follow the terms of this agreement and help preserve free future access to Project Gutenberg-tm electronic works. See paragraph 1.E below. 1.C. The Project Gutenberg Literary Archive Foundation ("the Foundation" or PGLAF), owns a compilation copyright in the collection of Project Gutenberg-tm electronic works. Nearly all the individual works in the collection are in the public domain in the United States. If an individual work is in the public domain in the United States and you are located in the United States, we do not claim a right to prevent you from copying, distributing, performing, displaying or creating derivative works based on the work as long as all references to Project Gutenberg are removed. Of course, we hope that you will support the Project Gutenberg-tm mission of promoting free access to electronic works by freely sharing Project Gutenberg-tm works in compliance with the terms of this agreement for keeping the Project Gutenberg-tm name associated with the work. You can easily comply with the terms of this agreement by keeping this work in the same format with its attached full Project Gutenberg-tm License when you share it without charge with others. 1.D. The copyright laws of the place where you are located also govern what you can do with this work. Copyright laws in most countries are in a constant state of change. If you are outside the United States, check the laws of your country in addition to the terms of this agreement before downloading, copying, displaying, performing, distributing or creating derivative works based on this work or any other Project Gutenberg-tm work. The Foundation makes no representations concerning the copyright status of any work in any country outside the United States. 1.E. Unless you have removed all references to Project Gutenberg: 1.E.1. The following sentence, with active links to, or other immediate access to, the full Project Gutenberg-tm License must appear prominently whenever any copy of a Project Gutenberg-tm work (any work on which the phrase "Project Gutenberg" appears, or with which the phrase "Project Gutenberg" is associated) is accessed, displayed, performed, viewed, copied or distributed: This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at book.klll.cc 1.E.2. If an individual Project Gutenberg-tm electronic work is derived from the public domain (does not contain a notice indicating that it is posted with permission of the copyright holder), the work can be copied and distributed to anyone in the United States without paying any fees or charges. If you are redistributing or providing access to a work with the phrase "Project Gutenberg" associated with or appearing on the work, you must comply either with the requirements of paragraphs 1.E.1 through 1.E.7 or obtain permission for the use of the work and the Project Gutenberg-tm trademark as set forth in paragraphs 1.E.8 or 1.E.9. 1.E.3. If an individual Project Gutenberg-tm electronic work is posted with the permission of the copyright holder, your use and distribution must comply with both paragraphs 1.E.1 through 1.E.7 and any additional terms imposed by the copyright holder. Additional terms will be linked to the Project Gutenberg-tm License for all works posted with the permission of the copyright holder found at the beginning of this work. 1.E.4. Do not unlink or detach or remove the full Project Gutenberg-tm License terms from this work, or any files containing a part of this work or any other work associated with Project Gutenberg-tm. 1.E.5. Do not copy, display, perform, distribute or redistribute this electronic work, or any part of this electronic work, without prominently displaying the sentence set forth in paragraph 1.E.1 with active links or immediate access to the full terms of the Project Gutenberg-tm License. 1.E.6. You may convert to and distribute this work in any binary, compressed, marked up, nonproprietary or proprietary form, including any word processing or hypertext form. However, if you provide access to or distribute copies of a Project Gutenberg-tm work in a format other than "Plain Vanilla ASCII" or other format used in the official version posted on the official Project Gutenberg-tm web site (book.klll.cc), you must, at no additional cost, fee or expense to the user, provide a copy, a means of exporting a copy, or a means of obtaining a copy upon request, of the work in its original "Plain Vanilla ASCII" or other form. Any alternate format must include the full Project Gutenberg-tm License as specified in paragraph 1.E.1. 1.E.7. Do not charge a fee for access to, viewing, displaying, performing, copying or distributing any Project Gutenberg-tm works unless you comply with paragraph 1.E.8 or 1.E.9. 1.E.8. You may charge a reasonable fee for copies of or providing access to or distributing Project Gutenberg-tm electronic works provided that - You pay a royalty fee of 20% of the gross profits you derive from the use of Project Gutenberg-tm works calculated using the method you already use to calculate your applicable taxes. The fee is owed to the owner of the Project Gutenberg-tm trademark, but he has agreed to donate royalties under this paragraph to the Project Gutenberg Literary Archive Foundation. Royalty payments must be paid within 60 days following each date on which you prepare (or are legally required to prepare) your periodic tax returns. Royalty payments should be clearly marked as such and sent to the Project Gutenberg Literary Archive Foundation at the address specified in Section 4, "Information about donations to the Project Gutenberg Literary Archive Foundation." - You provide a full refund of any money paid by a user who notifies you in writing (or by e-mail) within 30 days of receipt that s/he does not agree to the terms of the full Project Gutenberg-tm License. You must require such a user to return or destroy all copies of the works possessed in a physical medium and discontinue all use of and all access to other copies of Project Gutenberg-tm works. - You provide, in accordance with paragraph 1.F.3, a full refund of any money paid for a work or a replacement copy, if a defect in the electronic work is discovered and reported to you within 90 days of receipt of the work. - You comply with all other terms of this agreement for free distribution of Project Gutenberg-tm works. 1.E.9. If you wish to charge a fee or distribute a Project Gutenberg-tm electronic work or group of works on different terms than are set forth in this agreement, you must obtain permission in writing from both the Project Gutenberg Literary Archive Foundation and Michael Hart, the owner of the Project Gutenberg-tm trademark. Contact the Foundation as set forth in Section 3 below. 1.F. 1.F.1. Project Gutenberg volunteers and employees expend considerable effort to identify, do copyright research on, transcribe and proofread public domain works in creating the Project Gutenberg-tm collection. Despite these efforts, Project Gutenberg-tm electronic works, and the medium on which they may be stored, may contain "Defects," such as, but not limited to, incomplete, inaccurate or corrupt data, transcription errors, a copyright or other intellectual property infringement, a defective or damaged disk or other medium, a computer virus, or computer codes that damage or cannot be read by your equipment. 1.F.2. LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the "Right of Replacement or Refund" described in paragraph 1.F.3, the Project Gutenberg Literary Archive Foundation, the owner of the Project Gutenberg-tm trademark, and any other party distributing a Project Gutenberg-tm electronic work under this agreement, disclaim all liability to you for damages, costs and expenses, including legal fees. YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE, STRICT LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT THOSE PROVIDED IN PARAGRAPH F3. YOU AGREE THAT THE FOUNDATION, THE TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT BE LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL, PUNITIVE OR INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH DAMAGE. 1.F.3. LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a defect in this electronic work within 90 days of receiving it, you can receive a refund of the money (if any) you paid for it by sending a written explanation to the person you received the work from. If you received the work on a physical medium, you must return the medium with your written explanation. The person or entity that provided you with the defective work may elect to provide a replacement copy in lieu of a refund. If you received the work electronically, the person or entity providing it to you may choose to give you a second opportunity to receive the work electronically in lieu of a refund. If the second copy is also defective, you may demand a refund in writing without further opportunities to fix the problem. 1.F.4. Except for the limited right of replacement or refund set forth in paragraph 1.F.3, this work is provided to you 'AS-IS' WITH NO OTHER WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY PURPOSE. 1.F.5. Some states do not allow disclaimers of certain implied warranties or the exclusion or limitation of certain types of damages. If any disclaimer or limitation set forth in this agreement violates the law of the state applicable to this agreement, the agreement shall be interpreted to make the maximum disclaimer or limitation permitted by the applicable state law. The invalidity or unenforceability of any provision of this agreement shall not void the remaining provisions. 1.F.6. INDEMNITY - You agree to indemnify and hold the Foundation, the trademark owner, any agent or employee of the Foundation, anyone providing copies of Project Gutenberg-tm electronic works in accordance with this agreement, and any volunteers associated with the production, promotion and distribution of Project Gutenberg-tm electronic works, harmless from all liability, costs and expenses, including legal fees, that arise directly or indirectly from any of the following which you do or cause to occur: (a) distribution of this or any Project Gutenberg-tm work, (b) alteration, modification, or additions or deletions to any Project Gutenberg-tm work, and (c) any Defect you cause. Section 2. Information about the Mission of Project Gutenberg-tm Project Gutenberg-tm is synonymous with the free distribution of electronic works in formats readable by the widest variety of computers including obsolete, old, middle-aged and new computers. It exists because of the efforts of hundreds of volunteers and donations from people in all walks of life. Volunteers and financial support to provide volunteers with the assistance they need, are critical to reaching Project Gutenberg-tm's goals and ensuring that the Project Gutenberg-tm collection will remain freely available for generations to come. In 2001, the Project Gutenberg Literary Archive Foundation was created to provide a secure and permanent future for Project Gutenberg-tm and future generations. To learn more about the Project Gutenberg Literary Archive Foundation and how your efforts and donations can help, see Sections 3 and 4 and the Foundation web page at http://www.pglaf.org. Section 3. Information about the Project Gutenberg Literary Archive Foundation The Project Gutenberg Literary Archive Foundation is a non profit 501(c)(3) educational corporation organized under the laws of the state of Mississippi and granted tax exempt status by the Internal Revenue Service. The Foundation's EIN or federal tax identification number is 64-6221541. Its 501(c)(3) letter is posted at http://pglaf.org/fundraising. Contributions to the Project Gutenberg Literary Archive Foundation are tax deductible to the full extent permitted by U.S. federal laws and your state's laws. The Foundation's principal office is located at 4557 Melan Dr. S. Fairbanks, AK, 99712., but its volunteers and employees are scattered throughout numerous locations. Its business office is located at 809 North 1500 West, Salt Lake City, UT 84116, (801) 596-1887, email business@pglaf.org. Email contact links and up to date contact information can be found at the Foundation's web site and official page at http://pglaf.org For additional contact information: Dr. Gregory B. Newby Chief Executive and Director gbnewby@pglaf.org Section 4. Information about Donations to the Project Gutenberg Literary Archive Foundation Project Gutenberg-tm depends upon and cannot survive without wide spread public support and donations to carry out its mission of increasing the number of public domain and licensed works that can be freely distributed in machine readable form accessible by the widest array of equipment including outdated equipment. Many small donations ($1 to $5,000) are particularly important to maintaining tax exempt status with the IRS. The Foundation is committed to complying with the laws regulating charities and charitable donations in all 50 states of the United States. Compliance requirements are not uniform and it takes a considerable effort, much paperwork and many fees to meet and keep up with these requirements. We do not solicit donations in locations where we have not received written confirmation of compliance. To SEND DONATIONS or determine the status of compliance for any particular state visit http://pglaf.org While we cannot and do not solicit contributions from states where we have not met the solicitation requirements, we know of no prohibition against accepting unsolicited donations from donors in such states who approach us with offers to donate. International donations are gratefully accepted, but we cannot make any statements concerning tax treatment of donations received from outside the United States. U.S. laws alone swamp our small staff. Please check the Project Gutenberg Web pages for current donation methods and addresses. Donations are accepted in a number of other ways including checks, online payments and credit card donations. To donate, please visit: http://pglaf.org/donate Section 5. General Information About Project Gutenberg-tm electronic works. Professor Michael S. Hart is the originator of the Project Gutenberg-tm concept of a library of electronic works that could be freely shared with anyone. For thirty years, he produced and distributed Project Gutenberg-tm eBooks with only a loose network of volunteer support. Project Gutenberg-tm eBooks are often created from several printed editions, all of which are confirmed as Public Domain in the U.S. unless a copyright notice is included. Thus, we do not necessarily keep eBooks in compliance with any particular paper edition. Most people start at our Web site which has the main PG search facility: http://book.klll.cc This Web site includes information about Project Gutenberg-tm, including how to make donations to the Project Gutenberg Literary Archive Foundation, how to help produce our new eBooks, and how to subscribe to our email newsletter to hear about new eBooks. \end{PGtext} % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % % % End of Project Gutenberg's Plane Geometry, by George Albert Wentworth % % % % *** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY *** % % % % ***** This file should be named 33063-t.tex or 33063-t.zip ***** % % This and all associated files of various formats will be found in: % % http://book.klll.cc/3/3/0/6/33063/ % % % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % \end{document} ### @ControlwordReplace = ( ); @MathEnvironments = ( ['\\begin{supertabular}','\\end{supertabular}',''] ); @ControlwordArguments = ( ['\\part', 1, 1, '', ''], ['\\chapter', 0, 0, '', '', 1, 1, '', ''], ['\\section', 0, 0, '', '', 1, 1, '', ''], ['\\subsection', 1, 1, '', ''], ['\\scanpage', 1, 0, '', ''], ['\\figc', 1, 0, '
', '', 1, 0, '', ''], ['\\figcc', 1, 0, '
', '', 1, 0, '', ''], ['\\figccc', 1, 0, '
', '', 1, 0, '', '', 1, 0, '', ''], ['\\figcccc', 1, 0, '
', '', 1, 0, '', '', 1, 0, '', '', 1, 0, '', ''], ['\\phantom', 1, 0, '', ''], ['\\ex', 1, 1, 'Ex. ', ''], ['\\exheader', 1, 1, '', ''], ['\\defn', 1, 1, 'Def. ', ''], ['\\ax', 1, 1, 'Axiom. ', ''], ['\\thm', 1, 1, 'Theorem. ', ''], ['\\cor', 0, 0, '', '', 1, 1, 'Cor. ', ''], ['\\note', 0, 0, '', '', 1, 1, 'Note. ', ''], ['\\pp', 1, 1, '', ''], ['\\step', 0, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ''], ['\\eq', 0, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ' ', 1, 1, '', ''], ['\\pnote', 1, 1, '', ''], ['\\tablehead', 1, 0, '', ''], ['\\multicolumn', 1, 0, '', '', 1, 0, '', '', 1, 1, '', ''] ); ### This is pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) (format=pdflatex 2010.5.6) 3 JUL 2010 09:28 entering extended mode %&-line parsing enabled. **33063-t.tex (./33063-t.tex LaTeX2e <2005/12/01> Babel and hyphenation patterns for english, usenglishmax, dumylang, noh yphenation, arabic, farsi, croatian, ukrainian, russian, bulgarian, czech, slov ak, danish, dutch, finnish, basque, french, german, ngerman, ibycus, greek, mon ogreek, ancientgreek, hungarian, italian, latin, mongolian, norsk, icelandic, i nterlingua, turkish, coptic, romanian, welsh, serbian, slovenian, estonian, esp eranto, uppersorbian, indonesian, polish, portuguese, spanish, catalan, galicia n, swedish, ukenglish, pinyin, loaded. (/usr/share/texmf-texlive/tex/latex/base/book.cls Document Class: book 2005/09/16 v1.4f Standard LaTeX document class (/usr/share/texmf-texlive/tex/latex/base/bk12.clo File: bk12.clo 2005/09/16 v1.4f Standard LaTeX file (size option) ) \c@part=\count79 \c@chapter=\count80 \c@section=\count81 \c@subsection=\count82 \c@subsubsection=\count83 \c@paragraph=\count84 \c@subparagraph=\count85 \c@figure=\count86 \c@table=\count87 \abovecaptionskip=\skip41 \belowcaptionskip=\skip42 \bibindent=\dimen102 ) (/usr/share/texmf-texlive/tex/latex/base/inputenc.sty Package: inputenc 2006/05/05 v1.1b Input encoding file \inpenc@prehook=\toks14 \inpenc@posthook=\toks15 (/usr/share/texmf-texlive/tex/latex/base/latin1.def File: latin1.def 2006/05/05 v1.1b Input encoding file )) (/usr/share/texmf-texlive/tex/latex/base/ifthen.sty Package: ifthen 2001/05/26 v1.1c Standard LaTeX ifthen package (DPC) ) (/usr/share/texmf-texlive/tex/latex/amsmath/amsmath.sty Package: amsmath 2000/07/18 v2.13 AMS math features \@mathmargin=\skip43 For additional information on amsmath, use the `?' option. (/usr/share/texmf-texlive/tex/latex/amsmath/amstext.sty Package: amstext 2000/06/29 v2.01 (/usr/share/texmf-texlive/tex/latex/amsmath/amsgen.sty File: amsgen.sty 1999/11/30 v2.0 \@emptytoks=\toks16 \ex@=\dimen103 )) (/usr/share/texmf-texlive/tex/latex/amsmath/amsbsy.sty Package: amsbsy 1999/11/29 v1.2d \pmbraise@=\dimen104 ) (/usr/share/texmf-texlive/tex/latex/amsmath/amsopn.sty Package: amsopn 1999/12/14 v2.01 operator names ) \inf@bad=\count88 LaTeX Info: Redefining \frac on input line 211. \uproot@=\count89 \leftroot@=\count90 LaTeX Info: Redefining \overline on input line 307. \classnum@=\count91 \DOTSCASE@=\count92 LaTeX Info: Redefining \ldots on input line 379. LaTeX Info: Redefining \dots on input line 382. LaTeX Info: Redefining \cdots on input line 467. \Mathstrutbox@=\box26 \strutbox@=\box27 \big@size=\dimen105 LaTeX Font Info: Redeclaring font encoding OML on input line 567. LaTeX Font Info: Redeclaring font encoding OMS on input line 568. \macc@depth=\count93 \c@MaxMatrixCols=\count94 \dotsspace@=\muskip10 \c@parentequation=\count95 \dspbrk@lvl=\count96 \tag@help=\toks17 \row@=\count97 \column@=\count98 \maxfields@=\count99 \andhelp@=\toks18 \eqnshift@=\dimen106 \alignsep@=\dimen107 \tagshift@=\dimen108 \tagwidth@=\dimen109 \totwidth@=\dimen110 \lineht@=\dimen111 \@envbody=\toks19 \multlinegap=\skip44 \multlinetaggap=\skip45 \mathdisplay@stack=\toks20 LaTeX Info: Redefining \[ on input line 2666. LaTeX Info: Redefining \] on input line 2667. ) (/usr/share/texmf-texlive/tex/latex/amsfonts/amssymb.sty Package: amssymb 2002/01/22 v2.2d (/usr/share/texmf-texlive/tex/latex/amsfonts/amsfonts.sty Package: amsfonts 2001/10/25 v2.2f \symAMSa=\mathgroup4 \symAMSb=\mathgroup5 LaTeX Font Info: Overwriting math alphabet `\mathfrak' in version `bold' (Font) U/euf/m/n --> U/euf/b/n on input line 132. )) (/usr/share/texmf-texlive/tex/latex/base/alltt.sty Package: alltt 1997/06/16 v2.0g defines alltt environment ) (/usr/share/texmf-texlive/tex/latex/footmisc/footmisc.sty Package: footmisc 2005/03/17 v5.3d a miscellany of footnote facilities \FN@temptoken=\toks21 \footnotemargin=\dimen112 \c@pp@next@reset=\count100 Package footmisc Info: Declaring symbol style bringhurst on input line 817. Package footmisc Info: Declaring symbol style chicago on input line 818. Package footmisc Info: Declaring symbol style wiley on input line 819. Package footmisc Info: Declaring symbol style lamport-robust on input line 823. Package footmisc Info: Declaring symbol style lamport* on input line 831. Package footmisc Info: Declaring symbol style lamport*-robust on input line 840 . ) (/usr/share/texmf-texlive/tex/latex/tools/indentfirst.sty Package: indentfirst 1995/11/23 v1.03 Indent first paragraph (DPC) ) (/usr/share/texmf-texlive/tex/latex/fancyhdr/fancyhdr.sty \fancy@headwidth=\skip46 \f@ncyO@elh=\skip47 \f@ncyO@erh=\skip48 \f@ncyO@olh=\skip49 \f@ncyO@orh=\skip50 \f@ncyO@elf=\skip51 \f@ncyO@erf=\skip52 \f@ncyO@olf=\skip53 \f@ncyO@orf=\skip54 ) (/usr/share/texmf-texlive/tex/latex/graphics/graphicx.sty Package: graphicx 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR) (/usr/share/texmf-texlive/tex/latex/graphics/keyval.sty Package: keyval 1999/03/16 v1.13 key=value parser (DPC) \KV@toks@=\toks22 ) (/usr/share/texmf-texlive/tex/latex/graphics/graphics.sty Package: graphics 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR) (/usr/share/texmf-texlive/tex/latex/graphics/trig.sty Package: trig 1999/03/16 v1.09 sin cos tan (DPC) ) (/etc/texmf/tex/latex/config/graphics.cfg File: graphics.cfg 2007/01/18 v1.5 graphics configuration of teTeX/TeXLive ) Package graphics Info: Driver file: pdftex.def on input line 90. (/usr/share/texmf-texlive/tex/latex/pdftex-def/pdftex.def File: pdftex.def 2007/01/08 v0.04d Graphics/color for pdfTeX \Gread@gobject=\count101 )) \Gin@req@height=\dimen113 \Gin@req@width=\dimen114 ) (/usr/share/texmf-texlive/tex/latex/tools/enumerate.sty Package: enumerate 1999/03/05 v3.00 enumerate extensions (DPC) \@enLab=\toks23 ) (/usr/share/texmf-texlive/tex/latex/supertabular/supertabular.sty Package: supertabular 2004/02/20 v4.1e the supertabular environment \c@tracingst=\count102 \ST@wd=\dimen115 \ST@rightskip=\skip55 \ST@leftskip=\skip56 \ST@parfillskip=\skip57 \ST@pageleft=\dimen116 \ST@headht=\dimen117 \ST@tailht=\dimen118 \ST@pagesofar=\dimen119 \ST@pboxht=\dimen120 \ST@lineht=\dimen121 \ST@stretchht=\dimen122 \ST@prevht=\dimen123 \ST@toadd=\dimen124 \ST@dimen=\dimen125 \ST@pbox=\box28 ) (/usr/share/texmf-texlive/tex/latex/geometry/geometry.sty Package: geometry 2002/07/08 v3.2 Page Geometry \Gm@cnth=\count103 \Gm@cntv=\count104 \c@Gm@tempcnt=\count105 \Gm@bindingoffset=\dimen126 \Gm@wd@mp=\dimen127 \Gm@odd@mp=\dimen128 \Gm@even@mp=\dimen129 \Gm@dimlist=\toks24 (/usr/share/texmf-texlive/tex/xelatex/xetexconfig/geometry.cfg)) (/usr/share/te xmf-texlive/tex/latex/hyperref/hyperref.sty Package: hyperref 2007/02/07 v6.75r Hypertext links for LaTeX \@linkdim=\dimen130 \Hy@linkcounter=\count106 \Hy@pagecounter=\count107 (/usr/share/texmf-texlive/tex/latex/hyperref/pd1enc.def File: pd1enc.def 2007/02/07 v6.75r Hyperref: PDFDocEncoding definition (HO) ) (/etc/texmf/tex/latex/config/hyperref.cfg File: hyperref.cfg 2002/06/06 v1.2 hyperref configuration of TeXLive ) (/usr/share/texmf-texlive/tex/latex/oberdiek/kvoptions.sty Package: kvoptions 2006/08/22 v2.4 Connects package keyval with LaTeX options ( HO) ) Package hyperref Info: Option `hyperfootnotes' set `false' on input line 2238. Package hyperref Info: Option `bookmarks' set `true' on input line 2238. Package hyperref Info: Option `linktocpage' set `false' on input line 2238. Package hyperref Info: Option `pdfdisplaydoctitle' set `true' on input line 223 8. Package hyperref Info: Option `pdfpagelabels' set `true' on input line 2238. Package hyperref Info: Option `bookmarksopen' set `true' on input line 2238. Package hyperref Info: Option `colorlinks' set `true' on input line 2238. Package hyperref Info: Hyper figures OFF on input line 2288. Package hyperref Info: Link nesting OFF on input line 2293. Package hyperref Info: Hyper index ON on input line 2296. Package hyperref Info: Plain pages OFF on input line 2303. Package hyperref Info: Backreferencing OFF on input line 2308. Implicit mode ON; LaTeX internals redefined Package hyperref Info: Bookmarks ON on input line 2444. (/usr/share/texmf-texlive/tex/latex/ltxmisc/url.sty \Urlmuskip=\muskip11 Package: url 2005/06/27 ver 3.2 Verb mode for urls, etc. ) LaTeX Info: Redefining \url on input line 2599. \Fld@menulength=\count108 \Field@Width=\dimen131 \Fld@charsize=\dimen132 \Choice@toks=\toks25 \Field@toks=\toks26 Package hyperref Info: Hyper figures OFF on input line 3102. Package hyperref Info: Link nesting OFF on input line 3107. Package hyperref Info: Hyper index ON on input line 3110. Package hyperref Info: backreferencing OFF on input line 3117. Package hyperref Info: Link coloring ON on input line 3120. \Hy@abspage=\count109 \c@Item=\count110 ) *hyperref using driver hpdftex* (/usr/share/texmf-texlive/tex/latex/hyperref/hpdftex.def File: hpdftex.def 2007/02/07 v6.75r Hyperref driver for pdfTeX \Fld@listcount=\count111 ) \TmpLen=\skip58 \c@proposition=\count112 \c@point=\count113 \c@exnum=\count114 \dentwidth=\skip59 \eqalign=\skip60 \c@Lcount=\count115 (./33063-t.aux) \openout1 = `33063-t.aux'. LaTeX Font Info: Checking defaults for OML/cmm/m/it on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for T1/cmr/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for OT1/cmr/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for OMS/cmsy/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for OMX/cmex/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for U/cmr/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. LaTeX Font Info: Checking defaults for PD1/pdf/m/n on input line 485. LaTeX Font Info: ... okay on input line 485. (/usr/share/texmf/tex/context/base/supp-pdf.tex [Loading MPS to PDF converter (version 2006.09.02).] \scratchcounter=\count116 \scratchdimen=\dimen133 \scratchbox=\box29 \nofMPsegments=\count117 \nofMParguments=\count118 \everyMPshowfont=\toks27 \MPscratchCnt=\count119 \MPscratchDim=\dimen134 \MPnumerator=\count120 \everyMPtoPDFconversion=\toks28 ) -------------------- Geometry parameters paper: class default landscape: -- twocolumn: -- twoside: true asymmetric: -- h-parts: 18.0675pt, 397.48499pt, 18.0675pt v-parts: 9.07278pt, 591.61302pt, 13.60919pt hmarginratio: 1:1 vmarginratio: 2:3 lines: -- heightrounded: -- bindingoffset: 0.0pt truedimen: -- includehead: true includefoot: true includemp: -- driver: pdftex -------------------- Page layout dimensions and switches \paperwidth 433.62pt \paperheight 614.295pt \textwidth 397.48499pt \textheight 529.73923pt \oddsidemargin -54.20248pt \evensidemargin -54.20248pt \topmargin -63.1972pt \headheight 14.5pt \headsep 19.8738pt \footskip 30.0pt \marginparwidth 98.0pt \marginparsep 7.0pt \columnsep 10.0pt \skip\footins 10.8pt plus 4.0pt minus 2.0pt \hoffset 0.0pt \voffset 0.0pt \mag 1000 \@twosidetrue \@mparswitchtrue (1in=72.27pt, 1cm=28.45pt) ----------------------- (/usr/share/texmf-texlive/tex/latex/graphics/color.sty Package: color 2005/11/14 v1.0j Standard LaTeX Color (DPC) (/etc/texmf/tex/latex/config/color.cfg File: color.cfg 2007/01/18 v1.5 color configuration of teTeX/TeXLive ) Package color Info: Driver file: pdftex.def on input line 130. ) Package hyperref Info: Link coloring ON on input line 485. (/usr/share/texmf-texlive/tex/latex/hyperref/nameref.sty Package: nameref 2006/12/27 v2.28 Cross-referencing by name of section (/usr/share/texmf-texlive/tex/latex/oberdiek/refcount.sty Package: refcount 2006/02/20 v3.0 Data extraction from references (HO) ) \c@section@level=\count121 ) LaTeX Info: Redefining \ref on input line 485. LaTeX Info: Redefining \pageref on input line 485. (./33063-t.out) (./33063-t.out) \@outlinefile=\write3 \openout3 = `33063-t.out'. LaTeX Font Info: Try loading font information for U+msa on input line 521. (/usr/share/texmf-texlive/tex/latex/amsfonts/umsa.fd File: umsa.fd 2002/01/19 v2.2g AMS font definitions ) LaTeX Font Info: Try loading font information for U+msb on input line 521. (/usr/share/texmf-texlive/tex/latex/amsfonts/umsb.fd File: umsb.fd 2002/01/19 v2.2g AMS font definitions ) [1 {/var/lib/texmf/fonts/map/pdftex/updmap/pdftex.map}] [2 ] LaTeX Font Info: Try loading font information for OMS+cmr on input line 575. (/usr/share/texmf-texlive/tex/latex/base/omscmr.fd File: omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions ) LaTeX Font Info: Font shape `OMS/cmr/m/n' in size <12> not available (Font) Font shape `OMS/cmsy/m/n' tried instead on input line 575. [1 ] Underfull \hbox (badness 1199) in paragraph at lines 610--610 []\OT1/cmr/m/n/12 GINN & COM-PANY \OMS/cmsy/m/n/12 ^^A \OT1/cmr/m/n/12 PRO- [] [2] [3 ] [4] [5 ] [6] (./33063-t.toc [7 ]) \tf@toc=\write4 \openout4 = `33063-t.toc'. [8] Package hyperref Warning: Difference (2) between bookmark levels is greater (hyperref) than one, level fixed on input line 776. <./images/woodcutsmall.pdf, id=325, 171.64125pt x 144.54pt> File: ./images/woodcutsmall.pdf Graphic file (type pdf) [1 <./images/woodcutsmall.pdf>] <./images/012aaZ10.pdf, id=334, 112.42pt x 66.247 5pt> File: ./images/012aaZ10.pdf Graphic file (type pdf) [2 <./images/012aaZ10.pdf>] <./images/012bbZ14.pdf, id=350, 135.50626pt x 94.3525pt> File: ./images/012bbZ14.pdf Graphic file (type pdf) [3 <./images/012bbZ14.pdf>] [4] [5] [6] . Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 1099. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 1099. <./images/016aaZ37.pdf, id=379, 113.42375pt x 83.31125pt> File: ./images/016aaZ37.pdf Graphic file (type pdf) [7 <./images/016aaZ37.pdf>] <./images/018aaZ55.pdf, id=392, 169.63374pt x 36.135p t> File: ./images/018aaZ55.pdf Graphic file (type pdf) [8] <./images/018bbZ56.pdf, id=399, 174.6525pt x 29 .10875pt> File: ./images/018bbZ56.pdf Graphic file (type pdf) <./images/018ccZ57.pdf, id=400, 86.3225pt x 80.3pt> File: ./images/018ccZ57.pdf Graphic file (type pdf) [9 <./images/018aaZ55.pdf> <./images/018bbZ56.pdf>] <./images/019aaZ58.pdf, id=420, 67.25125pt x 66.2475pt> File: ./images/019aaZ58.pdf Graphic file (type pdf) <./images/019bbZ58.pdf, id=421, 91.34125pt x 61.228 74pt> File: ./images/019bbZ58.pdf Graphic file (type pdf) [10 <./images/018ccZ57.pdf> <./images/019aaZ58.pdf> <./images/019bbZ58.pdf>] <./images/019ccZ62.pdf, id=451, 92.345pt x 69.25874pt > File: ./images/019ccZ62.pdf Graphic file (type pdf) <./images/020aaZ63.pdf, id=452, 82.3075pt x 77.2887 6pt> File: ./images/020aaZ63.pdf Graphic file (type pdf) <./images/020bbZ66.pdf, id=453, 169.63374pt x 36.13 5pt> File: ./images/020bbZ66.pdf Graphic file (type pdf) <./images/020ccZ68.pdf, id=454, 78.2925pt x 54.2025 pt> File: ./images/020ccZ68.pdf Graphic file (type pdf) <./images/020ddZ69.pdf, id=455, 108.405pt x 56.21pt > File: ./images/020ddZ69.pdf Graphic file (type pdf) [11 <./images/019ccZ62.pdf> <./images/020aaZ63.pdf> <./images/020bbZ66.pdf>] <./images/021aaZ72.pdf, id=481, 105.39375pt x 114.427 5pt> File: ./images/021aaZ72.pdf Graphic file (type pdf) [12 <./images/020ccZ68.pdf> <./images/020ddZ69.pdf> ] <./images/022aaZ75.pdf, id=500, 62.2325pt x 56.21pt> File: ./images/022aaZ75.pdf Graphic file (type pdf) <./images/022bbZ76.pdf, id=501, 53.19875pt x 65.243 74pt> File: ./images/022bbZ76.pdf Graphic file (type pdf) <./images/022ccZ77.pdf, id=502, 92.345pt x 69.25874 pt> File: ./images/022ccZ77.pdf Graphic file (type pdf) [13 <./images/021aaZ72.pdf>] [14 <./images/022aaZ75 .pdf> <./images/022bbZ76.pdf> <./images/022ccZ77.pdf>] <./images/023aaZ79.pdf, id=546, 187.70125pt x 81.30376pt> File: ./images/023aaZ79.pdf Graphic file (type pdf) <./images/023bcZ80.pdf, id=547, 378.41376pt x 94.35 25pt> File: ./images/023bcZ80.pdf Graphic file (type pdf) [15 <./images/023aaZ79.pdf>] [16 <./images/023bcZ80 .pdf>] <./images/024aaZ81.pdf, id=571, 169.63374pt x 42.1575pt> File: ./images/024aaZ81.pdf Graphic file (type pdf) LaTeX Font Info: Font shape `OMS/cmr/m/n' in size <10.95> not available (Font) Font shape `OMS/cmsy/m/n' tried instead on input line 1512. <./images/024bbZ83.pdf, id=572, 79.29625pt x 70.2625pt> File: ./images/024bbZ83.pdf Graphic file (type pdf) [17 <./images/024aaZ81.pdf> <./images/024bbZ83.pdf> ] <./images/025aaZ86.pdf, id=586, 188.705pt x 81.30376pt> File: ./images/025aaZ86.pdf Graphic file (type pdf) [18 <./images/025aaZ86.pdf>] <./images/026aaZ90.pdf , id=596, 202.7575pt x 83.31125pt> File: ./images/026aaZ90.pdf Graphic file (type pdf) <./images/027aaZ93.pdf, id=597, 169.63374pt x 85.31 876pt> File: ./images/027aaZ93.pdf Graphic file (type pdf) [19 <./images/026aaZ90.pdf>] <./images/028aaZ95.pdf , id=607, 188.705pt x 131.49126pt> File: ./images/028aaZ95.pdf Graphic file (type pdf) [20 <./images/027aaZ93.pdf>] <./images/029aaZ96.pdf , id=617, 94.3525pt x 130.4875pt> File: ./images/029aaZ96.pdf Graphic file (type pdf) [21 <./images/028aaZ95.pdf>] [22 <./images/029aaZ96 .pdf>] <./images/030aaZ97.pdf, id=640, 94.3525pt x 130.4875pt> File: ./images/030aaZ97.pdf Graphic file (type pdf) [23 <./images/030aaZ97.pdf>] <./images/031aa100.pdf , id=653, 197.73875pt x 111.41624pt> File: ./images/031aa100.pdf Graphic file (type pdf) [24 <./images/031aa100.pdf>] <./images/032aa101.pdf , id=663, 149.55875pt x 131.49126pt> File: ./images/032aa101.pdf Graphic file (type pdf) [25 <./images/032aa101.pdf>] <./images/033aa104.pdf , id=673, 179.67125pt x 106.3975pt> File: ./images/033aa104.pdf Graphic file (type pdf) <./images/034aa107.pdf, id=674, 207.77625pt x 112.4 2pt> File: ./images/034aa107.pdf Graphic file (type pdf) [26 <./images/033aa104.pdf>] <./images/034bb109.pdf , id=684, 170.6375pt x 132.495pt> File: ./images/034bb109.pdf Graphic file (type pdf) [27 <./images/034aa107.pdf>] <./images/035aa110.pdf , id=695, 264.99pt x 95.35625pt> File: ./images/035aa110.pdf Graphic file (type pdf) [28 <./images/034bb109.pdf>] [29 <./images/035aa110 .pdf>] <./images/036aa111.pdf, id=716, 228.855pt x 130.4875pt> File: ./images/036aa111.pdf Graphic file (type pdf) [30 <./images/036aa111.pdf>] <./images/037aa112.pdf , id=726, 235.88126pt x 160.6pt> File: ./images/037aa112.pdf Graphic file (type pdf) [31 <./images/037aa112.pdf>] <./images/038aa115.pdf , id=736, 235.88126pt x 160.6pt> File: ./images/038aa115.pdf Graphic file (type pdf) [32 <./images/038aa115.pdf>] <./images/039aa117.pdf , id=746, 95.35625pt x 85.31876pt> File: ./images/039aa117.pdf Graphic file (type pdf) <./images/039bb118.pdf, id=747, 132.495pt x 85.3187 6pt> File: ./images/039bb118.pdf Graphic file (type pdf) <./images/039ce119.pdf, id=748, 285.065pt x 91.3412 5pt> File: ./images/039ce119.pdf Graphic file (type pdf) <./images/039fi120.pdf, id=749, 385.44pt x 97.36375 pt> File: ./images/039fi120.pdf Graphic file (type pdf) [33 <./images/039aa117.pdf> <./images/039bb118.pdf> ] [34 <./images/039ce119.pdf> <./images/039fi120.pdf>] <./images/041aa129.pdf, id=785, 208.78pt x 104.39pt> File: ./images/041aa129.pdf Graphic file (type pdf) [35] [36 <./images/041aa129.pdf>] <./images/042aa13 8.pdf, id=799, 164.615pt x 66.2475pt> File: ./images/042aa138.pdf Graphic file (type pdf) <./images/043ab139.pdf, id=800, 330.23375pt x 90.33 75pt> File: ./images/043ab139.pdf Graphic file (type pdf) [37 <./images/042aa138.pdf>] <./images/044ab143.pdf , id=810, 330.23375pt x 90.3375pt> File: ./images/044ab143.pdf Graphic file (type pdf) [38 <./images/043ab139.pdf>] <./images/045aa145.pdf , id=820, 158.5925pt x 113.42375pt> File: ./images/045aa145.pdf Graphic file (type pdf) [39 <./images/044ab143.pdf>] <./images/046aa147.pdf , id=831, 158.5925pt x 113.42375pt> File: ./images/046aa147.pdf Graphic file (type pdf) [40 <./images/045aa145.pdf>] <./images/047ab150.pdf , id=841, 323.2075pt x 150.5625pt> File: ./images/047ab150.pdf Graphic file (type pdf) [41 <./images/046aa147.pdf>] [42 <./images/047ab150 .pdf>] <./images/048ac151.pdf, id=864, 338.26375pt x 115.43124pt> File: ./images/048ac151.pdf Graphic file (type pdf) <./images/049aa152.pdf, id=865, 206.7725pt x 85.318 76pt> File: ./images/049aa152.pdf Graphic file (type pdf) [43 <./images/048ac151.pdf>] <./images/050aa153.pdf , id=878, 160.6pt x 95.35625pt> File: ./images/050aa153.pdf Graphic file (type pdf) [44 <./images/049aa152.pdf>] <./images/051ac154.pdf , id=888, 378.41376pt x 131.49126pt> File: ./images/051ac154.pdf Graphic file (type pdf) [45 <./images/050aa153.pdf>] [46 <./images/051ac154 .pdf>] <./images/052ab155.pdf, id=908, 322.20375pt x 112.42pt> File: ./images/052ab155.pdf Graphic file (type pdf) [47 <./images/052ab155.pdf>] <./images/054aa160.pdf , id=918, 133.49875pt x 112.42pt> File: ./images/054aa160.pdf Graphic file (type pdf) [48] [49 <./images/054aa160.pdf>] <./images/055aa16 2.pdf, id=932, 121.45375pt x 101.37875pt> File: ./images/055aa162.pdf Graphic file (type pdf) [50 <./images/055aa162.pdf>] <./images/056ac166.pdf , id=942, 355.3275pt x 75.28125pt> File: ./images/056ac166.pdf Graphic file (type pdf) <./images/056dg170.pdf, id=943, 383.4325pt x 66.247 5pt> File: ./images/056dg170.pdf Graphic file (type pdf) <./images/057aa174.pdf, id=944, 132.495pt x 68.255p t> File: ./images/057aa174.pdf Graphic file (type pdf) [51 <./images/056ac166.pdf> <./images/056dg170.pdf> ] <./images/057bb176.pdf, id=960, 190.7125pt x 87.32625pt> File: ./images/057bb176.pdf Graphic file (type pdf) [52 <./images/057aa174.pdf>] <./images/058aa178.pdf , id=970, 222.8325pt x 103.38625pt> File: ./images/058aa178.pdf Graphic file (type pdf) [53 <./images/057bb176.pdf>] <./images/058bb181.pdf , id=989, 192.72pt x 67.25125pt> File: ./images/058bb181.pdf Graphic file (type pdf) [54 <./images/058aa178.pdf> <./images/058bb181.pdf> ] <./images/059aa182.pdf, id=1004, 222.8325pt x 103.38625pt> File: ./images/059aa182.pdf Graphic file (type pdf) [55 <./images/059aa182.pdf>] <./images/060aa183.pdf , id=1014, 222.8325pt x 103.38625pt> File: ./images/060aa183.pdf Graphic file (type pdf) <./images/061aa184.pdf, id=1015, 222.8325pt x 103.3 8625pt> File: ./images/061aa184.pdf Graphic file (type pdf) [56 <./images/060aa183.pdf>] [57 <./images/061aa184 .pdf>] <./images/062ab185.pdf, id=1035, 336.25626pt x 75.28125pt> File: ./images/062ab185.pdf Graphic file (type pdf) <./images/063aa187.pdf, id=1036, 234.8775pt x 169.6 3374pt> File: ./images/063aa187.pdf Graphic file (type pdf) [58 <./images/062ab185.pdf>] <./images/064aa188.pdf , id=1049, 123.46124pt x 95.35625pt> File: ./images/064aa188.pdf Graphic file (type pdf) [59 <./images/063aa187.pdf>] <./images/064bb190.pdf , id=1059, 162.6075pt x 86.3225pt> File: ./images/064bb190.pdf Graphic file (type pdf) [60 <./images/064aa188.pdf> <./images/064bb190.pdf> ] <./images/065ac192.pdf, id=1074, 388.45125pt x 112.42pt> File: ./images/065ac192.pdf Graphic file (type pdf) [61 <./images/065ac192.pdf>] <./images/066ad203.pdf , id=1086, 355.3275pt x 78.2925pt> File: ./images/066ad203.pdf Graphic file (type pdf) <./images/067aa205.pdf, id=1087, 188.705pt x 160.6p t> File: ./images/067aa205.pdf Graphic file (type pdf) [62 <./images/066ad203.pdf>] <./images/068aa207.pdf , id=1097, 157.58875pt x 172.645pt> File: ./images/068aa207.pdf Graphic file (type pdf) [63 <./images/067aa205.pdf>] [64 <./images/068aa207 .pdf>] <./images/069ac208.pdf, id=1119, 379.4175pt x 123.46124pt> File: ./images/069ac208.pdf Graphic file (type pdf) <./images/069dd211.pdf, id=1120, 124.465pt x 113.42 375pt> File: ./images/069dd211.pdf Graphic file (type pdf) [65 <./images/069ac208.pdf>] <./images/070aa212.pdf , id=1133, 218.8175pt x 151.56625pt> File: ./images/070aa212.pdf Graphic file (type pdf) [66 <./images/069dd211.pdf>] <./images/071aa213.pdf , id=1146, 226.8475pt x 151.56625pt> File: ./images/071aa213.pdf Graphic file (type pdf) [67 <./images/070aa212.pdf>] [68 <./images/071aa213 .pdf>] [69] [70] <./images/074aaZ19.pdf, id=1208, 89.33376pt x 96.36pt> File: ./images/074aaZ19.pdf Graphic file (type pdf) [71] [72 <./images/074aaZ19.pdf>] <./images/074bbZ2 0.pdf, id=1224, 77.28876pt x 76.285pt> File: ./images/074bbZ20.pdf Graphic file (type pdf) <./images/075aaZ21.pdf, id=1225, 112.42pt x 76.285p t> File: ./images/075aaZ21.pdf Graphic file (type pdf) <./images/075bbZ22.pdf, id=1226, 112.42pt x 76.285p t> File: ./images/075bbZ22.pdf Graphic file (type pdf) [73 <./images/074bbZ20.pdf>] <./images/075ceZ23.pdf , id=1236, 270.00874pt x 57.21375pt> File: ./images/075ceZ23.pdf Graphic file (type pdf) [74 <./images/075aaZ21.pdf> <./images/075bbZ22.pdf> ] <./images/076aaZ24.pdf, id=1257, 95.35625pt x 70.2625pt> File: ./images/076aaZ24.pdf Graphic file (type pdf) <./images/076bbZ25.pdf, id=1258, 86.3225pt x 67.251 25pt> File: ./images/076bbZ25.pdf Graphic file (type pdf) [75 <./images/075ceZ23.pdf>] <./images/076ccZ26.pdf , id=1272, 126.4725pt x 86.3225pt> File: ./images/076ccZ26.pdf Graphic file (type pdf) <./images/076ddZ27.pdf, id=1273, 93.34875pt x 83.31 125pt> File: ./images/076ddZ27.pdf Graphic file (type pdf) [76 <./images/076aaZ24.pdf> <./images/076bbZ25.pdf> ] <./images/077adZ28.pdf, id=1291, 337.26pt x 65.24374pt> File: ./images/077adZ28.pdf Graphic file (type pdf) [77 <./images/076ccZ26.pdf> <./images/076ddZ27.pdf> <./images/077adZ28.pdf>] <./images/077eeZ34.pdf, id=1314, 54.2025pt x 56.21pt> File: ./images/077eeZ34.pdf Graphic file (type pdf) <./images/077ffZ35.pdf, id=1315, 72.27pt x 85.31876 pt> File: ./images/077ffZ35.pdf Graphic file (type pdf) <./images/078aaZ41.pdf, id=1316, 88.33pt x 80.3pt> File: ./images/078aaZ41.pdf Graphic file (type pdf) [78 <./images/077eeZ34.pdf> <./images/077ffZ35.pdf> ] <./images/078bbZ43.pdf, id=1331, 58.2175pt x 75.28125pt> File: ./images/078bbZ43.pdf Graphic file (type pdf) <./images/078ccZ46.pdf, id=1332, 95.35625pt x 67.25 125pt> File: ./images/078ccZ46.pdf Graphic file (type pdf) [79 <./images/078aaZ41.pdf> <./images/078bbZ43.pdf> ] <./images/078ddZ49.pdf, id=1348, 62.2325pt x 65.24374pt> File: ./images/078ddZ49.pdf Graphic file (type pdf) <./images/079adZ51.pdf, id=1349, 400.49625pt x 95.3 5625pt> File: ./images/079adZ51.pdf Graphic file (type pdf) [80 <./images/078ccZ46.pdf> <./images/078ddZ49.pdf> ] <./images/079eeZ57.pdf, id=1364, 53.19875pt x 67.25125pt> File: ./images/079eeZ57.pdf Graphic file (type pdf) [81 <./images/079adZ51.pdf>] <./images/080aaZ59.pdf , id=1375, 111.41624pt x 87.32625pt> File: ./images/080aaZ59.pdf Graphic file (type pdf) <./images/080bbZ60.pdf, id=1376, 92.345pt x 93.3487 5pt> File: ./images/080bbZ60.pdf Graphic file (type pdf) [82 <./images/079eeZ57.pdf> <./images/080aaZ59.pdf> <./images/080bbZ60.pdf>] <./images/080ccZ61.pdf, id=1396, 87.32625pt x 84.315p t> File: ./images/080ccZ61.pdf Graphic file (type pdf) <./images/080ddZ62.pdf, id=1397, 66.2475pt x 66.247 5pt> File: ./images/080ddZ62.pdf Graphic file (type pdf) <./images/080eeZ63.pdf, id=1398, 83.31125pt x 75.28 125pt> File: ./images/080eeZ63.pdf Graphic file (type pdf) <./images/080ffZ64.pdf, id=1399, 53.19875pt x 67.25 125pt> File: ./images/080ffZ64.pdf Graphic file (type pdf) <./images/081aaZ65.pdf, id=1400, 79.29625pt x 91.34 125pt> File: ./images/081aaZ65.pdf Graphic file (type pdf) <./images/081bbZ66.pdf, id=1401, 72.27pt x 67.25125 pt> File: ./images/081bbZ66.pdf Graphic file (type pdf) [83 <./images/080ccZ61.pdf> <./images/080ddZ62.pdf> <./images/080eeZ63.pdf> <./images/080ffZ64.pdf>] <./images/081ccZ67.pdf, id=14 26, 81.30376pt x 67.25125pt> File: ./images/081ccZ67.pdf Graphic file (type pdf) <./images/081ddZ68.pdf, id=1427, 88.33pt x 58.2175p t> File: ./images/081ddZ68.pdf Graphic file (type pdf) <./images/081eeZ69.pdf, id=1428, 95.35625pt x 49.18 375pt> File: ./images/081eeZ69.pdf Graphic file (type pdf) [84 <./images/081aaZ65.pdf> <./images/081bbZ66.pdf> <./images/081ccZ67.pdf> <./images/081ddZ68.pdf>] <./images/081ffZ73.pdf, id=14 53, 92.345pt x 57.21375pt> File: ./images/081ffZ73.pdf Graphic file (type pdf) <./images/082adZ74.pdf, id=1454, 380.42125pt x 94.3 525pt> File: ./images/082adZ74.pdf Graphic file (type pdf) [85 <./images/081eeZ69.pdf> <./images/081ffZ73.pdf> ] <./images/082ehZ78.pdf, id=1469, 382.42876pt x 94.3525pt> File: ./images/082ehZ78.pdf Graphic file (type pdf) <./images/082iiZ82.pdf, id=1470, 91.34125pt x 84.31 5pt> File: ./images/082iiZ82.pdf Graphic file (type pdf) [86 <./images/082adZ74.pdf> <./images/082ehZ78.pdf> ] <./images/083adZ83.pdf, id=1485, 417.56pt x 84.315pt> File: ./images/083adZ83.pdf Graphic file (type pdf) [87 <./images/082iiZ82.pdf> <./images/083adZ83.pdf> ] <./images/083eeZ88.pdf, id=1501, 109.40875pt x 94.3525pt> File: ./images/083eeZ88.pdf Graphic file (type pdf) <./images/083ffZ89.pdf, id=1502, 63.23625pt x 67.25 125pt> File: ./images/083ffZ89.pdf Graphic file (type pdf) [88 <./images/083eeZ88.pdf> <./images/083ffZ89.pdf> ] . Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 4330. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 4330. <./images/084aa220.pdf, id=1517, 123.46124pt x 115.43124pt> File: ./images/084aa220.pdf Graphic file (type pdf) [89 <./images/084aa220.pdf>] <./images/085ac224.pdf, id=1529, 373.395pt x 124.465p t> File: ./images/085ac224.pdf Graphic file (type pdf) [90 <./images/085ac224.pdf>] <./images/086aa235.pdf , id=1545, 108.405pt x 124.465pt> File: ./images/086aa235.pdf Graphic file (type pdf) [91 <./images/086aa235.pdf>] <./images/087ab236.pdf , id=1555, 263.98625pt x 104.39pt> File: ./images/087ab236.pdf Graphic file (type pdf) [92 <./images/087ab236.pdf>] [93] <./images/089ab24 1.pdf, id=1573, 250.9375pt x 98.3675pt> File: ./images/089ab241.pdf Graphic file (type pdf) [94] [95 <./images/089ab241.pdf>] <./images/090ab24 3.pdf, id=1590, 250.9375pt x 98.3675pt> File: ./images/090ab243.pdf Graphic file (type pdf) [96] [97 <./images/090ab243.pdf>] <./images/091aa24 5.pdf, id=1607, 108.405pt x 123.46124pt> File: ./images/091aa245.pdf Graphic file (type pdf) [98 <./images/091aa245.pdf>] <./images/092aa249.pdf , id=1617, 114.4275pt x 112.42pt> File: ./images/092aa249.pdf Graphic file (type pdf) [99 <./images/092aa249.pdf>] <./images/093aa250.pdf , id=1628, 101.37875pt x 98.3675pt> File: ./images/093aa250.pdf Graphic file (type pdf) [100 <./images/093aa250.pdf>] <./images/094aa251.pd f, id=1638, 101.37875pt x 98.3675pt> File: ./images/094aa251.pdf Graphic file (type pdf) [101 <./images/094aa251.pdf>] <./images/095aa253.pd f, id=1648, 169.63374pt x 115.43124pt> File: ./images/095aa253.pdf Graphic file (type pdf) [102 <./images/095aa253.pdf>] <./images/096ac257.pd f, id=1658, 412.54124pt x 104.39pt> File: ./images/096ac257.pdf Graphic file (type pdf) [103] [104 <./images/096ac257.pdf>] <./images/097aa 258.pdf, id=1677, 121.45375pt x 117.43875pt> File: ./images/097aa258.pdf Graphic file (type pdf) [105 <./images/097aa258.pdf>] <./images/098aa261.pd f, id=1688, 188.705pt x 111.41624pt> File: ./images/098aa261.pdf Graphic file (type pdf) [106 <./images/098aa261.pdf>] <./images/099aa264.pd f, id=1698, 147.55125pt x 93.34875pt> File: ./images/099aa264.pdf Graphic file (type pdf) [107 <./images/099aa264.pdf>] <./images/100aa265.pd f, id=1716, 159.59625pt x 108.405pt> File: ./images/100aa265.pdf Graphic file (type pdf) [108 <./images/100aa265.pdf>] [109] [110] <./images /103aa272.pdf, id=1742, 269.005pt x 24.09pt> File: ./images/103aa272.pdf Graphic file (type pdf) <./images/103bb273.pdf, id=1743, 145.54375pt x 145. 54375pt> File: ./images/103bb273.pdf Graphic file (type pdf) [111 <./images/103aa272.pdf>] [112 <./images/103bb2 73.pdf>] [113] [114] [115] [116] <./images/108aa286.pdf, id=1786, 255.95625pt x 63.23625pt> File: ./images/108aa286.pdf Graphic file (type pdf) [117 <./images/108aa286.pdf>] [118] <./images/109ac 287.pdf, id=1801, 353.32pt x 108.405pt> File: ./images/109ac287.pdf Graphic file (type pdf) [119 <./images/109ac287.pdf>] [120] <./images/111ac289.pdf, id=1822, 366.36874pt x 120.45pt> File: ./images/111ac289.pdf Graphic file (type pdf) [121] [122 <./images/111ac289.pdf>] <./images/112ac 290.pdf, id=1838, 370.38374pt x 107.40125pt> File: ./images/112ac290.pdf Graphic file (type pdf) [123 <./images/112ac290.pdf>] <./images/113aa294.pd f, id=1851, 114.4275pt x 98.3675pt> File: ./images/113aa294.pdf Graphic file (type pdf) [124 <./images/113aa294.pdf>] <./images/114aa295.pd f, id=1861, 151.56625pt x 106.3975pt> File: ./images/114aa295.pdf Graphic file (type pdf) <./images/115ac296.pdf, id=1862, 340.27126pt x 114. 4275pt> File: ./images/115ac296.pdf Graphic file (type pdf) [125 <./images/114aa295.pdf>] <./images/115dd297.pd f, id=1872, 93.34875pt x 95.35625pt> File: ./images/115dd297.pdf Graphic file (type pdf) <./images/115ee297.pdf, id=1873, 105.39375pt x 118. 4425pt> File: ./images/115ee297.pdf Graphic file (type pdf) [126 <./images/115ac296.pdf>] <./images/116ag299.pd f, id=1883, 423.5825pt x 78.2925pt> File: ./images/116ag299.pdf Graphic file (type pdf) [127 <./images/115dd297.pdf> <./images/115ee297.pdf >] [128 <./images/116ag299.pdf>] [129] <./images/117aa104.pdf, id=1942, 68.255p t x 66.2475pt> File: ./images/117aa104.pdf Graphic file (type pdf) <./images/117bb105.pdf, id=1943, 70.2625pt x 56.21p t> File: ./images/117bb105.pdf Graphic file (type pdf) <./images/118aa107.pdf, id=1944, 105.39375pt x 65.2 4374pt> File: ./images/118aa107.pdf Graphic file (type pdf) <./images/118be108.pdf, id=1945, 410.53375pt x 93.3 4875pt> File: ./images/118be108.pdf Graphic file (type pdf) [130 <./images/117aa104.pdf> <./images/117bb105.pdf > <./images/118aa107.pdf>] <./images/119aa120.pdf, id=1966, 108.405pt x 74.2775 pt> File: ./images/119aa120.pdf Graphic file (type pdf) <./images/119bb121.pdf, id=1967, 69.25874pt x 50.18 75pt> File: ./images/119bb121.pdf Graphic file (type pdf) [131 <./images/118be108.pdf>] <./images/119cc122.pd f, id=1977, 75.28125pt x 52.195pt> File: ./images/119cc122.pdf Graphic file (type pdf) <./images/119dd123.pdf, id=1978, 105.39375pt x 78.2 925pt> File: ./images/119dd123.pdf Graphic file (type pdf) <./images/119ee125.pdf, id=1979, 74.2775pt x 74.277 5pt> File: ./images/119ee125.pdf Graphic file (type pdf) <./images/119ff126.pdf, id=1980, 92.345pt x 64.24pt > File: ./images/119ff126.pdf Graphic file (type pdf) [132 <./images/119aa120.pdf> <./images/119bb121.pdf > <./images/119cc122.pdf> <./images/119dd123.pdf>] <./images/120aa130.pdf, id=2 008, 80.3pt x 80.3pt> File: ./images/120aa130.pdf Graphic file (type pdf) <./images/120bb131.pdf, id=2009, 64.24pt x 73.27374 pt> File: ./images/120bb131.pdf Graphic file (type pdf) [133 <./images/119ee125.pdf> <./images/119ff126.pdf >] <./images/120cd132.pdf, id=2024, 307.1475pt x 130.4875pt> File: ./images/120cd132.pdf Graphic file (type pdf) [134 <./images/120aa130.pdf> <./images/120bb131.pdf >] <./images/121aa300.pdf, id=2042, 206.7725pt x 123.46124pt> File: ./images/121aa300.pdf Graphic file (type pdf) [135 <./images/120cd132.pdf>] <./images/122ab301.pd f, id=2053, 340.27126pt x 128.48pt> File: ./images/122ab301.pdf Graphic file (type pdf) [136 <./images/121aa300.pdf>] <./images/123aa302.pd f, id=2065, 160.6pt x 142.5325pt> File: ./images/123aa302.pdf Graphic file (type pdf) [137 <./images/122ab301.pdf>] <./images/123bb303.pd f, id=2079, 150.5625pt x 131.49126pt> File: ./images/123bb303.pdf Graphic file (type pdf) [138 <./images/123aa302.pdf>] <./images/124aa304.pd f, id=2089, 152.57pt x 178.6675pt> File: ./images/124aa304.pdf Graphic file (type pdf) [139 <./images/123bb303.pdf>] <./images/125ab305.pd f, id=2099, 292.09125pt x 90.3375pt> File: ./images/125ab305.pdf Graphic file (type pdf) [140 <./images/124aa304.pdf>] <./images/125cc306.pd f, id=2111, 206.7725pt x 127.47626pt> File: ./images/125cc306.pdf Graphic file (type pdf) [141 <./images/125ab305.pdf>] <./images/126aa307.pd f, id=2122, 273.02pt x 113.42375pt> File: ./images/126aa307.pdf Graphic file (type pdf) [142 <./images/125cc306.pdf>] <./images/127ab308.pd f, id=2132, 246.9225pt x 123.46124pt> File: ./images/127ab308.pdf Graphic file (type pdf) [143 <./images/126aa307.pdf>] <./images/127cd309.pd f, id=2142, 355.3275pt x 131.49126pt> File: ./images/127cd309.pdf Graphic file (type pdf) [144 <./images/127ab308.pdf>] <./images/128ab310.pd f, id=2154, 377.41pt x 98.3675pt> File: ./images/128ab310.pdf Graphic file (type pdf) [145 <./images/127cd309.pdf>] <./images/129ab311.pd f, id=2166, 305.14pt x 111.41624pt> File: ./images/129ab311.pdf Graphic file (type pdf) [146 <./images/128ab310.pdf>] <./images/129cc311.pd f, id=2178, 148.555pt x 86.3225pt> File: ./images/129cc311.pdf Graphic file (type pdf) <./images/129dd311.pdf, id=2179, 155.58125pt x 90.3 375pt> File: ./images/129dd311.pdf Graphic file (type pdf) <./images/130aa311.pdf, id=2180, 101.37875pt x 54.2 025pt> File: ./images/130aa311.pdf Graphic file (type pdf) [147 <./images/129ab311.pdf> <./images/129cc311.pdf > <./images/129dd311.pdf>] <./images/130bb311.pdf, id=2208, 113.42375pt x 54.20 25pt> File: ./images/130bb311.pdf Graphic file (type pdf) <./images/130cc311.pdf, id=2209, 88.33pt x 53.19875 pt> File: ./images/130cc311.pdf Graphic file (type pdf) <./images/130dd311.pdf, id=2210, 112.42pt x 51.1912 5pt> File: ./images/130dd311.pdf Graphic file (type pdf) <./images/131aa312.pdf, id=2211, 274.02374pt x 93.3 4875pt> File: ./images/131aa312.pdf Graphic file (type pdf) [148 <./images/130aa311.pdf> <./images/130bb311.pdf > <./images/130cc311.pdf> <./images/130dd311.pdf>] <./images/131bb313.pdf, id=2 245, 352.31625pt x 98.3675pt> File: ./images/131bb313.pdf Graphic file (type pdf) [149 <./images/131aa312.pdf>] <./images/132aa314.pd f, id=2255, 117.43875pt x 120.45pt> File: ./images/132aa314.pdf Graphic file (type pdf) [150 <./images/131bb313.pdf>] <./images/133aa315.pd f, id=2265, 285.065pt x 86.3225pt> File: ./images/133aa315.pdf Graphic file (type pdf) [151 <./images/132aa314.pdf>] <./images/133bb316.pd f, id=2275, 157.58875pt x 147.55125pt> File: ./images/133bb316.pdf Graphic file (type pdf) [152 <./images/133aa315.pdf>] <./images/134aa317.pd f, id=2285, 258.9675pt x 93.34875pt> File: ./images/134aa317.pdf Graphic file (type pdf) [153 <./images/133bb316.pdf>] <./images/135aa318.pd f, id=2296, 137.51375pt x 100.375pt> File: ./images/135aa318.pdf Graphic file (type pdf) [154 <./images/134aa317.pdf>] [155 <./images/135aa3 18.pdf>] <./images/137aa147.pdf, id=2315, 119.44624pt x 91.34125pt> File: ./images/137aa147.pdf Graphic file (type pdf) [156] <./images/137bb148.pdf, id=2321, 210.7875pt x 78.2925pt> File: ./images/137bb148.pdf Graphic file (type pdf) [157 <./images/137aa147.pdf>] [158 <./images/137bb1 48.pdf>] <./images/138aa158.pdf, id=2340, 50.1875pt x 52.195pt> File: ./images/138aa158.pdf Graphic file (type pdf) <./images/138bb165.pdf, id=2341, 71.26625pt x 55.20 625pt> File: ./images/138bb165.pdf Graphic file (type pdf) [159 <./images/138aa158.pdf>] <./images/139aa167.pd f, id=2355, 80.3pt x 44.165pt> File: ./images/139aa167.pdf Graphic file (type pdf) <./images/139bb168.pdf, id=2356, 60.225pt x 49.1837 5pt> File: ./images/139bb168.pdf Graphic file (type pdf) <./images/139cc169.pdf, id=2357, 54.2025pt x 42.157 5pt> File: ./images/139cc169.pdf Graphic file (type pdf) [160 <./images/138bb165.pdf> <./images/139aa167.pdf > <./images/139bb168.pdf> <./images/139cc169.pdf>] <./images/139dd174.pdf, id=2 382, 138.5175pt x 62.2325pt> File: ./images/139dd174.pdf Graphic file (type pdf) <./images/139ee175.pdf, id=2383, 103.38625pt x 58.2 175pt> File: ./images/139ee175.pdf Graphic file (type pdf) <./images/139gg176.pdf, id=2384, 68.255pt x 52.195p t> File: ./images/139gg176.pdf Graphic file (type pdf) <./images/139ff177.pdf, id=2385, 97.36375pt x 91.34 125pt> File: ./images/139ff177.pdf Graphic file (type pdf) <./images/140aa178.pdf, id=2386, 64.24pt x 57.21375 pt> File: ./images/140aa178.pdf Graphic file (type pdf) <./images/140bb182.pdf, id=2387, 100.375pt x 54.202 5pt> File: ./images/140bb182.pdf Graphic file (type pdf) [161 <./images/139dd174.pdf> <./images/139ee175.pdf > <./images/139gg176.pdf> <./images/139ff177.pdf> <./images/140aa178.pdf>] <./i mages/140cc191.pdf, id=2417, 82.3075pt x 87.32625pt> File: ./images/140cc191.pdf Graphic file (type pdf) [162 <./images/140bb182.pdf> <./images/140cc191.pdf >] <./images/141aa206.pdf, id=2432, 72.27pt x 51.19125pt> File: ./images/141aa206.pdf Graphic file (type pdf) <./images/141bb207.pdf, id=2433, 102.3825pt x 121.4 5375pt> File: ./images/141bb207.pdf Graphic file (type pdf) [163] [164 <./images/141aa206.pdf> <./images/141bb2 07.pdf>] <./images/142aa223.pdf, id=2452, 59.22125pt x 53.19875pt> File: ./images/142aa223.pdf Graphic file (type pdf) [165 <./images/142aa223.pdf>] <./images/143aa241.pd f, id=2463, 60.225pt x 49.18375pt> File: ./images/143aa241.pdf Graphic file (type pdf) <./images/143bb242.pdf, id=2464, 92.345pt x 66.2475 pt> File: ./images/143bb242.pdf Graphic file (type pdf) <./images/143cc243.pdf, id=2465, 91.34125pt x 76.28 5pt> File: ./images/143cc243.pdf Graphic file (type pdf) <./images/143dd244.pdf, id=2466, 96.36pt x 62.2325p t> File: ./images/143dd244.pdf Graphic file (type pdf) [166 <./images/143aa241.pdf> <./images/143bb242.pdf > <./images/143cc243.pdf> <./images/143dd244.pdf>] <./images/143ee245.pdf, id=2 491, 238.8925pt x 99.37125pt> File: ./images/143ee245.pdf Graphic file (type pdf) [167 <./images/143ee245.pdf>] . Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 7214. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 7214. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 7214. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 7214. [168 ] [169] [170] [171] [172] [173] [174] [175] <./images/151aa342.pdf, id=2537, 35 3.32pt x 103.38625pt> File: ./images/151aa342.pdf Graphic file (type pdf) [176] [177 <./images/151aa342.pdf>] <./images/152aa 344.pdf, id=2552, 90.3375pt x 75.28125pt> File: ./images/152aa344.pdf Graphic file (type pdf) [178 <./images/152aa344.pdf>] <./images/153aa345.pd f, id=2562, 197.73875pt x 98.3675pt> File: ./images/153aa345.pdf Graphic file (type pdf) [179 <./images/153aa345.pdf>] <./images/154aa346.pd f, id=2572, 236.885pt x 20.075pt> File: ./images/154aa346.pdf Graphic file (type pdf) <./images/154bb346.pdf, id=2573, 221.82875pt x 18.0 675pt> File: ./images/154bb346.pdf Graphic file (type pdf) <./images/155aa348.pdf, id=2574, 174.6525pt x 83.31 125pt> File: ./images/155aa348.pdf Graphic file (type pdf) [180 <./images/154aa346.pdf> <./images/154bb346.pdf >] [181 <./images/155aa348.pdf>] <./images/156aa349.pdf, id=2604, 170.6375pt x 93.34875pt> File: ./images/156aa349.pdf Graphic file (type pdf) <./images/157aa351.pdf, id=2607, 193.72375pt x 76.2 85pt> File: ./images/157aa351.pdf Graphic file (type pdf) [182 <./images/156aa349.pdf>] <./images/157bb353.pd f, id=2621, 313.17pt x 36.135pt> File: ./images/157bb353.pdf Graphic file (type pdf) <./images/158aa354.pdf, id=2622, 218.8175pt x 88.33 pt> File: ./images/158aa354.pdf Graphic file (type pdf) [183 <./images/157aa351.pdf> <./images/157bb353.pdf >] <./images/159aa357.pdf, id=2646, 218.8175pt x 88.33pt> File: ./images/159aa357.pdf Graphic file (type pdf) [184 <./images/158aa354.pdf>] <./images/160aa358.pd f, id=2659, 218.8175pt x 88.33pt> File: ./images/160aa358.pdf Graphic file (type pdf) [185 <./images/159aa357.pdf>] <./images/161aa359.pd f, id=2672, 283.0575pt x 100.375pt> File: ./images/161aa359.pdf Graphic file (type pdf) [186 <./images/160aa358.pdf>] [187 <./images/161aa3 59.pdf>] <./images/162aa361.pdf, id=2697, 282.05376pt x 80.3pt> File: ./images/162aa361.pdf Graphic file (type pdf) <./images/163aa362.pdf, id=2698, 275.0275pt x 106.3 975pt> File: ./images/163aa362.pdf Graphic file (type pdf) [188 <./images/162aa361.pdf>] <./images/164aa363.pd f, id=2711, 101.37875pt x 112.42pt> File: ./images/164aa363.pdf Graphic file (type pdf) [189 <./images/163aa362.pdf>] <./images/165aa364.pd f, id=2725, 278.03876pt x 135.50626pt> File: ./images/165aa364.pdf Graphic file (type pdf) [190 <./images/164aa363.pdf>] [191 <./images/165aa3 64.pdf>] <./images/166aa365.pdf, id=2744, 278.03876pt x 135.50626pt> File: ./images/166aa365.pdf Graphic file (type pdf) [192] [193 <./images/166aa365.pdf>] <./images/167aa 366.pdf, id=2758, 278.03876pt x 135.50626pt> File: ./images/167aa366.pdf Graphic file (type pdf) [194 <./images/167aa366.pdf>] <./images/168aa263.pd f, id=2768, 63.23625pt x 68.255pt> File: ./images/168aa263.pdf Graphic file (type pdf) [195] [196 <./images/168aa263.pdf>] <./images/169aa 367.pdf, id=2783, 153.57375pt x 83.31125pt> File: ./images/169aa367.pdf Graphic file (type pdf) [197 <./images/169aa367.pdf>] <./images/170aa370.pdf, id=2799, 109.40875pt x 64.24p t> File: ./images/170aa370.pdf Graphic file (type pdf) [198 <./images/170aa370.pdf>] <./images/171aa371.pd f, id=2809, 152.57pt x 83.31125pt> File: ./images/171aa371.pdf Graphic file (type pdf) <./images/171bb373.pdf, id=2810, 60.225pt x 70.2625 pt> File: ./images/171bb373.pdf Graphic file (type pdf) <./images/171cc374.pdf, id=2811, 132.495pt x 69.258 74pt> File: ./images/171cc374.pdf Graphic file (type pdf) [199 <./images/171aa371.pdf> <./images/171bb373.pdf > <./images/171cc374.pdf>] <./images/172aa375.pdf, id=2831, 287.0725pt x 112.42 pt> File: ./images/172aa375.pdf Graphic file (type pdf) <./images/173aa376.pdf, id=2832, 155.58125pt x 112. 42pt> File: ./images/173aa376.pdf Graphic file (type pdf) [200 <./images/172aa375.pdf>] [201 <./images/173aa3 76.pdf>] <./images/174aa377.pdf, id=2852, 206.7725pt x 75.28125pt> File: ./images/174aa377.pdf Graphic file (type pdf) <./images/175aa378.pdf, id=2853, 114.4275pt x 101.3 7875pt> File: ./images/175aa378.pdf Graphic file (type pdf) [202 <./images/174aa377.pdf>] [203 <./images/175aa3 78.pdf>] <./images/176aa381.pdf, id=2875, 105.39375pt x 126.4725pt> File: ./images/176aa381.pdf Graphic file (type pdf) [204 <./images/176aa381.pdf>] <./images/177aa383.pd f, id=2888, 111.41624pt x 108.405pt> File: ./images/177aa383.pdf Graphic file (type pdf) [205 <./images/177aa383.pdf>] <./images/178aa384.pd f, id=2901, 111.41624pt x 99.37125pt> File: ./images/178aa384.pdf Graphic file (type pdf) [206 <./images/178aa384.pdf>] <./images/179aa270.pd f, id=2914, 95.35625pt x 64.24pt> File: ./images/179aa270.pdf Graphic file (type pdf) [207 <./images/179aa270.pdf>] <./images/180aa280.pd f, id=2925, 48.18pt x 51.19125pt> File: ./images/180aa280.pdf Graphic file (type pdf) <./images/180bb281.pdf, id=2926, 69.25874pt x 50.18 75pt> File: ./images/180bb281.pdf Graphic file (type pdf) [208 <./images/180aa280.pdf> <./images/180bb281.pdf >] [209] <./images/181aa385.pdf, id=2945, 174.6525pt x 107.40125pt> File: ./images/181aa385.pdf Graphic file (type pdf) <./images/182aa386.pdf, id=2946, 159.59625pt x 106. 3975pt> File: ./images/182aa386.pdf Graphic file (type pdf) [210 <./images/181aa385.pdf>] <./images/183aa387.pdf, id=2956, 192.72pt x 113.42375 pt> File: ./images/183aa387.pdf Graphic file (type pdf) [211 <./images/182aa386.pdf>] <./images/184aa388.pd f, id=2966, 324.21124pt x 101.37875pt> File: ./images/184aa388.pdf Graphic file (type pdf) [212 <./images/183aa387.pdf>] <./images/185aa390.pd f, id=2976, 336.25626pt x 115.43124pt> File: ./images/185aa390.pdf Graphic file (type pdf) [213 <./images/184aa388.pdf>] <./images/186aa391.pd f, id=2987, 268.00125pt x 125.46875pt> File: ./images/186aa391.pdf Graphic file (type pdf) [214 <./images/185aa390.pdf>] [215 <./images/186aa3 91.pdf>] <./images/187aa291.pdf, id=3009, 118.4425pt x 72.27pt> File: ./images/187aa291.pdf Graphic file (type pdf) <./images/187bb292.pdf, id=3010, 64.24pt x 56.21pt> File: ./images/187bb292.pdf Graphic file (type pdf) <./images/187cc293.pdf, id=3011, 82.3075pt x 74.277 5pt> File: ./images/187cc293.pdf Graphic file (type pdf) <./images/187dd294.pdf, id=3012, 76.285pt x 73.2737 4pt> File: ./images/187dd294.pdf Graphic file (type pdf) <./images/187ee295.pdf, id=3013, 106.3975pt x 76.28 5pt> File: ./images/187ee295.pdf Graphic file (type pdf) <./images/187ff296.pdf, id=3014, 89.33376pt x 79.29 625pt> File: ./images/187ff296.pdf Graphic file (type pdf) <./images/187gg297.pdf, id=3015, 71.26625pt x 80.3p t> File: ./images/187gg297.pdf Graphic file (type pdf) <./images/187hh298.pdf, id=3016, 108.405pt x 71.266 25pt> File: ./images/187hh298.pdf Graphic file (type pdf) [216 <./images/187aa291.pdf> <./images/187bb292.pdf > <./images/187cc293.pdf> <./images/187dd294.pdf> <./images/187ee295.pdf> <./im ages/187ff296.pdf> <./images/187gg297.pdf> <./images/187hh298.pdf>] <./images/1 88aa303.pdf, id=3064, 65.24374pt x 57.21375pt> File: ./images/188aa303.pdf Graphic file (type pdf) [217 <./images/188aa303.pdf>] <./images/189aa312.pd f, id=3074, 71.26625pt x 74.2775pt> File: ./images/189aa312.pdf Graphic file (type pdf) [218] [219 <./images/189aa312.pdf>] <./images/189bb 313.pdf, id=3090, 71.26625pt x 61.22874pt> File: ./images/189bb313.pdf Graphic file (type pdf) <./images/190aa314.pdf, id=3091, 104.39pt x 97.3637 5pt> File: ./images/190aa314.pdf Graphic file (type pdf) [220 <./images/189bb313.pdf>] <./images/190bb315.pd f, id=3101, 104.39pt x 90.3375pt> File: ./images/190bb315.pdf Graphic file (type pdf) [221 <./images/190aa314.pdf>] [222 <./images/190bb3 15.pdf>] [223] [224] [225] . Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 9509. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 9509. <./images/193aa395.pdf, id=3133, 357.335pt x 71.26625pt> File: ./images/193aa395.pdf Graphic file (type pdf) [226 <./images/193aa395.pdf>] <./images/194aa395.pdf, id=3143, 290.08376pt x 67.251 25pt> File: ./images/194aa395.pdf Graphic file (type pdf) [227 <./images/194aa395.pdf>] <./images/195aa397.pd f, id=3153, 341.275pt x 77.28876pt> File: ./images/195aa397.pdf Graphic file (type pdf) <./images/196aa398.pdf, id=3154, 180.675pt x 90.337 5pt> File: ./images/196aa398.pdf Graphic file (type pdf) [228 <./images/195aa397.pdf>] <./images/196bb399.pd f, id=3167, 136.51pt x 80.3pt> File: ./images/196bb399.pdf Graphic file (type pdf) <./images/197aa400.pdf, id=3168, 365.365pt x 75.281 25pt> File: ./images/197aa400.pdf Graphic file (type pdf) [229 <./images/196aa398.pdf> <./images/196bb399.pdf >] <./images/198aa403.pdf, id=3183, 170.6375pt x 81.30376pt> File: ./images/198aa403.pdf Graphic file (type pdf) [230 <./images/197aa400.pdf>] <./images/199aa407.pd f, id=3193, 177.66376pt x 107.40125pt> File: ./images/199aa407.pdf Graphic file (type pdf) [231 <./images/198aa403.pdf>] <./images/199bb409.pd f, id=3204, 172.645pt x 105.39375pt> File: ./images/199bb409.pdf Graphic file (type pdf) [232 <./images/199aa407.pdf>] <./images/200aa410.pd f, id=3217, 129.48375pt x 132.495pt> File: ./images/200aa410.pdf Graphic file (type pdf) [233 <./images/199bb409.pdf>] [234 <./images/200aa4 10.pdf>] <./images/201ab411.pdf, id=3236, 150.5625pt x 147.55125pt> File: ./images/201ab411.pdf Graphic file (type pdf) [235 <./images/201ab411.pdf>] <./images/202ab412.pdf, id=3249, 140.525pt x 240.9pt> File: ./images/202ab412.pdf Graphic file (type pdf) [236 <./images/202ab412.pdf>] <./images/203aa415.pd f, id=3262, 126.4725pt x 170.6375pt> File: ./images/203aa415.pdf Graphic file (type pdf) [237] [238 <./images/203aa415.pdf>] <./images/204aa 356.pdf, id=3277, 81.30376pt x 79.29625pt> File: ./images/204aa356.pdf Graphic file (type pdf) <./images/204bb357.pdf, id=3278, 81.30376pt x 106.3 975pt> File: ./images/204bb357.pdf Graphic file (type pdf) <./images/204cc358.pdf, id=3279, 80.3pt x 95.35625p t> File: ./images/204cc358.pdf Graphic file (type pdf) [239 <./images/204aa356.pdf> <./images/204bb357.pdf > <./images/204cc358.pdf>] [240] [241] <./images/206aa417.pdf, id=3307, 317.185 pt x 77.28876pt> File: ./images/206aa417.pdf Graphic file (type pdf) <./images/207aa418.pdf, id=3308, 285.065pt x 74.277 5pt> File: ./images/207aa418.pdf Graphic file (type pdf) [242 <./images/206aa417.pdf>] [243 <./images/207aa418.pdf>] <./images/208aa419.pdf, id=3334, 380.42125pt x 110.4125pt> File: ./images/208aa419.pdf Graphic file (type pdf) [244 <./images/208aa419.pdf>] <./images/209aa420.pd f, id=3347, 188.705pt x 112.42pt> File: ./images/209aa420.pdf Graphic file (type pdf) <./images/210aa421.pdf, id=3348, 385.44pt x 82.3075 pt> File: ./images/210aa421.pdf Graphic file (type pdf) [245 <./images/209aa420.pdf>] <./images/211aa424.pd f, id=3358, 217.81375pt x 73.27374pt> File: ./images/211aa424.pdf Graphic file (type pdf) [246 <./images/210aa421.pdf>] <./images/212aa425.pd f, id=3368, 355.3275pt x 116.435pt> File: ./images/212aa425.pdf Graphic file (type pdf) [247 <./images/211aa424.pdf>] <./images/213aa426.pd f, id=3378, 293.095pt x 87.32625pt> File: ./images/213aa426.pdf Graphic file (type pdf) [248 <./images/212aa425.pdf>] <./images/214aa427.pd f, id=3391, 301.125pt x 123.46124pt> File: ./images/214aa427.pdf Graphic file (type pdf) [249 <./images/213aa426.pdf>] [250 <./images/214aa4 27.pdf>] <./images/215aa428.pdf, id=3414, 270.00874pt x 87.32625pt> File: ./images/215aa428.pdf Graphic file (type pdf) [251 <./images/215aa428.pdf>] [252] <./images/217aa 404.pdf, id=3431, 80.3pt x 86.3225pt> File: ./images/217aa404.pdf Graphic file (type pdf) <./images/217bb405.pdf, id=3432, 95.35625pt x 66.24 75pt> File: ./images/217bb405.pdf Graphic file (type pdf) <./images/217cc406.pdf, id=3433, 113.42375pt x 99.3 7125pt> File: ./images/217cc406.pdf Graphic file (type pdf) [253 <./images/217aa404.pdf> <./images/217bb405.pdf > <./images/217cc406.pdf>] [254] [255] [256] [257] . Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\spacefactor' on input line 10726. Package hyperref Warning: Token not allowed in a PDFDocEncoded string, (hyperref) removing `\@m' on input line 10726. <./images/220aa430.pdf, id=3473, 109.40875pt x 103.38625pt> File: ./images/220aa430.pdf Graphic file (type pdf) <./images/221aa431.pdf, id=3474, 109.40875pt x 105. 39375pt> File: ./images/221aa431.pdf Graphic file (type pdf) [258 <./images/220aa430.pdf>] [259 <./images/221aa431.pdf>] <./images/223aa439.pdf, id=3493, 149.55875pt x 141.52875pt> File: ./images/223aa439.pdf Graphic file (type pdf) [260] <./images/224aa441.pdf, id=3498, 112.42pt x 1 03.38625pt> File: ./images/224aa441.pdf Graphic file (type pdf) <./images/224bb442.pdf, id=3499, 94.3525pt x 94.352 5pt> File: ./images/224bb442.pdf Graphic file (type pdf) <./images/224cc443.pdf, id=3500, 96.36pt x 92.345pt > File: ./images/224cc443.pdf Graphic file (type pdf) [261 <./images/223aa439.pdf>] <./images/225aa445.pd f, id=3511, 193.72375pt x 103.38625pt> File: ./images/225aa445.pdf Graphic file (type pdf) [262 <./images/224aa441.pdf> <./images/224bb442.pdf > <./images/224cc443.pdf>] <./images/226aa447.pdf, id=3538, 193.72375pt x 103.3 8625pt> File: ./images/226aa447.pdf Graphic file (type pdf) [263 <./images/225aa445.pdf>] [264 <./images/226aa4 47.pdf>] <./images/227aa449.pdf, id=3563, 80.3pt x 84.315pt> File: ./images/227aa449.pdf Graphic file (type pdf) [265 <./images/227aa449.pdf>] <./images/228aa451.pd f, id=3573, 100.375pt x 66.2475pt> File: ./images/228aa451.pdf Graphic file (type pdf) <./images/229aa454.pdf, id=3574, 117.43875pt x 125. 46875pt> File: ./images/229aa454.pdf Graphic file (type pdf) [266 <./images/228aa451.pdf>] [267 <./images/229aa4 54.pdf>] <./images/231aa455.pdf, id=3597, 230.8625pt x 118.4425pt> File: ./images/231aa455.pdf Graphic file (type pdf) [268] [269 <./images/231aa455.pdf>] <./images/232aa 459.pdf, id=3614, 131.49126pt x 117.43875pt> File: ./images/232aa459.pdf Graphic file (type pdf) <./images/233aa461.pdf, id=3615, 107.40125pt x 103. 38625pt> File: ./images/233aa461.pdf Graphic file (type pdf) [270 <./images/232aa459.pdf>] [271 <./images/233aa4 61.pdf>] <./images/234aa466.pdf, id=3634, 139.52126pt x 83.31125pt> File: ./images/234aa466.pdf Graphic file (type pdf) [272] [273 <./images/234aa466.pdf>] <./images/235aa 467.pdf, id=3652, 99.37125pt x 99.37125pt> File: ./images/235aa467.pdf Graphic file (type pdf) <./images/236aa469.pdf, id=3653, 113.42375pt x 99.3 7125pt> File: ./images/236aa469.pdf Graphic file (type pdf) [274 <./images/235aa467.pdf>] <./images/237aa472.pdf, id=3663, 104.39pt x 106.3975p t> File: ./images/237aa472.pdf Graphic file (type pdf) [275 <./images/236aa469.pdf>] [276 <./images/237aa4 72.pdf>] <./images/239aa475.pdf, id=3682, 104.39pt x 106.3975pt> File: ./images/239aa475.pdf Graphic file (type pdf) [277] <./images/240aa477.pdf, id=3687, 222.8325pt x 101.37875pt> File: ./images/240aa477.pdf Graphic file (type pdf) [278 <./images/239aa475.pdf>] <./images/241aa478.pd f, id=3697, 99.37125pt x 112.42pt> File: ./images/241aa478.pdf Graphic file (type pdf) [279 <./images/240aa477.pdf>] [280 <./images/241aa4 78.pdf>] <./images/242aa480.pdf, id=3720, 99.37125pt x 99.37125pt> File: ./images/242aa480.pdf Graphic file (type pdf) [281 <./images/242aa480.pdf>] <./images/243aa484.pd f, id=3728, 158.5925pt x 84.315pt> File: ./images/243aa484.pdf Graphic file (type pdf) [282 <./images/243aa484.pdf>] <./images/244aa485.pd f, id=3738, 153.57375pt x 153.57375pt> File: ./images/244aa485.pdf Graphic file (type pdf) [283 <./images/244aa485.pdf>] <./images/245aa486.pd f, id=3750, 228.855pt x 103.38625pt> File: ./images/245aa486.pdf Graphic file (type pdf) <./images/246aa487.pdf, id=3751, 229.85875pt x 113. 42375pt> File: ./images/246aa487.pdf Graphic file (type pdf) [284 <./images/245aa486.pdf>] <./images/247aa488.pd f, id=3761, 117.43875pt x 124.465pt> File: ./images/247aa488.pdf Graphic file (type pdf) [285 <./images/246aa487.pdf>] <./images/248aa490.pd f, id=3775, 192.72pt x 84.315pt> File: ./images/248aa490.pdf Graphic file (type pdf) [286 <./images/247aa488.pdf>] <./images/249aa491.pd f, id=3785, 300.12125pt x 86.3225pt> File: ./images/249aa491.pdf Graphic file (type pdf) [287 <./images/248aa490.pdf>] [288 <./images/249aa4 91.pdf>] [289] [290] [291] [292] [293] [294] [295] [296] [297] [298] [299] [300 ] [301] . [302 ] [303] [304] . [305 ] [306] [307] [308 ] [309 ] [310] [311] [312] [313] [314] [315] [316] (./33063-t.aux) *File List* book.cls 2005/09/16 v1.4f Standard LaTeX document class bk12.clo 2005/09/16 v1.4f Standard LaTeX file (size option) inputenc.sty 2006/05/05 v1.1b Input encoding file latin1.def 2006/05/05 v1.1b Input encoding file ifthen.sty 2001/05/26 v1.1c Standard LaTeX ifthen package (DPC) amsmath.sty 2000/07/18 v2.13 AMS math features amstext.sty 2000/06/29 v2.01 amsgen.sty 1999/11/30 v2.0 amsbsy.sty 1999/11/29 v1.2d amsopn.sty 1999/12/14 v2.01 operator names amssymb.sty 2002/01/22 v2.2d amsfonts.sty 2001/10/25 v2.2f alltt.sty 1997/06/16 v2.0g defines alltt environment footmisc.sty 2005/03/17 v5.3d a miscellany of footnote facilities indentfirst.sty 1995/11/23 v1.03 Indent first paragraph (DPC) fancyhdr.sty graphicx.sty 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR) keyval.sty 1999/03/16 v1.13 key=value parser (DPC) graphics.sty 2006/02/20 v1.0o Standard LaTeX Graphics (DPC,SPQR) trig.sty 1999/03/16 v1.09 sin cos tan (DPC) graphics.cfg 2007/01/18 v1.5 graphics configuration of teTeX/TeXLive pdftex.def 2007/01/08 v0.04d Graphics/color for pdfTeX enumerate.sty 1999/03/05 v3.00 enumerate extensions (DPC) supertabular.sty 2004/02/20 v4.1e the supertabular environment geometry.sty 2002/07/08 v3.2 Page Geometry geometry.cfg hyperref.sty 2007/02/07 v6.75r Hypertext links for LaTeX pd1enc.def 2007/02/07 v6.75r Hyperref: PDFDocEncoding definition (HO) hyperref.cfg 2002/06/06 v1.2 hyperref configuration of TeXLive kvoptions.sty 2006/08/22 v2.4 Connects package keyval with LaTeX options (HO ) url.sty 2005/06/27 ver 3.2 Verb mode for urls, etc. hpdftex.def 2007/02/07 v6.75r Hyperref driver for pdfTeX supp-pdf.tex color.sty 2005/11/14 v1.0j Standard LaTeX Color (DPC) color.cfg 2007/01/18 v1.5 color configuration of teTeX/TeXLive nameref.sty 2006/12/27 v2.28 Cross-referencing by name of section refcount.sty 2006/02/20 v3.0 Data extraction from references (HO) 33063-t.out 33063-t.out umsa.fd 2002/01/19 v2.2g AMS font definitions umsb.fd 2002/01/19 v2.2g AMS font definitions omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions ./images/woodcutsmall.pdf ./images/012aaZ10.pdf ./images/012bbZ14.pdf ./images/016aaZ37.pdf ./images/018aaZ55.pdf ./images/018bbZ56.pdf ./images/018ccZ57.pdf ./images/019aaZ58.pdf ./images/019bbZ58.pdf ./images/019ccZ62.pdf ./images/020aaZ63.pdf ./images/020bbZ66.pdf ./images/020ccZ68.pdf ./images/020ddZ69.pdf ./images/021aaZ72.pdf ./images/022aaZ75.pdf ./images/022bbZ76.pdf ./images/022ccZ77.pdf ./images/023aaZ79.pdf ./images/023bcZ80.pdf ./images/024aaZ81.pdf ./images/024bbZ83.pdf ./images/025aaZ86.pdf ./images/026aaZ90.pdf ./images/027aaZ93.pdf ./images/028aaZ95.pdf ./images/029aaZ96.pdf ./images/030aaZ97.pdf ./images/031aa100.pdf ./images/032aa101.pdf ./images/033aa104.pdf ./images/034aa107.pdf ./images/034bb109.pdf ./images/035aa110.pdf ./images/036aa111.pdf ./images/037aa112.pdf ./images/038aa115.pdf ./images/039aa117.pdf ./images/039bb118.pdf ./images/039ce119.pdf ./images/039fi120.pdf ./images/041aa129.pdf ./images/042aa138.pdf ./images/043ab139.pdf ./images/044ab143.pdf ./images/045aa145.pdf ./images/046aa147.pdf ./images/047ab150.pdf ./images/048ac151.pdf ./images/049aa152.pdf ./images/050aa153.pdf ./images/051ac154.pdf ./images/052ab155.pdf ./images/054aa160.pdf ./images/055aa162.pdf ./images/056ac166.pdf ./images/056dg170.pdf ./images/057aa174.pdf ./images/057bb176.pdf ./images/058aa178.pdf ./images/058bb181.pdf ./images/059aa182.pdf ./images/060aa183.pdf ./images/061aa184.pdf ./images/062ab185.pdf ./images/063aa187.pdf ./images/064aa188.pdf ./images/064bb190.pdf ./images/065ac192.pdf ./images/066ad203.pdf ./images/067aa205.pdf ./images/068aa207.pdf ./images/069ac208.pdf ./images/069dd211.pdf ./images/070aa212.pdf ./images/071aa213.pdf ./images/074aaZ19.pdf ./images/074bbZ20.pdf ./images/075aaZ21.pdf ./images/075bbZ22.pdf ./images/075ceZ23.pdf ./images/076aaZ24.pdf ./images/076bbZ25.pdf ./images/076ccZ26.pdf ./images/076ddZ27.pdf ./images/077adZ28.pdf ./images/077eeZ34.pdf ./images/077ffZ35.pdf ./images/078aaZ41.pdf ./images/078bbZ43.pdf ./images/078ccZ46.pdf ./images/078ddZ49.pdf ./images/079adZ51.pdf ./images/079eeZ57.pdf ./images/080aaZ59.pdf ./images/080bbZ60.pdf ./images/080ccZ61.pdf ./images/080ddZ62.pdf ./images/080eeZ63.pdf ./images/080ffZ64.pdf ./images/081aaZ65.pdf ./images/081bbZ66.pdf ./images/081ccZ67.pdf ./images/081ddZ68.pdf ./images/081eeZ69.pdf ./images/081ffZ73.pdf ./images/082adZ74.pdf ./images/082ehZ78.pdf ./images/082iiZ82.pdf ./images/083adZ83.pdf ./images/083eeZ88.pdf ./images/083ffZ89.pdf ./images/084aa220.pdf ./images/085ac224.pdf ./images/086aa235.pdf ./images/087ab236.pdf ./images/089ab241.pdf ./images/090ab243.pdf ./images/091aa245.pdf ./images/092aa249.pdf ./images/093aa250.pdf ./images/094aa251.pdf ./images/095aa253.pdf ./images/096ac257.pdf ./images/097aa258.pdf ./images/098aa261.pdf ./images/099aa264.pdf ./images/100aa265.pdf ./images/103aa272.pdf ./images/103bb273.pdf ./images/108aa286.pdf ./images/109ac287.pdf ./images/111ac289.pdf ./images/112ac290.pdf ./images/113aa294.pdf ./images/114aa295.pdf ./images/115ac296.pdf ./images/115dd297.pdf ./images/115ee297.pdf ./images/116ag299.pdf ./images/117aa104.pdf ./images/117bb105.pdf ./images/118aa107.pdf ./images/118be108.pdf ./images/119aa120.pdf ./images/119bb121.pdf ./images/119cc122.pdf ./images/119dd123.pdf ./images/119ee125.pdf ./images/119ff126.pdf ./images/120aa130.pdf ./images/120bb131.pdf ./images/120cd132.pdf ./images/121aa300.pdf ./images/122ab301.pdf ./images/123aa302.pdf ./images/123bb303.pdf ./images/124aa304.pdf ./images/125ab305.pdf ./images/125cc306.pdf ./images/126aa307.pdf ./images/127ab308.pdf ./images/127cd309.pdf ./images/128ab310.pdf ./images/129ab311.pdf ./images/129cc311.pdf ./images/129dd311.pdf ./images/130aa311.pdf ./images/130bb311.pdf ./images/130cc311.pdf ./images/130dd311.pdf ./images/131aa312.pdf ./images/131bb313.pdf ./images/132aa314.pdf ./images/133aa315.pdf ./images/133bb316.pdf ./images/134aa317.pdf ./images/135aa318.pdf ./images/137aa147.pdf ./images/137bb148.pdf ./images/138aa158.pdf ./images/138bb165.pdf ./images/139aa167.pdf ./images/139bb168.pdf ./images/139cc169.pdf ./images/139dd174.pdf ./images/139ee175.pdf ./images/139gg176.pdf ./images/139ff177.pdf ./images/140aa178.pdf ./images/140bb182.pdf ./images/140cc191.pdf ./images/141aa206.pdf ./images/141bb207.pdf ./images/142aa223.pdf ./images/143aa241.pdf ./images/143bb242.pdf ./images/143cc243.pdf ./images/143dd244.pdf ./images/143ee245.pdf ./images/151aa342.pdf ./images/152aa344.pdf ./images/153aa345.pdf ./images/154aa346.pdf ./images/154bb346.pdf ./images/155aa348.pdf ./images/156aa349.pdf ./images/157aa351.pdf ./images/157bb353.pdf ./images/158aa354.pdf ./images/159aa357.pdf ./images/160aa358.pdf ./images/161aa359.pdf ./images/162aa361.pdf ./images/163aa362.pdf ./images/164aa363.pdf ./images/165aa364.pdf ./images/166aa365.pdf ./images/167aa366.pdf ./images/168aa263.pdf ./images/169aa367.pdf ./images/170aa370.pdf ./images/171aa371.pdf ./images/171bb373.pdf ./images/171cc374.pdf ./images/172aa375.pdf ./images/173aa376.pdf ./images/174aa377.pdf ./images/175aa378.pdf ./images/176aa381.pdf ./images/177aa383.pdf ./images/178aa384.pdf ./images/179aa270.pdf ./images/180aa280.pdf ./images/180bb281.pdf ./images/181aa385.pdf ./images/182aa386.pdf ./images/183aa387.pdf ./images/184aa388.pdf ./images/185aa390.pdf ./images/186aa391.pdf ./images/187aa291.pdf ./images/187bb292.pdf ./images/187cc293.pdf ./images/187dd294.pdf ./images/187ee295.pdf ./images/187ff296.pdf ./images/187gg297.pdf ./images/187hh298.pdf ./images/188aa303.pdf ./images/189aa312.pdf ./images/189bb313.pdf ./images/190aa314.pdf ./images/190bb315.pdf ./images/193aa395.pdf ./images/194aa395.pdf ./images/195aa397.pdf ./images/196aa398.pdf ./images/196bb399.pdf ./images/197aa400.pdf ./images/198aa403.pdf ./images/199aa407.pdf ./images/199bb409.pdf ./images/200aa410.pdf ./images/201ab411.pdf ./images/202ab412.pdf ./images/203aa415.pdf ./images/204aa356.pdf ./images/204bb357.pdf ./images/204cc358.pdf ./images/206aa417.pdf ./images/207aa418.pdf ./images/208aa419.pdf ./images/209aa420.pdf ./images/210aa421.pdf ./images/211aa424.pdf ./images/212aa425.pdf ./images/213aa426.pdf ./images/214aa427.pdf ./images/215aa428.pdf ./images/217aa404.pdf ./images/217bb405.pdf ./images/217cc406.pdf ./images/220aa430.pdf ./images/221aa431.pdf ./images/223aa439.pdf ./images/224aa441.pdf ./images/224bb442.pdf ./images/224cc443.pdf ./images/225aa445.pdf ./images/226aa447.pdf ./images/227aa449.pdf ./images/228aa451.pdf ./images/229aa454.pdf ./images/231aa455.pdf ./images/232aa459.pdf ./images/233aa461.pdf ./images/234aa466.pdf ./images/235aa467.pdf ./images/236aa469.pdf ./images/237aa472.pdf ./images/239aa475.pdf ./images/240aa477.pdf ./images/241aa478.pdf ./images/242aa480.pdf ./images/243aa484.pdf ./images/244aa485.pdf ./images/245aa486.pdf ./images/246aa487.pdf ./images/247aa488.pdf ./images/248aa490.pdf ./images/249aa491.pdf *********** ) Here is how much of TeX's memory you used: 7259 strings out of 94074 109629 string characters out of 1165154 167599 words of memory out of 1500000 9671 multiletter control sequences out of 10000+50000 19537 words of font info for 73 fonts, out of 1200000 for 2000 645 hyphenation exceptions out of 8191 29i,14n,43p,264b,486s stack positions out of 5000i,500n,6000p,200000b,5000s Output written on 33063-t.pdf (326 pages, 3009904 bytes). PDF statistics: 4377 PDF objects out of 5155 (max. 8388607) 474 named destinations out of 1000 (max. 131072) 2001 words of extra memory for PDF output out of 10000 (max. 10000000)