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\begin{document}
\frontmatter
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\begin{alltt}
* A Distributed Proofreaders Canada eBook *
This eBook is made available at no cost and with very few
restrictions. These restrictions apply only if (1) you make
a change in the eBook (other than alteration for different
display devices), or (2) you are making commercial use of
the eBook. If either of these conditions applies, please
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This work is in the Canadian public domain, but may be under
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country's copyright laws. IF THE BOOK IS UNDER COPYRIGHT
IN YOUR COUNTRY, DO NOT DOWNLOAD OR REDISTRIBUTE THIS FILE.
Title: Geometry for Technical Students
Date of first publication: 1904
Author: E. H. (Ernest Headly) Sprague
Date first posted: Aug. 29, 2018
Date last updated: Aug. 29, 2018
Faded Page eBook #20180889
This eBook was produced by: Ron McBeth, David T. Jones, Howard Ross
\& the online Distributed Proofreaders Canada team at http://www.pgdpcanada.net
\end{alltt}
\clearpage
\subsection*{Transcriber's Note}
Minor typographical corrections and presentational changes have been
made without comment.
\clearpage
\begin{titlepage}
\centering
\vfill
{\bfseries{\LARGE GEOMETRY FOR TECHNICAL STUDENTS}\\
{\small BEING}\\
{\normalsize AN INTRODUCTION TO PURE AND APPLIED GEOMETRY AND THE MENSURATION OF SURFACES AND SOLIDS IN SIMPLE PROPOSITIONS} \\
{\small TO WHICH ARE ADDED}\\
{\normalsize PROBLEMS IN PLANE GEOMETRY FOUND
USEFUL IN DRAWING}\\
\vskip0.5cm
{\small BY} \\
{\large E. H. SPRAGUE, \textsc{Assoc.M.Inst.C.E.}} \\
{\small ASSISTANT IN THE ENGINEERING DEPARTMENT, UNIVERSITY COLLEGE, LONDON} \\
{\small LECTURER ON APPLIED MATHEMATICS AND MECHANICS AT} \\
{\small THE WESTMINSTER TECHNICAL INSTITUTE} \\
{\small LATE PRINCIPAL OF THE IMPERIAL CHINESE RAILWAY COLLEGE} \\
{\small SHAN-HAI-KUAN} \\
\vskip0.5cm
\includegraphics[width=0.3\linewidth]{../images/illus_02_1} \\
\vskip0.5cm
{\normalsize LONDON} \\
{\large CROSBY LOCKWOOD AND SON} \\
{\normalsize 7, STATIONERS' HALL COURT, LUDGATE HILL}}\\
1904
\end{titlepage}
\mainmatter
\chapter*{PREFACE}
If an apology be expected for putting forth yet another
Text-book for the use of Students of Geometry, it must
be based on the Author's experience\textemdash when training
engineering students for the Chinese Government and
in the positions he now holds\textemdash that a small volume
upon the plan attempted to be carried out in the following
pages is required by many students, more especially
those who have to take up Geometry as part of their
professional training as engineers.
The aim and scope of the work will be found explained
in the Introduction (pp. 1, 2).
It is right that the Author should acknowledge here his
great indebtedness to the German treatise of Schlomilch.
\vspace{2cm}
\textsc{University College, London. }
\textit{October, 1903}
%% Can't figure how to have 3rd column bottom aligned with 2nd column
\chapter*{LIST OF PROPOSITIONS}
\begin{longtable}{r p{0.85\textwidth} r}
& & \textsc{PAGE} \\
1.&When two lines cut one another, the opposite angles are equal & \pageref{prop1} \\
2.&If one line meet two parallel lines (i.) the corresponding angles are equal; (ii.) the alternate angle are equal; (iii.) the exterior angles, and (iv.) the interior angles, on the same side of the line, are together equal to two right angles & \pageref{prop2} \\
3.&In any triangle, an exterior angle is equal to the two interior opposite angles & \pageref{prop3} \\
4.&The interior angles of any triangle are together equal to two right angles & \pageref{prop4} \\
5.&A triangle is determined when two angles are given, and a side which is known to be either opposite or adjacent to these angles & \pageref{prop5} \\
6.&A triangle is determined when one angle and two sides are given; except when the given angle lies opposite the smaller of the given sides; in which case there are two triangles having supplementary angles & \pageref{prop6} \\
7.&A triangle is determined when the three sides are given & \pageref{prop7} \\
8.&A parallelogram is bisected by its diagonal & \pageref{prop8} \\
9.&Parallelograms on equal bases and between the same parallels are equal in area & \pageref{prop9} \\
10.&Parallelograms about the diagonal of any parallelogram are equal & \pageref{prop10} \\
11.&Parallelograms of equal altitude are to one another as their bases & \pageref{prop11} \\
12.&The areas of equiangular parallelograms are as the products of their sides & \pageref{prop12} \\
13.&Equiangular triangles are also similar & \pageref{prop13} \\
14.&In any right−angled triangle, the perpendicular on the hypothenuse from the opposite vertex makes triangles which are similar to the whole triangle and to one another & \pageref{prop14} \\
15.&In any triangle the square on one side is less than the sum of the squares on the other two sides, by twice the area of the rectangle contained by either of these and the projection on it of the other & \pageref{prop15} \\
16.&The areas of similar triangles are in the ratio of the squares of their corresponding sides & \pageref{prop16} \\
17.&The areas of similar polygons are in the ratio of the squares of their corresponding sides & \pageref{prop17} \\
18.&The area of any figure described on the hypothenuse of a right−angled triangle is equal to the similar and similarly described figures on the sides about the right angle & \pageref{prop18} \\
19.&If any two similar figures are placed with their corresponding sides parallel, the lines joining corresponding points in the two figures are concurrent & \pageref{prop19} \\
20.&The straight line drawn from the centre at a circle to the middle point of a chord is perpendicular to the chord & \pageref{prop20} \\
21.&The angle at the centre of a circle is double the angle at the circumference standing on the same arc & \pageref{prop21} \\
22.&The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles & \pageref{prop22} \\
23.&In any circle the product of the segments made by the intersection of two chords with the circumference are equal & \pageref{prop23} \\
24.&If a straight line touch a circle, and from the point of contact another straight line be drawn cutting the circle, the angles which this straight line makes with the first at the point of contact are equal to the angles in the adjacent segments & \pageref{prop24} \\
25.&To find the ratio of the circumference of a circle to its diameter & \pageref{prop25} \\
26.&To find the length of a circular arc & \pageref{prop26} \\
27.&To find the area of a rectangle & \pageref{prop27} \\
28.&To find the area of a parallelogram & \pageref{prop28} \\
29.&To find the area of a triangle & \pageref{prop29} \\
30.&To find the area of a trapezium & \pageref{prop30} \\
31.&To find the area of a quadrilateral & \pageref{prop31} \\
32.&To find the area of a polygon & \pageref{prop32} \\
33.&To find the area of a circle and its sector & \pageref{prop33} \\
34.&To find the area of a circular segment & \pageref{prop34} \\
35.&To find areas by a sum−curve & \pageref{prop35} \\
36.&The areas of the sections of a pyramid made by planes parallel to the base, are proportional to the squares of their distances from the vertex & \pageref{prop36} \\
37.&The volume of a right prism is equal to the area of its base multiplied by the height & \pageref{prop37} \\
38.&The volume of an oblique prism is equal to the area of its right section multiplied by its length & \pageref{prop38} \\
39.&Pyramids on equal bases and of equal altitude are equal in volume & \pageref{prop39} \\
40.&The volume of a pyramid is one−third of the prism standing on the same base & \pageref{prop40} \\
41.&To find the volume of a frustum of a triangular pyramid between parallel planes, in terms of its altitude and the areas of its bases & \pageref{prop41} \\
42.&To find the volume of the frustum of a triangular prism & \pageref{prop42} \\
43.&To find the volume of a wedge & \pageref{prop43} \\
44.&To find the lateral surface and volume of a right circular cylinder & \pageref{prop44} \\
45.&To find the lateral surface and volume of a right circular cone & \pageref{prop45} \\
46.&To find the lateral surface of the frustum of a cone & \pageref{prop46} \\
47.&To find the surface of a sphere & \pageref{prop47} \\
48.&To find the volume of a sphere & \pageref{prop48} \\
\end{longtable}
\chapter*{PROBLEMS IN PLANE GEOMETRY FOUND USEFUL IN DRAWING.}
\begin{tabular}{r p{0.85\textwidth} r} %% See comment on previous table
& & \textsc{PAGE} \\
1. & To divide a straight line into two equal parts & \pageref{p1} \\
2. & To divide an angle into two equal parts& \pageref{p2} \\
3. & To divide a line into any number of equal parts& \pageref{p3} \\
4. & To draw a triangle, whose sides are of known length& \pageref{p4} \\
5. & To inscribe a circle in a given triangle& \pageref{p5} \\
6. & To circumscribe a circle about a given triangle& \pageref{p6} \\
7. & To inscribe a hexagon in a given circle& \pageref{p7} \\
8. & To draw a circular arc through three given points& \pageref{p8} \\
9. & To inscribe in a given angle a circle of given radius& \pageref{p9} \\
10.& To describe a circle of given radius to touch a given line and a given circle& \pageref{p10} \\
11.& To describe a circle, whose radius is given, to touch two given circles& \pageref{p11} \\
12.& To describe a circle tangent to a given line at a given point, and touching a given circle & \pageref{p12} \\
13.& To describe a circle tangent to a given line and touching a given circle in a given point& \pageref{p13} \\
14.& To draw a circle to touch three given straight lines & \pageref{p14} \\
\end{tabular}
\chapter*{INTRODUCTION.}
Etymologically the word \textit{geometry} signifies the
measurement of the earth.
Geometry had its origin in the practical needs of the
Egyptians, but Thales of Miletus, about 600 \textsc{B.C.}, first
dealt with the subject in an \textit{abstract} manner. Pythagoras
and his school greatly added to the science, and Euclid,
who taught at Alexandria about 300 \textsc{B.C.} collected
together and arranged the labours of his predecessors
in the famous ``Elements,'' which consists of thirteen
books. This treatise has been in use up to the present
time, with the exception of Books VII., VIII., IX., and
X., which treat of Greek arithmetic and incommensurable
magnitudes, and Book XIII., which treats of the
regular solids.
The first six Books contain 164 propositions, and the
XIth and XIIth Books 58 propositions. The greater
number of these are merely links in the chain of reasoning
by which the more important results are deduced. By
starting with the theory of parallel lines, continental
mathematicians have shown that a large number of the
less important propositions can be omitted, and the same
results obtained, without affecting the precision of the
method; and this is the course adopted in the present
work.
Moreover, by introducing early the idea of ratio, many
of Euclid's proofs are materially simplified, so that
although, in the words of Euclid to Ptolemy, there may
be no royal road to geometry, it is believed that the
student may find in the following pages an easier guide
to his requirements than our ordinary text-books.
It will be noted that the definitions of terms are
distributed throughout the work as required for the
elucidation of successive Propositions, and that in the
case of many Propositions corollaries are stated, and
examples appended, where deemed advisable.
\textbf{Symbols.}\textemdash The following symbols are commonly used
for the sake of abbreviation:\textemdash
\begin{tabular}{cclcccl}
$\therefore$ & meaning & \textit{therefore.} & & $ \angle $ & meaning & \textit{angle.} \\
$ \because $ & " & \textit{because. }& & $ \triangle $ & " & \textit{triangle.} \\
$ = $ & " & \textit{equal to.} & & $\parallelogram$ & " & \textit{parallelogram.} \\
$ \parallel $ & " & \textit{parallel to.} & & $ \bigcirc $ & " & \textit{circle.} \\
$ + $ & " & \textit{addition.} & & $ - $ & " & \textit{subtraction.} \\
\end{tabular}
\chapter*{GEOMETRY FOR TECHNICAL STUDENTS.}
\section*{Preliminary Definitions.}
1. \textbf{Geometry.}\textemdash Geometry is the science which treats
of the properties of space.
The object of geometry is, starting from facts whose
truth is universally recognised, to deduce therefrom
results, the truth of which, being less apparent, can only
be established by a chain of connected reasoning.
2. \textbf{Axiom.}\textemdash Self-evident facts are in geometry called
\textit{Axioms}\textemdash for example, ``The whole of a thing is greater
than a part of it;'' ``If equals be added to equals the
wholes are equal;'' and so on. Such axioms must be
admitted as fundamental truths, for no proof can make
them clearer.
3. \textbf{Proposition.}\textemdash A \textit{Proposition} is the statement of
something which it is required to do. Propositions are
divided into Problems and Theorems.
4. \textbf{Problem.}\textemdash A \textit{Problem} is a proposition which states
that a certain thing is required to be done; \textit{e.g.}, ``To
bisect a given angle.''
5. \textbf{Theorem.}\textemdash A \textit{Theorem} is a proposition which states
that a certain assertion is to be proved true; \textit{e.g.}, ``Any
two sides of a triangle are together greater than the third.''
6. \textbf{Hypothesis.}\textemdash A Theorem consists of two parts; viz.,
the \textit{Hypothesis}, or assumption; and the \textit{Conclusion}, or
that which follows from the reasoning based on the
hypothesis. Thus,
\begin{quotation}
If two sides of a triangle are equal (hypothesis)
The angles opposite to those sides are equal (conclusion).
\end{quotation}
7. \textbf{Converse.}\textemdash One Theorem is said to be the \textit{Converse}
of another, when the hypothesis and conclusion are
interchanged. Thus, the converse of the above theorem
would be
\begin{quotation}
If two angles of a triangle are equal (hypothesis)
The sides opposite to those angles are equal (conclusion).
\end{quotation}
8. \textbf{Corollary.}\textemdash \textit{Corollary} is a deduction which follows
easily from a proposition already established.
\section*{The Elements of Geometrical Form.}
9. \textbf{Point.}\textemdash A \textit{Point} is the smallest magnitude that can
be imagined. It has \textit{no dimensions}, that is, it has no
measurement in any direction.
If a point be represented by a dot, this must only be
regarded as a picture to show that the point has a certain
position roughly indicated by the dot.
\begin{wrapfigure}[4]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_13_1} \\
\end{center}
\end{wrapfigure}
10. \textbf{Line.}\textemdash When a point moves it generates a \textit{Line}.
A line has \textit{one dimension}\textemdash length;
thus, $AB$ is a line generated by
a point which starts from $A$,
and moves to $B$, along the path
indicated by the line.
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_13_2} \\
\end{center}
\end{wrapfigure}
11. \textbf{Surface.}\textemdash In general, when a straight line moves,
it generates a \textit{Surface}. A surface
has \textit{two dimensions} at right angles
to each other\textemdash length and breadth;
thus, if the line $AB$ move to $CD$
along the path indicated, it will
generate the surface $ABCD$
12. \textbf{Solid.}\textemdash In general, when a surface moves it generates
\textit{a solid}. A solid has \textit{three
dimensions} at right angles to one
another\textemdash length, breadth, and
thickness; thus, if a surface
$ABCD$ move to $A'B'C'D'$, it will
generate the solid shown.
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_14_1} \\
\end{center}
\end{wrapfigure}
When a solid moves it generates another solid; consequently
geometrical forms may be divided into \textit{points},
\textit{lines}, \textit{surfaces}, and \textit{solids}.
\section*{The Line.}
13. \textbf{Straight Line}\textemdash 14. \textbf{Curved Line.}\textemdash Lines are
either \textit{straight} or \textit{curved}. A straight line is a line which
is generated by a point moving always in the same direction,
and is therefore the shortest distance between its
extreme points. When a point continually changes its
direction of motion, it generates a \textit{curved} line.
A line may be affected in four ways; it may have
\begin{enumerate}
\item \textit{Sense}; that is, it may be generated by the movement
of a point from one extremity to the other, or \textit{vice versâ},
as, from $A$ to $B$ in the one sense, or from
$B$ to $A$ in the other.
\item \textit{Direction}; that is, it has a definite inclination
relatively to some fixed standard.
\item \textit{Position}; that is, it has a definite place.
\item \textit{Magnitude}; that is, it has a certain length.
\end{enumerate}
The word \textit{line} will, when used alone, signify a straight
line.
\section*{Two Lines.}
15. \textbf{Angle}\textemdash \textbf{Right Angle}\textemdash \textbf{Perpendicular}\textemdash \textbf{Acute
Angle}\textemdash \textbf{Obtuse Angle}\textemdash \textbf{Reflex Angle.}\textemdash An \textit{Angle} is
the inclination of two lines to one another. Angles may
be measured by the rotation of one of these lines relatively
to the other. One-fourth of a complete revolution is
called a right angle, and the lines are then said to be
\textit{perpendicular} to one another.
An \textit{acute} angle is an angle less than a right angle.
An \textit{obtuse} angle is an angle greater than one and less
than two right angles.
An angle greater than two right angles is called a \textit{reflex},
or \textit{convex}, angle.
In practice, it is found convenient to measure angles
by dividing a complete revolution into 360 equal parts, or
\textit{degrees}; these are sub-divided into 60 equal parts, or
\textit{minutes}; and these again into 60 equal parts, or \textit{seconds}.
Thus $35$\textdegree $4' 22''$ denotes an angle of 35 degrees 4 minutes
22 seconds. A right angle is, of course, 90\textdegree.
16. \textbf{Complement.}\textemdash When two angles together make
a right angle, either of them is said to be the \textit{complement}
of the other.
17. \textbf{Supplement.}\textemdash When two angles together make
two right angles, either of them is said to be the \textit{supplement}
of the other.
18. \textbf{Parallel.}\textemdash When the angle between two lines is
zero, the lines are \textit{parallel}.
\subsection*{PROPOSITION 1.}\label{prop1}
\textbf{When two lines cut one another, the opposite
angles are equal.}
\medskip
Let $AB$, $CD$ cut one another in $E$.
\begin{wrapfigure}[3]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_15_1} \\
\end{center}
\end{wrapfigure}
Now, if $AB$ be supposed to rotate about $E$, until it
coincide with $CD$, the parts $AE$ and
$EB$, since they turn together, must
move through equal angles.
Consequently the angles $AEC$,
$BED$, generated by rotation contra-clockwise, must be
equal.
So also must the angles $AED$, $BEC$, which are generated
when the coincidence is effected by clockwise rotation.
\section*{Three Lines.}
19.\textemdash When a line intersects two other lines, it makes
with them eight angles, 1, 2, 3,
4, 5, 6, 7, 8.
Of these\textemdash
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_16_1} \\
\end{center}
\end{wrapfigure}
1 and 5, 2 and 6, 3 and 7,
4 and 8 are called \textit{corresponding}
$\angle $s.
4 and 6, 3 and 5 are called
\textit{alternate} $\angle $s.
1, 2, 7 and 8 are called \textit{exterior} $\angle $s.
3, 4, 5 and 6 are called \textit{interior} $\angle $s.
\subsection*{PROPOSITION 2.}\label{prop2}
\textbf{If one line meet two parallel lines (i.) the corresponding
angles are equal; (ii.) the alternate angles
are equal; (iii.) the exterior angles, and (iv.) the
interior angles, on the same side of the line, are
together equal to two right angles.}
\begin{center}
\includegraphics[width=0.25\linewidth]{../images/illus_16_2} \\
\end{center}
Let $AB$, $A'B'$ be any two parallel lines, and let $CD$
meet them. Then since $CD$ is
equally inclined to both of
them\textemdash
$\angle1 = \angle5; \angle2 = \angle6; \angle3 = \angle7; \angle4 = \angle8$ (by def. 19).
\begin{tabbing}
I.e., \textit{the corresponding} $\angle $s \textit{are equal.} \` (i.) \\
Again, since $ \angle4 = \angle8, and \angle6 = \angle8 $ (Prop. 1)
\end{tabbing}
\[ \therefore \angle4 = \angle6 \]
Similarly, it may be proved that $ \angle3 = \angle5 $.
\begin{tabbing}
I.e., \textit{the alternate} $ \angle $s \textit{are equal}. \`(ii.)
\end{tabbing}
\begin{tabular}{rrcl}
Again, since & $\angle1 + \angle4$ & $=$ & $2 \text{ rt. } \angle $s \\
and &$\angle4$ &$=$ & $ \angle8$ (by (i.)) \\
$ \therefore$ &$\angle1 + \angle8$ & $=$ & $2 \text{ rt. } \angle $s
\end{tabular}
Similarly, it may be proved that $ \angle2 + \angle7 = 2 \text{ rt. } \angle $s.
\begin{tabbing}
I.e., \textit{the exterior} $ \angle $s \textit{on the same side of the line are
together equal to} $ 2 \text{ rt. } \angle $s. \`(iii.)
\end{tabbing}
\begin{tabular}{rrcl}
Finally, since& $ \angle3 + \angle4 $ &$ = $& $ 2 \text{ rt. } \angle $s \\
and& $ \angle4 $ &$ = $& $ \angle6 $ (by (ii.)) \\
$ \therefore $& $ \angle3 + \angle6 $ &$ = $& $ 2 \text{ rt. } \angle $s.
\end{tabular}
Similarly, it may be proved that $ \angle4 + \angle5 = 2 \text{ rt. } \angle $s.
\begin{tabbing}
I.e., \textit{the interior}$ \angle $s \textit{on the same side of the line are
together equal to} $ 2 \text{ rt. } \angle $s. \`(iv.)
\end{tabbing}
\textit{Corollary}.\textemdash The converse of this proposition is obviously
true also, viz., that if any of the statements (i.), (ii.),
(iii.), or (iv.) be true, the lines $ AB $, $ A'B' $ must be parallel;
for if they be not parallel, those statements cannot be
true.
\subsubsection*{Example 1.}
\begin{wrapfigure}[7]{l}{0.5\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_17_1} \\
\end{center}
\end{wrapfigure}
(i.) In calculating the horizontal distance $ AS $ of a ship
$ S $ from the point $ A $, vertically beneath the point of observation
$ T $, \textit{the angle of depression}
$ HTS $ is observed
through which it is necessary
to depress the instrument
in order to sight on $ S $.
Then, since $ TH $ and $ AS $ are
parallel, $ \angle AST = \angle STH $
(by Prop. 2 (ii.)), and this, with the measurement of $ AT $,
enables the triangle to be drawn, or calculated. (See
Prop. 5.)
\subsubsection*{Example 2.}
\begin{wrapfigure}[5]{r}{0.15\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.15\textwidth]{../images/illus_18_1} \\
\end{center}
\end{wrapfigure}
The fact that when corresponding $ \angle $s are
equal the lines are parallel, is made use of
by draughtsmen in drawing parallel lines;
a straight-edge $ AB $ and set-square are employed.
A line being drawn at $ ED $, the
square $ CED $ may be moved along the
straight-edge to any position $ C'E'D' $, and a
line $ E'D' $ drawn. This line will be parallel
to $ ED $, for the corresponding $ \angle $s $CED $,
$ C'E'D' $ are equal. (By Prop. 2 (i.).)
20. \textbf{Triangle.}\textemdash When no two of three given lines are
parallel they intersect in pairs, and form a \textit{Triangle}, whose
properties we next proceed to investigate.
\section*{Triangles.}
21. \textbf{Equilateral triangle.}\textemdash A triangle is \textit{equilateral}
when its sides are of equal length.
22. \textbf{Isosceles triangle}.\textemdash A triangle is \textit{isosceles} when
any two of its sides are of equal length.
\subsection*{PROPOSITION 3.}\label{prop3}
\textbf{In any triangle, an exterior angle is equal to the
two interior opposite angles.}
\medskip
\begin{wrapfigure}[3]{r}{0.3\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_18_2} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be any $\triangle$, and let
any side $ CB $ be produced to $ D $.
Then if $ BE $ be a line parallel
to $ CA $, $ BE $ and $ CA $ are equally inclined to $ CD $.
\begin{align*}
\therefore\: \angle DBE &= \angle BCA \quad \text{(Prop. 2 (i.).)} \\
also \angle EBA &= \angle BAC \quad \text{(Prop. 2 (ii.).)} \\
\therefore \; \text{the whole } \angle DBA &= \angle BCA + \angle BAC \text{.}
\end{align*}
\textsc{Corollary.}\textemdash An exterior angle of a triangle is greater
than either of the interior opposite angles.
\subsubsection*{Example.}
To calculate the height $ AT $ of an inaccessible object, on
level ground, a base-line $ BC $ is
measured, and the angles $ ABT $,
$ BCT $ are observed.
\begin{wrapfigure}[3]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_19_1} \\
\end{center}
\end{wrapfigure}
In order to calculate $ AT $ it is
necessary to know the angle
$ BTC $. But by Prop. 3
$ \angle ABT = \angle BTC + \angle BCT $
$ \therefore\: \angle BTC = \angle ABT - \angle BCT $.
\subsection*{PROPOSITION 4.}\label{prop4}
\textbf{The interior angles of any triangle are together
equal to two right angles.}
\medskip
Let $ ABC $ be any triangle, and let any side $ CB $ be
produced to $ D $, then $ \angle DBA = \angle BCA + \angle BAC $. (Prop. 3.)
\begin{wrapfigure}[4]{l}{0.25\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_19_2} \\
\end{center}
\end{wrapfigure}
Add to each the $ \angle ABC $,
then $ \angle DBA + \angle ABC = \angle BCA + \angle BAC + \angle ABC $
but $ \angle DBA + \angle ABC = 2 \text{ rt. } \angle $s
$ \therefore\: \angle \text{ s } BCA $, $ BAC $, $ ABC = 2 \text{ rt. } \angle $s.
\textsc{Cor. 1.}\textemdash Any two angles of a triangle are together less
than two right angles.
\textsc{Cor. 2.}\textemdash All the interior angles of any polygon = twice
as many right angles as the polygon has sides, less four
right angles.
\begin{center}
\includegraphics[width=0.25\linewidth]{../images/illus_19_3} \\
\end{center}
For since any polygon $ ABCDE $ of $ n $ sides may be
divided into two less $ \triangle $s than there
are sides, that is $ n-2 \triangle $s, and
since each $ \triangle $ contains two rt. $ \angle $s, the
sum of all the $ \angle $s must be $ 2n-4 \text{ rt. } \angle $s.
\subsubsection*{Example 1.}
If the interior angles of a quadrilateral are 60\textdegree, 125\textdegree
and 160\textdegree, find the remaining angle.
\begin{tabbing}
\`Answer 15\textdegree.
\end{tabbing}
\subsubsection*{Example 2.}
The interior angles of a six-sided polygon were observed
to be 80\textdegree, 160\textdegree, 125\textdegree, 82\textdegree, 150\textdegree and 122\textdegree. What was the
total error in the observations?
\begin{tabbing}
\`Answer 1\textdegree.
\end{tabbing}
\subsection*{Conditions which determine a Triangle.}
23. A triangle is said to be \textit{determined}, when any other
triangle constructed from the same data is \textit{congruent} with
it, that is to say, identical in every respect with it.
A triangle cannot be determined unless three parts at least are given.
Hence there are four cases to consider:\textemdash
\begin{enumerate}[(i)]
\item When three angles are given.
\item When two angles and a side are given.
\item When one angle and two sides are given.
\item When three sides are given.
\end{enumerate}
In the first case it is evident that any number of
triangles can be drawn, having the angles in each respectively
equal, and therefore, in this case, the triangle is
not determined.
\subsection*{PROPOSITION 5.}\label{prop5}
\textbf{A triangle is determined when two angles are
given, and a side which is known to be either
opposite or adjacent to these angles.}
\medskip
Suppose any side $ AB $ to be given,
which is adjacent to the given angles
$ A $ and $ B $.
\begin{wrapfigure}[3]{4}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_20_1} \\
\end{center}
\end{wrapfigure}
The directions of $ AC $ and $ BC $ are
determined, because the angles $ BAC $,
$ ABC $ are known. Hence the point $ C $, where $ AC $ and $ BC $
intersect, is determined.
Again, if the $ \angle $s $ A $ and $ C $ are given and the side
$ AB $ opposite to one of these $ \angle $s; then since $ \angle B$ can be
found (Prop. 4), the $ \angle $s $ A $ and $ B $ are known, and this case
becomes the same as the last, and the $ \triangle $ is therefore
determined.
\textsc{Cor. 1.}\textemdash Two triangles are congruent, when two angles
in the one are equal to two angles in the other, and a side
in each, either opposite or adjacent to one of the equal
angles, are equal. For these triangles, being determined
by the same data, must be identical.
\textsc{Cor. 2.}\textemdash If any two angles $ CAB $, $ CBA $ in a triangle
are equal, it follows, on supposing the angle at $ C $ bisected
by $ CD $, that the sides $ CB $, $ CA $ opposite the equal angles
are equal. For in this case the angles $ CAD $, $ CBD $ have two
angles and a side in each equal. Therefore by Cor. 1
they are congruent, and $ AC = BC $.
\textsc{Cor. 3.}\textemdash If a triangle be equiangular, it is also equilateral.
This follows immediately from Cor. 2.
\subsubsection*{Example.}
\begin{wrapfigure}[5]{l}{0.5\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_21_1} \\
\end{center}
\end{wrapfigure}
Suppose it required to measure the width between $ A $
and $ C $ on opposite banks of a river.
Measure a base-line $ AB $ on one
side of the river, and at its extremities
measure the angles $ ABC $, $ CAB $. Then the width $ AC $ is determined
either by drawing to scale or by calculation (Prop. 5).
\subsection*{PROPOSITION 6.}\label{prop6}
\textbf{A triangle is determined when one angle and two
sides are given; except when the given angle lies
opposite the smaller of the given sides; in which case
there are two triangles having supplementary angles.}
\medskip
\textit{Case} 1.\textemdash When the given angle is included by the given
sides.
\begin{wrapfigure}[6]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_22_1} \\
\end{center}
\end{wrapfigure}
Let $ BAC $ be the given angle, and $ AB $,
$ AC $ the given sides.
Then since only one straight line can
be drawn between $ B $ and $ C $, the triangle
is determined.
\medskip
\textsc{Cor. 1.}\textemdash If the given side $ BA = $ the
given side $ CA $, and $ AD $ be the line which
by supposition bisects the angle $ BAC $,
then in the $ \triangle $s $ ABD $, $ ACD $, the sides $ AB $, $ AD $ in the one
are $ = $ the sides $ AC $, $ AD $ in the other, and the included
$ \angle $s are equal; the $ \triangle $ s are therefore congruent (by the
preceding), and $ \angle ABC = \angle ACB $.
\textit{I.e.}, the angles at the base of an isosceles triangle are
equal.
\medskip
\textsc{Cor. 2.}\textemdash An equilateral triangle is also equiangular.
This follows immediately from the preceding.
\textit{Case 2.}\textemdash When the given angle is not included by the
given sides.
Let the sides $ AB $, $ BC $ and the $ \angle A $ be given; and first,
let the greater side $ BC $ be opposite the given $ \angle A $.
\begin{wrapfigure}[7]{r}{0.5\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_22_2} \\
\end{center}
\end{wrapfigure}
Then since the point $ C $
must lie on $ AC $ or $ AC $
produced, and must also
be at a distance from
$ B = BC $, it must lie somewhere
on the circle whose
centre is $ B $ and radius $ BC $,
and must therefore be
either at $ C $ or $ C' $ where
the line $ AC $ cuts the circle. In the first case we have the
$ \triangle ABC $, and in the second case we have the $ \triangle ABC' $.
But of these, the $ \triangle ABC $ alone contains the given $ \angle A $.
Therefore the $ \triangle $ is determined.
Next, let $ BC $ opposite the given$ \angle A $, be the smaller
side.
In this case the point $ C' $ lies between $ A $ and $ C $, and we
get two $ \triangle $s $ ABC $, $ ABC' $, both having their sides $ AB $, $ BC $
and $ AB $, $ BC' $ equal to the given sides, and containing the
given $ \angle A $.
\begin{wrapfigure}[4]{l}{0.3\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_23_1} \\
\end{center}
\end{wrapfigure}
In this case, therefore, there are two $ \triangle $s satisfying the
data, but the $ \angle $s $ BCA $, $ BC'A $ are
supplementary.
This case is known as the
``ambiguous case.''
\medskip
\textsc{Cor. 3.}\textemdash Triangles are congruent
when one angle and two
sides in each are equal, except when the given angle
lies opposite the smaller of the given sides, in which case
the triangles may be congruent or may have supplementary
angles.
\subsection*{PROPOSITION 7.}\label{prop7}
\textbf{A triangle is determined when the three sides
are given.}
\medskip
Let $ AB $ be one of the sides.
\begin{wrapfigure}[7]{l}{0.3\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_23_2} \\
\end{center}
\end{wrapfigure}
Then, since the vertex of the $ \triangle $ must lie on the circumference
of a circle whose centre is $ A $ and radius $ AC $, equal
to one of the remaining sides, and
likewise on the circumference of a
circle whose centre is $ B $ and radius
$ BC $, equal to the third side, it must
be either at $ C $ or $ C' $, the points where
these circles cut one another. But,
from the nature of the construction,
$ AB $ is an axis of symmetry, and
therefore the $ \triangle $s $ ABC $, $ ABC' $ are
congruent.
\textsc{Cor. 1.}\textemdash Triangles which have the three sides of the
one equal respectively to the three sides of the other, are
congruent.
\textbf{Hence, triangles are congruent, if}
(i) two angles and a corresponding side in each, are equal;
(ii) one angle and two sides in the one are equal to
one angle and two sides in the other, except when the
given angle is opposite the smaller side; in which case
two triangles can be formed, one of which is congruent
with the first, and the other not;
(iii) the three sides in the one are equal to the three
sides in the other; that is to say, if in two triangles three
elements in the one are known to be equal to the three
corresponding elements in the other, the
triangles are congruent in all cases, except
\begin{wrapfigure}[7]{r}{0.15\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.15\textwidth]{../images/illus_24_1} \\
\end{center}
\end{wrapfigure}
\begin{enumerate}[(i)]
\item when three angles are equal.
\item in the ambiguous case.
\end{enumerate}
Practical applications of the congruence
of triangles:\textemdash
1. To determine the position of a point
by offsets from a straight line.
Suppose $ AB $ to be a survey line, and it
is required to fix the position of the corners
of a building $ D $, $ E $ with respect to it.
Take any convenient points $ A $, $ C $, $ B $ in
the line, and measure $ BD $, $ CD $, $ CE $, $ AE $.
Then the $ \triangle $s $ BCD $, $ ACE $ are determined. (Prop. 7.)
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_24_2} \\
\end{center}
\end{wrapfigure}
2. Let it be required to determine an angle $ ABC $ with
a chain only. Produce $ CB $ to
any convenient length $ BD $. Set
off the same or any other convenient
length $ BE $ along $ BA $, and
measure $ DE $. Then the $ \triangle DBE $
is determined, and therefore also the $ \angle CBE $, which is
supplementary to it. (Prop. 7.)
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_24_3} \\
\end{center}
\end{wrapfigure}
3. To measure an inaccessible distance
by means of a chain only.
(Prop. 6.)
Let $ AB $ be the inaccessible distance.
Take any convenient point $ C $, and
produce $ AC $, $ BC $ until $ CE = AC $ and
$ CD = BC $. Then the $ \triangle $s $ ABC $, $ CDE $
are congruent, and $ \therefore DE = AB $.
\subsubsection*{Exercises.}
1. To draw a straight line perpendicular to a given
straight line from a given point in it.
2. To draw a straight line perpendicular to a given
straight line from a given point without it.
3. Any two sides of a triangle are together greater than
the third side.
4. To trisect a right angle.
5. To construct a triangle, having given the base, an
angle at the base, and the sum of the sides.
6. The bisectors of the angles of a triangle are concurrent.
7. The perpendicular bisectors of the sides of a triangle
are concurrent.
8. The medians of a triangle are concurrent, and the point
of intersection is one-third of any median from the corresponding
side.
9. The straight line which joins the points of bisection
of the sides of a triangle is parallel to the base, and one-half
of it.
10. If the adjacent sides of a quadrilateral are bisected
and the points joined, the figure so formed is a parallelogram.
11. Given two straight lines and a given point between
them. To draw through the given point a straight line
which shall be bisected in that point.
12. Given two angles of a triangle and the perimeter;
to construct the triangle.
13. Through a given point, to draw a straight line that
shall make equal angles with two given straight lines.
\section*{Figures Consisting of Four Lines.}
24. A \textit{parallelogram} ($ \parallelogram $) is a four-sided figure having
its opposite sides parallel.
25. A \textit{rhombus} is an equilateral parallelogram.
26. A \textit{rectangle} is a right-angled parallelogram.
27. A \textit{square} is an equilateral rectangle.
28. A \textit{trapezium} is a four-sided figure having two of its
sides parallel.
29. A \textit{quadrilateral} is any plane four-sided figure.
30. A line which joins any two non-adjacent corners of
a polygon is called a \textit{diagonal}.
\subsection*{PROPOSITION 8.}\label{prop8}
\textbf{A parallelogram is bisected by its diagonal.}
\medskip
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_26_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be any $ \parallelogram $, and $ BD $ a
diagonal.
Then in the $ \triangle $s $ ABD $, $ BCD $, \\
$ \because\; \angle ABD = \angle BDC $ and $ \angle ADB = \angle DBC $ (Prop. 2), and the side $ BD $ is common,
$ \triangle ABD = \triangle BDC $. (Prop. 5, Cor. 1.)
\subsection*{PROPOSITION 9.}\label{prop9}
\textbf{Parallelograms on equal bases and between the
same parallels are equal in area.}
\medskip
Let $ ABCD $, $ A'B'C'D' $ be two
$ \parallelogram $s on equal bases $ AB $, $ A'B' $
and between the same parallels
$ DC' $, $ AB' $.
\begin{wrapfigure}[5]{r}{0.3\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_26_2} \\
\end{center}
\end{wrapfigure}
Then if the trapezium
$ AA'D'D $ be moved parallel to
$ AB' $, until $ A' $ coincides with $ B' $, it will coincide with the
trapezium $ BB'C'C $, since $ AB = A'B' $.
Hence $ AA'D'D $ and $ BB'C'C $ are equal in area.
From each take away the trapezium $ CD'A'B $.
Then the remaining $ \parallelogram ABCD = \text{ remaining } \parallelogram
A'B'C'D' $.
\medskip
\textsc{Cor. 1.}\textemdash Parallelograms on equal bases and of equal
altitude are equal in area.
\textsc{Cor. 2.}\textemdash Triangles on equal bases and of equal altitude
are equal in area.
\textsc{Cor. 3.}\textemdash A triangle has half the area of a parallelogram
on the same base. (Prop. 8 and Cor. 2.)
\subsubsection*{Example.}
\textit{To reduce a rectilinear figure to a triangle of equal area.}
Let $ 012340 $ be any rectilinear figure.
\begin{wrapfigure}[6]{l}{0.6\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.6\textwidth]{../images/illus_27_1} \\
\end{center}
\end{wrapfigure}
Join $ 14 $ and through $ 0 $
draw $ 00' $ parallel to $ 14 $ to
meet the side $ 34 $ produced
in $ 0' $.
Then $ \triangle 014 = \triangle 0'14 $,
and the quadrilateral $ 0'1230' $ is equal to the five-sided
figure $ 012340 $. (Prop. 9, Cor. 2.)
Similarly, by joining $ 0'2 $ and drawing $ 11' $ parallel $ 0'2 $
to meet $ 32 $ produced in $ 1' $, the quadrilateral is reduced to
the equal $ \triangle 0'1'3 $.
\subsection*{PROPOSITION 10.}\label{prop10}
\textbf{Parallelograms about the diagonal of any parallelogram
are equal.}
\medskip
Let $ AF $, $ FC $ be $ \parallelogram $s about the diagonal of any $ \parallelogram
ABCD $.
\begin{wrapfigure}[3]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_27_2} \\
\end{center}
\end{wrapfigure}
Then $ \because $ the $ \triangle ABD = \triangle BCD $ \quad (Prop. 7, Cor. 1)
$ \triangle DEF = \triangle DKF $ \quad (Prop. 7, Cor. 1)
and $ \triangle FHB = \triangle FGB $ \quad (Prop. 7, Cor. 1)
$ \therefore \triangle ABD-\triangle DEF -\triangle FHB = \triangle BCD-\triangle DKF -\triangle FGB $
\textit{i.e.} $ \parallelogram AF = \parallelogram FC $.
\subsubsection*{Exercises.}
1. When equal triangles stand on equal bases in one
straight line and on the same side of it, they are of equal
altitude, or lie between the same parallels.
2. To draw a triangle, the altitude and the base angles
being given.
\bigskip
31. \textbf{Ratio.}\textemdash The \textit{ratio} of one quantity to another is the
fraction which expresses the numerical relation between
their magnitudes. Thus $ \frac{A}{B} $ is the ratio of $ A $ to $ B $.
32. \textbf{Proportion.}\textemdash When two ratios are equal, the
quantities which constitute the ratios are said to be in
\textit{proportion}.
Thus, if $ \frac{A}{B} = \frac{C}{D} $, \quad $ A $, $ B $, $ C $, $ D $ are in \textit{proportion}.
\subsection*{PROPOSITION 11.}\label{prop11}
\textbf{Parallelograms of equal altitude are to one
another as their bases.}
\medskip
\begin{wrapfigure}[4]{r}{0.6\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.6\textwidth]{../images/illus_28_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $, $ EFGH $ be any two $ \parallelogram $s of equal altitude.
If $ EH' $, $ FG' $ be drawn $ \parallel AD $ or $ BC $ \\
the $ \parallelogram EFG'H' = \parallelogram EFGH $. (Prop. 9.)
But the $ \parallelogram ABCD $ may be divided into as many $ \parallelogram $s
equal to $ EFG'H' $ and parts of it as the base $ AB $ contains
the base $ EF $.
\medskip
$ \therefore \frac{\parallelogram ABCD}{\parallelogram EFGH} = \frac{AB}{EF} $, \textit{i.e.} the $ \parallelogram $s are proportional to their bases.
\medskip
\textsc{Cor.}\textemdash Triangles of equal altitude are to one another
as their bases.
\subsection*{PROPOSITION 12.}\label{prop12}
\textbf{The areas of equiangular parallelograms are as
the products of their sides.}
\medskip
\begin{wrapfigure}[5]{r}{0.5\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_29_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $, $ A'B'C'D' $ be two equiangular $ \parallelogram $s.
Construct a $ \parallelogram MNPQ $ having
the same
angles, and with
one side $ MN = AB $, and the other $ NP = B'C' $.
\[ \text{Then}\; \frac{\parallelogram ABCD}{\parallelogram MNPQ} = \frac{BC}{NP} \; \text{(Prop. 11.)}\]
\[ \text{and} \; \frac{\parallelogram A'B'C'D'}{\parallelogram MNPQ} = \frac{A'B'}{MN} \; \text{(Prop. 11.)} \]
\[ \therefore \; \text{by division,} \; \frac{\parallelogram ABCD}{\parallelogram A'B'C'D'} = \frac{MN\cdot BC}{A'B'\cdot NP} = \frac{AB\cdot BC}{A'B'\cdot B'C'} \]
%Not sure about the dot operator
\medskip
\textsc{Cor.}\textemdash The areas of triangles having an angle in each
equal, are as the products of the sides about that angle.
\subsubsection*{Example 1.}
$ ABCD $, $ A'B'C'D' $ are two equiangular $ \parallelogram $s, the area
of the first being three times that of the second; if $ AB = 2' $, $ BC = 3' $, and $ A'B' = 1' $, find $ B'C' $.
\begin{tabbing}
\`Answer $ 2' $.
\end{tabbing}
\subsubsection*{Example 2.}
In the preceding, if $ AB = 4' $, $ BC = 5' $, $ A'B' = 3' $ and
$ B'C' = 4' $, and the area of $ A'B'C'D' $ be 9 square feet, find
the area of $ ABCD $.
\begin{tabbing}
\`Answer 15 square feet.
\end{tabbing}
\subsubsection*{Example 3.}
\begin{wrapfigure}[7]{r}{0.4\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_30_1} \\
\end{center}
\end{wrapfigure}
Let it be required to
find the length $ EB $ of the
side of a level cutting
$ EBCF $ having the same
area as another cutting
$ ABCD $, the ground surface
of which is not level.
Produce $ AB $, $ DC $ to meet
in $ G $.
\[ \text{Then}\; \frac{\triangle EGF}{\triangle AGD} = \frac{EG\cdot GF}{AG\cdot GD} \]
But $ \triangle EGF = \triangle AGD $ (by Hyp.)
\[ \therefore \; \frac{EG \cdot GF}{AG\cdot GD} = 1 \; \text{ and } \; \therefore \; EG\cdot GF = AG\cdot GD. \]
But when the side-slopes are equal, as is generally the
case, $ EG = GF $
and $ \therefore \; EG = \sqrt{AG\cdot GD} $ and $ \therefore \; EB = \sqrt{AG\cdot GD}-\overline{BG} $.
\section*{Similar Figures.}
33. \textbf{Similar figures} are equiangular, and have their
corresponding sides proportional.
\subsection*{PROPOSITION 13.}\label{prop13}
\textbf{Equiangular triangles are also similar.}
\medskip
\begin{wrapfigure}[6]{r}{0.5\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_30_2} \\
\end{center}
\end{wrapfigure}
Let $ ABC $, $ A'B'C' $ be equiangular $ \triangle $s, having the
$ \angle A = \angle A' $.
Apply the $ \triangle A'B'C' $ to
the $ \triangle ABC $, so that $ A' $
falls on $ A $, and $ A'B' $ on
$ AB $; then will $ A'C' $ fall
on $ AC $, because $ \angle A' = \angle A $
and $ \because\; \angle A'B'C' = \angle ABC $ \; (by Hyp.)
$ B'C' $ is parallel to $ BC $. (Prop. 2.)
Join $ B'C $, $ BC' $.
\begin{align*}
\text{Then} \; \frac{\triangle ABC'}{\triangle AB'C} & = \frac{AB\cdot AC'}{AC\cdot AB'} \quad \text{(Prop. 12, Cor.)} \\
\text{But} \; \triangle ABC' &= \triangle AB'C' + \triangle BB'C' \\
\text{and} \; \triangle AB'C &= \triangle AB'C' + \triangle B'C'C \\
\text{But} \; \triangle BB'C' &= \triangle B'C'C \text{.} \quad \text{(Prop. 9, Cor. 2.)}\\
\therefore \; \triangle ABC' & = \triangle AB'C \\
\text{Consequently} \; \frac{AB\cdot AC'}{AC\cdot AB'} &= 1 \\
\text{\textit{i.e.}} \; \frac{AB}{AC} &= \frac{AB'}{AC'}= \frac{A'B'}{A'C'} \\
\text{Similarly} \; \frac{BC}{BA} &= \frac{B'C'}{B'A'} \\
\text{and} \; \frac{CA}{CB} &= \frac{C'A'}{C'B'}
\end{align*}
\subsubsection*{Example.}
\textbf{Diagonal Scale.}
\begin{wrapfigure}[4]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_31_1} \\
\end{center}
\end{wrapfigure}
The diagonal scale enables us to subdivide a small
distance very accurately. Thus, if $AB$ be a line, which
it is required to divide into 40 equal
parts, say. Divide $AB$ into 10 equal
parts, and set up a perpendicular $AC$
of any convenient length, and divide
it into four parts; then if horizontal
and vertical lines be drawn, and also
diagonals $C1$, $22'$, $33'$, \&c., each of these divisions will be
subdivided into four equal parts (by Prop. 13), and the
whole line therefore into 40 equal parts.
\subsubsection*{Example 1.}
Draw a diagonal scale 6 inches long, to read 1/100ths of
an inch.
\subsubsection*{Example 2.}
A mechanical drawing is made in terms of a unit
whose length is 1.25 inches. Draw a diagonal scale to
give tenths and hundredths of the unit.
\subsubsection*{Example 3.}
Draw a diagonal scale of 60 chains to an inch, to read
chains.
\subsubsection*{Conditions which Determine the Similarity of Triangles.}
It follows from the preceding propositions that two
triangles are similar when\textemdash
1. One angle in the one is equal to one angle in the
other, and the ratio of the sides about these angles are
equal;
2. One angle in the one is equal to one angle in the
other, and the ratio of any two sides in each are equal,
provided that the triangle is determined by the three
parts considered;
3. Any two side-ratios in each are equal;
4. Any two angles in each are equal.
\subsection*{PROPOSITION 14.}\label{prop14}
\textbf{In any right-angled triangle, the perpendicular
on the hypothenuse from the opposite vertex makes
triangles which are similar to the whole triangle
and to one another.}
\medskip
\begin{wrapfigure}[4]{r}{0.3\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_32_1} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be any rt.- $ \angle $d $ \triangle $,
having a rt. $ \angle $ at $ A $, and let
$ AA' = h $ be the perpendicular
on the hypothenuse $ BC $.
Then, in the $ \triangle $s $ ABC $, $ ABA' $,
the $ \angle $s $ BAC $, $ BA'A $ are rt. $ \angle $s,
and the $ \angle B $ is common, $ \therefore $ the $ \triangle $s are similar.
In the same way the $ \triangle $s $ ABC $, $ ACA' $ are similar.
$ \therefore $ the three $ \triangle $s $ ABC $, $ ABA' $, $ ACA' $ are similar.
\textsc{Cor. 1.}\textemdash Since the $ \triangle $s $ ABA' $, $ ACA' $ are similar
\[ \frac{c' }{h'} = \frac{h}{b'} \quad \text{(Prop. 13)} \qquad \therefore \; h^{2} = b'c' \]
\textit{i.e.} the square on the perpendicular is equal to the
rectangle contained by the segments into which it divides
the hypothenuse.
\medskip
\textsc{Cor. 2.}\textemdash Since the $ \triangle $s $ ABA', $ $ ABC $ are similar
\[ \frac{c}{a} = \frac{c'}{c} \quad \text{(Prop. 13)} \qquad \therefore \; c^{2} = ac' \]
Similarly since the $ \triangle $s $ ACA' $, $ ABC $ are similar
\[ \frac{b}{a} = \frac{b'}{b} \quad \text{(Prop. 13)} \qquad \therefore \; b^{2} = ab' \]
\textit{i.e.} the square on one side of a rt. $ \angle $d $ \triangle $ is equal to the
rectangle contained by the hypothenuse and the projection
of that side on the hypothenuse.
\textsc{Cor. 3.}\textemdash Hence
\begin{align*}
b^{2} + c^{2} &= ab' + ac' \quad \text{(Cor. 2.)} \\
&= a(b' + c') = a \; \times \; a = a^{2}
\end{align*}
I\textbf{mportant.}\textemdash \textit{I.e}. In any rt.- $ \angle $d $ \triangle $, the squares on the
sides about the rt. $ \angle $ are together equal to the square on
the hypothenuse.
\subsubsection*{Example.}
\begin{wrapfigure}[5]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_33_1} \\
\end{center}
\end{wrapfigure}
To measure the width of a river, \&c., indirectly.
Measure a distance $ AB $, and
make $ BD = 2AB $, say; $ BD $ being
at rt. $ \angle $s to $ AB $. Sight $ C $ in line
with $ AB $, and make $ ADC $ a rt. $ \angle $.
Then $ BD^{2} = AB\cdot BC $
\[ \therefore \; BC = \frac{BD^{2}}{AB} = \frac{4AB^{2}}{AB} = 4AB . \]
\subsection*{PROPOSITION 15.}\label{prop15}
\textbf{In any triangle the square on one side is less
than the sum of the squares on the other two sides,
by twice the area of the rectangle contained by
either of these and the projection on it of the other.}
\medskip
\begin{wrapfigure}[5]{l}{0.3\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_33_2} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be any $ \triangle $, and let
$ AD $ be perp. to $ BC $, then $ BD $ is
the projection of $ AB $ on $ BC $.
\begin{align*}
\text{Now} \; b^{2} &= h^{2}+(a-d)^{2}. \quad \text{(Prop. 14, Cor. 3).}\\
&= h^{2} + a^{2} - 2 ad + d^{2}. \quad \text{(By Algebra.)} \\
\text{but} \; h^{2} + d^{2} &= c^{2}. \quad \text{(Prop. 14, Cor. 3.)} \\
\therefore \; b^{2} &= a^{2} + c^{2} - 2 ad.\
\end{align*}
If the angle $ B $ be obtuse, then the projection $ d $ will be
on the side produced, and, being drawn in the opposite
direction to what it was before, must be regarded as a
negative quantity, so that the last term will then be added
instead of subtracted. With this convention, Prop. 15
holds for any triangle.
\subsection*{PROPOSITION 16.}\label{prop16}
\textbf{The areas of similar triangles are in the ratio of
the squares of their corresponding sides.}
\medskip
Let $ ABC $, $ A'B'C' $ be any
similar $ \triangle $s.
\begin{wrapfigure}[5]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_34_1} \\
\end{center}
\end{wrapfigure}
\[ \text{Then} \; \frac{\triangle ABC}{\triangle A'B'C'}=\frac{AB \cdot AC}{A'B' \cdot A'C'} \quad \text{(Prop. 12, Cor.)} \]
\[ \text{But} \; \frac{AC}{A'C'}=\frac{AB}{A'B'} \quad \text{(Prop. 13.)} \]
\[ \therefore \; \frac{\triangle ABC}{\triangle A'B'C'} = \frac{AB \cdot AB}{A'B \cdot A'B'} = \left(\frac{AB}{A'B'}\right)^2 = \left(\frac{AC}{A'C'}\right)^2 = \left(\frac{CB}{C'B'}\right)^2 \]
\subsection*{PROPOSITION 17.}\label{prop17}
\begin{wrapfigure}[5]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_34_2} \\
\end{center}
\end{wrapfigure}
\textbf{The areas of similar polygons are in the ratio
of the squares of
their corresponding
sides.}
\medskip
Let $ ABCDE $, $ A'B'C'D'E' $
be similar polygons,
which are made up of the
similar $ \triangle $s $ ABE $, $ A'B'E' $, \&c.
\[ \text{Then since} \; \frac{\triangle ABE}{\triangle A'B'E'} = \frac{AB^{2}}{A'B'^{2}} \]
\[ \text{and} \; \frac{\triangle BCE}{\triangle B'C'E'} = \frac{BC^{2}}{B'C'^{2}} \; \&c. \quad \text{(Prop. 16)} \]
\[ \text{and since} \; \frac{AB}{A'B'} = \frac{BC}{B'C'} = \text{\&c.} \quad \text{(Prop. 13.)} \]
\[ \therefore \; \frac{\triangle ABE}{\triangle A'B'E'} = \frac{\triangle BCE}{\triangle B'C'E'} = \text{\&c.} = \left(\frac{AB}{A'B'}\right)^2 \]
\[ \therefore \; \frac{\triangle ABE + \triangle BCE + \text{\&c.}}{\triangle A'B'E' + \triangle B'C'E' + \text{\&c.}} = \left(\frac{AB}{A'B'}\right)^2 \quad \text{(by Algebra).} \]
\[ \therefore \; \frac{ABCDE}{A'B'C'D'E'} = \left(\frac{AB}{A'B'}\right)^2 = \left(\frac{BC}{B'C'}\right)^2 = \text{\&c.} \]
and similarly for any other polygons.
\medskip
\textsc{Cor.}\textemdash The areas of similar figures are in the ratio of
the squares of their corresponding linear dimensions.
\subsection*{PROPOSITION 18.}\label{prop18}
\textbf{The area of any figure described on the hypothenuse
of a right-angled triangle is equal to the similar
and similarly described figures on the sides about
the right angle.}
\medskip
\begin{wrapfigure}[3]{l}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_35_1} \\
\end{center}
\end{wrapfigure}
Let ABC be any rt.- $ \angle $d $ \triangle $.
Then if $ M_{1} $ and $ M_{2} $ be the areas
of the figures on $ b $ and $ c $, the sides
about the rt. $ \angle A $, and $ M $ the
area of the similar figure on $ a $, then
\[ \frac{M_{1}}{M} = \frac{b^{2}}{a^{2}} \qquad \text{and} \qquad \frac{M_{2}}{M} = \frac{c^{2}}{a^{2}} \quad \text{(Prop. 17).} \]
Hence, by addition,
\[ \frac{M_{1} + M_{2}}{M} = \frac{b^{2} + c^{2}}{a^{2}} = \frac{a^{2}}{a^{2}} = 1 \quad \text{(Prop. 14, Cor. 3).} \]
\[ \therefore \; M_{1} + M_{2} = M \]
\subsubsection*{Practical Example.}
To substitute for a hollow round column a solid one of
equal area.
\begin{wrapfigure}[7]{r}{0.4\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_36_1} \\
\end{center}
\end{wrapfigure}
Let $ OA $ be the internal radius, and $ OB $ the external
radius of the hollow column.
Draw $ BA $ tangent to the inner
circle.
Then a circle described with
$ AB $ as radius will have the same
area as the difference between
the other two.
\textit{Proof.}\textemdash For since $ OAB $ is a
rt.- $ \angle \triangle $, by Prop. 18, the circles
described with the sides $ OA $, $ AB $, as radii, will be equal
in area to the circle described with the hypothenuse $ OB $
as radius.
\subsection*{PROPOSITION 19.}\label{prop19}
\textbf{If any two similar figures are placed with their
corresponding sides parallel, the lines joining corresponding
points in the two figures are concurrent.}
\medskip
\begin{wrapfigure}[7]{r}{0.5\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_36_2} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $,
$ A'B'C'D' $ be two such
figures, and let $ AA' $
meet $ BB' $ in $ S $.
Then, $ \because \; AB $ is par.
to $ A'B' $ (by Hyp.),
the $ \triangle $s $ SAB $, $ SA'B' $
are similar,
and $ \therefore \; \frac{SB}{SB'} = \frac{AB}{A'B'} $ \quad (Prop. 13.)
Again, if possible, let $ CC' $ meet $ BB' $ in the point $ S' $ not
coincident with $ S $.
Then $ \because \; BC $ is $ \parallel $ to $ B'C' $ (by Hyp.), the $ \triangle $s $ S'BC $, $ S'B'C' $
are similar,
\begin{align*}
\text{and} \; \therefore \; \frac{S'B}{S'B'} &= \frac{BC}{B'C'} \quad \text{(Prop. 13.)}\\
\text{But since} \; ABC\text{,} \; A'B'C' &\text{are similar} \; \triangle\text{s (Hyp.)} \\
\frac{AB}{BC} &= \frac{A'B'}{B'C'} \quad \text{(Prop. 13.)} \\
\text{and} \; \therefore \; \frac{AB}{A'B'} &= \frac{BC}{B'C'} \quad \text{(By Algebra.)} \\
\therefore \; \frac{SB}{SB'} &= \frac{ S'B}{S'B'}\\
\text{Hence,} \; \frac{SB-SB'}{SB'} &= \frac{S'B-S'B'}{S'B'} \quad \text{\textit{i.e.,}} \quad \frac{BB'}{SB'} = \frac{BB'}{S'B'}
\end{align*}
$ \therefore \; SB' = S'B' $, and $ \therefore \; S $ and $ S' $ must be coincident.
In the same way it may be shown that $ DD' $ must pass
through $ S $.
\subsubsection*{Practical Example.}
To draw a line through a given point to pass through
the inaccessible join of two given lines.
\begin{wrapfigure}[6]{l}{0.4\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_37_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be a drawing-board, and let $ EF $ and $ GH $
be two lines upon it which
intersect beyond the limits of
the board.
It is required to draw through
a given point $ P $ a straight line
which shall pass through the
intersection of the given lines
produced. Draw any $ \triangle PFH $,
and draw $ EK \parallel $ to $ FH $. Through $ E $ and $ K $ draw lines
$ EL $, $ KL $ par. to $ FP $, $ PH $ intersecting in $ L $. Then $ LP $
produced will pass through the intersection of the given
lines, as required (by Prop. 19).
\textsc{Exercise.}\textemdash Draw a line to pass through the inaccessible
points which are given by two pairs of lines.
\section*{The Circle.}
34. A \textit{circle} is a figure contained by the path of a
point which rotates about a fixed point or \textit{centre}, at a
constant distance from it, called the \textit{radius}. The path of
the point is the \textit{circumference} of the circle, and any line
through the centre is called a \textit{diameter}.
35. A \textit{chord} of a circle is the straight line joining any
two points on its circumference.
36. A \textit{tangent} is a line which touches a circle.
37. A \textit{secant} is a line which cuts a circle.
38. A \textit{sector} of a circle is the figure contained by an arc
and the radii at its extremities.
39. A \textit{segment} of a circle is the figure contained by a
chord and an arc of the circle.
40. \textit{Concentric} circles are circles having a common
centre.
41. An \textit{arc} of a circle is part of its circumference.
\subsection*{PROPOSITION 20.}\label{prop20}
\textbf{The straight line drawn from the centre of a
circle to the middle point of a chord is perpendicular
to the chord.}
\medskip
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_38_1} \\
\end{center}
\end{wrapfigure}
Let $ AB $ be any chord bisected in $ C $.
Join $ OA $, $ OB $.
Then in the $ \triangle $s $ OAC $, $ OBC $, since the
three sides in each are respectively equal,
the $ \angle OCA = \angle OCB $ (Prop. 7, Cor. 1), and
therefore each of them is a right angle.
$ \therefore \; OC $ is perpendicular to $ AB $.
\medskip
\textsc{Cor. 1.}\textemdash Conversely, the straight line drawn from the
point of bisection of a chord perpendicular to it, passes
through the centre.
\medskip
\textsc{Cor. 2.}\textemdash The line from the centre of a circle perpendicular
to a chord, bisects the chord.
\medskip
\textsc{Cor. 3.}\textemdash If the points $ A $ and $ B $ approach each other
indefinitely, the chord becomes a tangent, and the line
$ OC $, which bisects $ AB $, becomes the radius at the point of
contact, and therefore a radius and the tangent at its
extremity are at right angles.
\subsection*{PROPOSITION 21.}\label{prop21}
\textbf{The angle at the centre of a circle is double the
angle at the circumference standing on the same arc.}
\medskip
There are three cases to be considered, as in (i.), (ii.),
(iii.), where the $ \angle APB $ is the $ \angle $ at the circumference,
and $ ACB $ the $ \angle $ at the centre standing on the same
arc $ AB $.
\begin{center}
\includegraphics[width=0.75\linewidth]{../images/illus_39_1} \\
\end{center}
Then, in each case, $ \angle ACD = \angle CAP + \angle CPA =
2 \angle CPA $ \quad (Prop. 3 and Prop. 7, Cor. 1),
and $ \angle BCD = \angle CBP + \angle CPB = 2 \angle CPB $ \quad (Prop.
3 and Prop. 7, Cor. 1).
$ \therefore \; $ in Case (i.) and Case (iii.) by Addition $ \angle ACB =
2 \angle APB $,
and in Case (ii.) by Subtraction $ \angle ABC = 2 \angle APB $.
\medskip
\textsc{Cor. 1.}\textemdash All angles at the circumference standing on
the same arc are equal.
\medskip
\textsc{Cor. 2.}\textemdash When the angle at the centre is equal to two
right angles, the angle at the circumference is one right
angle, that is to say, \textit{the angle in a semicircle is a right angle}.
\subsubsection*{Example 1.}
\begin{wrapfigure}[7]{r}{0.5\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_40_1} \\
\end{center}
\end{wrapfigure}
To determine the position of a ship at sea by observations
on three known points ashore. (Three-point
Problem.) Let $ A $, $ B $, $ C $ be the known points on shore,
$ DE $ being the shore-line, and let $ S $ be the ship.
The angles $ ASB $,
$ BSC $ are observed.
If a $ \bigcirc $ be described
on $ AB $ containing
the angle equal to
the observed angle
$ ASB $, the point $ S $
must lie upon it
(Cor. 1). Similarly,
if a $\bigcirc$ be described
on $ BC $ containing an angle equal to the observed angle
$ BSC $, $ S $ must lie upon it also. Consequently $ S $ must lie
at their point of intersection. In order to construct these
circles, double the observed angle to find the angle at the
centre (Prop. 21), subtract this from two right angles,
and halve the remainder. The result will be the angles
at the base of the $ \triangle s AB1 $, $ BC2 $.
\subsubsection*{Example 2.}
To draw a perpendicular at $ B $ to a given line $ AB $ at its
extremity without producing it.
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_40_2} \\
\end{center}
\end{wrapfigure}
Let AB be the given line. Take a
convenient point $ O $, and with $ OB $ as
radius describe a circle cutting $ AB $ in
$ C $. Join $ CO $, producing it to cut the
circle in $ D $. Then $ BD $ is the perpendicular
required, for the angle $ CBD $,
being the angle in a semi-$\bigcirc$, is a
right angle. (Cor. 2.)
\subsection*{PROPOSITION 22.}\label{prop22}
\textbf{The opposite angles of a quadrilateral inscribed
in a circle are together equal to two right angles.}
\medskip
\begin{wrapfigure}[8]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_41_1} \\
\end{center}
\end{wrapfigure}
Let $ APBQ $ be a quadrilateral in a circle.
Then, since $ \angle APB = \frac{1}{2}$ concave angle $ ACD $. \quad (Prop. 21),
and $ \angle AQB = \frac{1}{2}$ convex angle $ ACB $ \quad (Prop. 21),
$ \therefore \; \angle APB + \angle AQB = \frac{1}{2} $ (sum
of the convex and concave angles)
$ = \frac{1}{2} $ (four rt. $ \angle $s)
$ = 2 $ rt.$ \angle $s.
\subsection*{PROPOSITION 23.}\label{prop23}
\textbf{In any circle the product of the segments made
by the intersection of two chords with the circumference
are equal.}
\medskip
\begin{wrapfigure}[6]{l}{0.6\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.6\textwidth]{../images/illus_41_2} \\
\end{center}
\end{wrapfigure}
Let $ AB $, $ CD $ be any two chords which intersect either
within or without
the circle at $ E $.
Join $ AC $, $ BD $.
Then $ \angle AEC = \angle BED $,
and $ \angle ACE = \angle DBE $. (Prop. 21, Cor. 1.)
Therefore the $ \triangle $s $ ACE $, $ BDE $ are equiangular (Prop. 4),
\[ \text{and} \; \therefore \; \frac{AE}{CE} = \frac{DE}{BE} \quad \text{(Prop. 13).} \]
or $ AE\cdot BE = CE\cdot DE $.
\medskip
\textsc{Cor. 1.}\textemdash Conversely, when the products of the segments
are equal, the points $ A $, $ B $, $ C $, $ D $ lie on a circle.
\medskip
\textsc{Cor. 2.}\textemdash In any circle the square on the tangent is
equal to the product of the segments cut off on the
secant.
\medskip
\textsc{Cor. 3.}\textemdash Tangents to a circle from the same point are
equal.
\subsubsection*{Example.}
\begin{wrapfigure}[7]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_42_1} \\
\end{center}
\end{wrapfigure}
To find the distance of the horizon at sea.
Let the observer be at $ H $ at an elevation
$ AH $ above the sea-level.
Then, if $ HT $ be a tangent to the surface
of the water,
\[ HT^{2} = HB\cdot HA \quad \text{(Prop. 23, Cor. 2).} \]
$ \therefore $ the distance of the horizon $ HT $
\begin{align*}
&=\sqrt{ HB\cdot HA} \\
&= \sqrt{(d + h)h} \; \text{,}
\end{align*}
if $ d $ is the Earth's diameter, and $ h $ is
the elevation of the observer above sea-level.
Since $ HT^{2} = (d + h)h $ , where $ d $ is about 8000 miles,
and $ h $ is usually measured in feet,
\begin{align*}
\text{we have} \; \overline{HT^{2}} &= \left(8000+\frac{h}{5280}\right) \frac{h}{5280} \; \text{in miles.}\\
&= \frac{8000h}{5280} \; \text{, since} \; \frac{h^{2}}{5280^{2}} \; \text{, being very small,}
\end{align*}
may be neglected without sensible error.
\[ \therefore \; \overline{HT^{2}} = \frac{100}{66}h = \frac{3}{2}h \; \text{approximately.} \]
Hence the rule:
Three times the height of the observer above the sea level
in feet is equal to twice the square of the distance
seen in miles.
\subsection*{PROPOSITION 24.}\label{prop24}
\textbf{If a straight line touch a circle, and from the
point of contact another straight line be drawn
cutting the circle, the angles which this straight
line makes with the first at the point of contact are
equal to the angles in the adjacent segments.}
\begin{wrapfigure}[6]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_43_1} \\
\end{center}
\end{wrapfigure}
If $ EF $ be a tangent, and $ BD $ a secant at the point $ B $,
Then $ \angle DBF = \angle BAD $,
and $ \angle DBE = \angle BCD $.
\textit{Proof.}\textemdash Draw $ BA $ perpr. to $ EF $
at the point $ B $,
Then $ \angle ADB $ is a rt. $ \angle $. (Prop.
21, Cor. 2.)
and $ \therefore \angle $s $ ABC $, $ BAD = $ a rt. $ \angle $
(Prop. 4.)
$ \therefore \; \angle ABF = \angle ABD + \angle BAD $.
Take away the common $ \angle ABD $.
Then the remaining angle $ DBF = $ the remaining angle
$ BAD $,
\textit{i.e.}, the $ \angle DBF $ is equal to the angle in the adjacent
segment $ BAD $.
\begin{align*}
\text{Again,} \; \because \; \text{ the } \angle \text{s} \; BAD, BCD &= 2 \text{ rt. } \angle \text{s} \quad \text{(Prop. 22)} \\
\text{and the} \; \angle \text{s } DBF, DBE &= 2 \text{ rt. } \angle \text{s} \\
\therefore \; \angle \text{s } BAD, BCD &= \angle \text{s } DBF, DBE. \\
\text{But} \; \angle BAD &= \angle DBF \quad \text{(By the preceding part),} \\
\therefore \; \angle BCD &= \angle DBE .
\end{align*}
\subsubsection*{Exercises on the Circle.}
1. To draw a triangle, having given the base, the vertical
angle, and the altitude.
2. To draw a rt.- $ \angle $d $ \triangle $ when the hypotenuse and one
side are given.
3. To draw a tangent to a circle from a given point
without it.
4. To inscribe an equilateral $ \triangle $ and a regular hexagon
in a circle.
5. To inscribe a square and a regular octagon in a
circle.
6. To find a mean proportional between two given
straight lines.
7. Divide a circle into two segments, so that the angle
contained in the one may be three times the angle contained
in the other.
8. If a quadrilateral figure be described about a circle,
the sums of the opposite sides will be equal to one another.
9. $ DF $ is a tangent to a circle, and terminated at $ D $ and
$ F $ by two tangents drawn at the extremities of a diameter
$ AB $; show that the segment $ DF $ subtends a rt. $ \angle $ at the
centre of the circle.
10. If a circle be inscribed, in a rt.- $ \angle $d $ \triangle $, the excess of
the two sides over the hypotenuse is equal to the
diameter of the circle.
11. Two circles touch one another in $ A $, and have a
common tangent $ BC $. Show that the angle $ BAC $ is a
rt. $ \angle $.
\subsection*{PROPOSITION 25.}\label{prop25}
\textbf{To find the ratio of the circumference of a circle
to its diameter.}
\medskip
\begin{wrapfigure}[7]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_44_1} \\
\end{center}
\end{wrapfigure}
If two similar polygons be drawn, one inscribed in a
circle, and the other circumscribed
about it, it is evident that the circumference
of the circle is greater
than the first and less than the
second. The following table gives
the lengths of the perimeters of regular
polygons inscribed said circumscribed
to a circle.
\medskip
\begin{center}
\begin{tabular}{ | r | c | c | }
\hline
& & Perimeter of \\
Number of & Perimeter of & Circumscribed\\
Sides. & Inscribed Polygon.& Polygon.\\
& $ d \; \times $ & $ d \; \times $ \\ \hline
6 & 3.00000 & 3.46410 \\
12 & 3.10583 & 3.21539 \\
24 & 3.13263 & 3.15966 \\
48 & 3.13935 & 3.14609 \\
96 & 3.14103 & 3.14271 \\
192 & 3.14145 & 3.14187 \\
384 & 3.14156 & 3.14166 \\
768 & 3.14158 & 3.14161 \\
1536 & 3.14159 & 3.14160 \\
3072 & 3.14159 & 3.14159 \\
\hline
\end{tabular}
\end{center}
\medskip
It is evident, therefore, that the circumference of a
circle is equal to 3.14159 times the diameter to five places
of decimals. The exact value is an incommensurable
number, that is to say, it cannot be expressed exactly in
figures. The Greek letter $\pi$ (pi) is for convenience used
to denote the \textit{true} value, and the circumference of a circle
is therefore $\pi$ times its diameter. For all ordinary work
3.1416 is sufficiently accurate, and the vulgar fraction
$\frac{355}{113} $ also expresses the value of $\pi$ correctly to six decimal
places. For rough calculations $ \frac{22}{7} $ is frequently used,
which is nearly correct to three decimal places.
\subsection*{PROPOSITION 26.}\label{prop26}
\textbf{To find the length of a circular arc.}
\medskip
To find the length of an arc of a circle of given radius r,
subtending a given angle of n degrees.
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_46_1} \\
\end{center}
\end{wrapfigure}
Since the arcs of a circle are in proportion to the angles
which they subtend,
\[ \frac{arc AB}{circumference} = \frac{\angle AOB}{4 \text{ rt. } \angle \text{s}} \]
\[ \text{whence arc} \; AB = 2 \pi r \; \times \; \frac{n \SIUnitSymbolDegree}{360 \SIUnitSymbolDegree} = \frac{\pi rn \SIUnitSymbolDegree}{180 \SIUnitSymbolDegree} \]
\textbf{Measurement of angles in radian
measure.}\textemdash In mathematical work it is
frequently convenient and necessary to measure angles
without reference to any arbitrary unit, such as a degree
or grade, by the ratio of the arc
subtending it to the radius, on any
circle.
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_46_2} \\
\end{center}
\end{wrapfigure}
Thus, the angle $ AOB $ is evidently
determined if the ratio of the arc $ AB $
to the radius $ OB $ is known.
If, then, angles are measured by the
ratio arc/radius, the unit angle will be that for which the
ratio arc/radius = 1, or for which arc = radius.
This angle is called a \textit{radian}. If arc $ BC = $ radius,
then $ \angle COB $ is a radian, and the length of any arc $ AB = \text{radius} \; \times \; \text{number of radians in} \; \angle AOB $.
\[ \text{For an angle of } \; 180 \SIUnitSymbolDegree, \frac{\text{arc}}{\text{radius}} = \frac{\pi r}{r} = \pi \; \text{units,} \]
\[ \text{\textit{i.e.,}} \; \pi \; \text{radians} = 2 \text{ rt. } \angle \text{s.} \]
\[ \text{Hence 1 radian } \; = \frac{180 \SIUnitSymbolDegree}{\pi} = \frac{180 \SIUnitSymbolDegree}{3.14159} = 57 \SIUnitSymbolDegree 2958 \;\text{nearly.} \]
\subsubsection*{Exercises.}
1. Turn into radians $ 60 \SIUnitSymbolDegree $, $ 90 \SIUnitSymbolDegree $, $ 300 \SIUnitSymbolDegree $, $ 52 \SIUnitSymbolDegree 30' $, $ 135 \SIUnitSymbolDegree 4' 57'' $.
\smallskip
\textsc{Answer}: $ \frac{\pi}{3} $, $ \frac{\pi}{2} $, $ \frac{ 5\pi}{3} $, $ \frac{7 \pi}{24} $, .75046 $\pi$.
\medskip
2. Turn into degrees $ \frac{\pi}{3} $, $ \frac{\pi}{2} $, $ \frac{2}{3} $, $ \frac{7}{15} $ radians.
\smallskip
\textsc{Answer}: $ 60 \SIUnitSymbolDegree $, $ 25 \frac{5 \SIUnitSymbolDegree}{7} $, $ \frac{120 \SIUnitSymbolDegree}{\pi} $, $ \frac{84 \SIUnitSymbolDegree}{\pi} $.
\medskip
3. An arc of a circle whose radius is $ 6'' $, subtends an
angle of 2 radians; how many degrees will be subtended
by the same arc in a circle of $ 4'' $ radius?
\smallskip
\textsc{Answer}: $ \frac{540}{\pi} $ degrees.
\medskip
\begin{wrapfigure}[7]{l}{0.4\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_47_1} \\
\end{center}
\end{wrapfigure}
Graphical determination of the length of a circular arc.
Let $ AB $ be the circular arc
whose length is required. Join
$ AB $, and produce it backward,
making $ BC = \frac{1}{2}AB $. With $ C $ as
centre, and radius $ CA $, describe
an arc of a circle cutting the
tangent at $ B $ in the point $ D $.
Then $ BD $ is approximately equal
to the arc $ AB $.
If $ \angle AOB $ exceed $ 45 \SIUnitSymbolDegree $, it is best to treat the arc in two
parts.
\section*{Numerical Value of Areas.}
\subsection*{PROPOSITION 27.}\label{prop27}
\textbf{To find the area of a rectangle.}
\medskip
\begin{wrapfigure}[4]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_47_2} \\
\end{center}
\end{wrapfigure}
The unit of area is a square, whose sides are equal to
the unit of length. Thus, if 1 inch is the linear unit,
1 sq. inch is the unit of area.
Let $ ABCD $ be a rectangle, and let there be $ l $units in
the length $ AB $, and $ b $ units in the breadth $ AD $. Then, if
$ AB $ be divided into $ l $ equal parts, and
lines be drawn through the points
parallel to $ AD $, the whole rectangle is
divided into $ l $ new rectangles of unit
width.
If now $ AD $ be divided into $ b $ equal
parts, and lines be drawn parallel to $ AD $, each of the former
rectangles is divided into $ b $ new rectangles, whose sides
are of unit length. Therefore, the whole number of units
of area is $ b \; \times \; l $, the product of the linear dimensions of
the length and breadth.
\subsection*{PROPOSITION 28.}\label{prop28}
\textbf{To find the area of a parallelogram.}
\medskip
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_48_2} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be a $ \parallelogram $. Then (by
Prop. 9), the area of the $ \parallelogram $ is equal
to the area of a rectangle, $ A'BCD' $,
having the same base and altitude.
$ \therefore $ the area of $ \parallelogram = \text{base} \; \times \; \text{the
altitude} $.
\subsection*{PROPOSITION 29.}\label{prop29}
\textbf{To find the area of a triangle.}
\medskip
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_48_2} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be a triangle.
Draw $ CD $ parallel to $ AB $ and $ AD $
parallel to M.
Then $ ABCD $ is a parallelogram.
$ \therefore \; \text{area} \; = \; \text{base} \; \times \; \text{altitute.} $
But $ \triangle ABC = \frac{1}{2} \; \parallelogram ABCD $ (by Prop. 8).
$ \therefore \; \text{area of a} \; \triangle = \frac{1}{2} \; \text{base} \; \times \; \text{the altitude.} $
\subsection*{PROPOSITION 30.}\label{prop30}
\textbf{To find the area of a trapezium.}
\medskip
Let $ ABCD $ be a trapezium. Join
$ AC $.
\begin{wrapfigure}[4]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_48_3} \\
\end{center}
\end{wrapfigure}
Then area of $ \triangle ADC = \frac{1}{2} \; DC \; \times \; h $,
and area of $ \triangle ABC = \frac{1}{2} \; AB \; \times \; h $.
\begin{align*}
\therefore \; \text{the area of the trapezium} &= \frac{1}{2} \; DC \; \times \; h + \frac{1}{2} \; AB \; \times \; h \\
&= \frac{1}{2} \; h (AB + DC)\text{.}
\end{align*}
\textit{I.e.}, $ \text{the area of a trapezium} = \text{the altitude} \; \times \; \frac{1}{2} \; \text{the
sum of the parallel sides.} $
\subsection*{PROPOSITION 31.}\label{prop31}
\textbf{To find the area of a quadrilateral.}
\medskip
\begin{wrapfigure}[6]{l}{0.3\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_49_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be a quadrilateral.
Join $ AC $.
Then area of $ \triangle ABC = \frac{1}{2} \; AC \; \times \; h_{1} $,
and area of $ \triangle ACD = \frac{1}{2} \; AC \; \times \; h_{2} $,
$ \therefore $ area of the quadrilateral
$ = \frac{1}{2} \; AC \; \times \; h_{1} + \frac{1}{2} \; AC \; \times \; h_{2} $
$ = \frac{1}{2} \; AC (h_{1} + h_{2}) $
\textit{I.e.}, the area of a quadrilateral $ = \frac{1}{2} $ product of a
diagonal into the sum of the perpendiculars on the
diagonal from the opposite vertices.
\subsection*{PROPOSITION 32.}\label{prop32}
\begin{wrapfigure}[8]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_49_2} \\
\end{center}
\end{wrapfigure}
\textbf{To find the area of a polygon.}
\medskip
Let $ ABCDE $ be any polygon.
Draw any line $ OG $, and drop
perps. upon it from the angles
of the polygon, \textit{viz}., $ p_{1 }$, $ p_{2} $, $ p_{3} $, $ p_{4} $, $ p_{5} $;
and let $ a_{1} $, $ a_{2} $, $ a_{3} $, $ a_{4} $, $ a_{5} $,
be the distances of these perpendiculars
from any point $ O $ in
the line $ OG $.
Then the area of the polygon
\begin{align*}
ABCDE &= \text{trap.} \; AA_{1}B_{1}B + \text{trap.} \; BB_{1}C_{1}C \\
& \qquad - \text{trapms} \; AA_{1}E_{1}E \text{,} \, EE_{1}D_{1}D \text{,} \, DD_{1}C_{1}C \\
&= \frac{1}{2} \; (p_{1} + p_{2}) (a_{2} - a_{1}) + \frac{1}{2} \; (p_{2} + p_{3}) (a_{3} - a_{2}) \\
& \qquad - \frac{1}{2} \; (p_{1} + p_{5}) (a_{5} - a_{1}) - \frac{1}{2} \; (p_{5} + p_{4}) (a_{4} - a_{5}) \\
& \qquad - \frac{1}{2} \; (p_{4} + p_{3}) (a_{3} - a_{4}) \; \text{,}
\end{align*}
which reduces to
\begin{align*}
\frac{1}{2} \; \{ & p_{1}a_{2} - p_{2}a_{1} + p_{2}a_{3} - p_{3}a_{2} + p_{3}a_{4} \\
&- p_{4}a_{3} + p_{4}a_{5} - p_{5}a_{4} + p_{5}a_{1} - p_{1}a_{5}\}
\end{align*}
an expression which can be easily written down, on
account of its symmetry.
This method is convenient in taking out areas in
surveys, \&c.
\subsection*{PROPOSITION 33.}\label{prop33}
\textbf{To find the area of a circle and its sector.}
\medskip
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_50_1} \\
\end{center}
\end{wrapfigure}
Let $ AOB $ be a sector of a circle.
Then if $ AOB $ be supposed divided
into $ n $ smaller equal sectors, such as
$ ODE $, and $ DE $ be joined, the area of
the $ \triangle ODE $ approximates more nearly
to the area of the sector $ ODE $, as the
number of $ \triangle $s is increased, and when,
therefore, the number is indefinitely great, the error is
indefinitely small.
\begin{align*}
\text{But the area of each} \; \triangle &= \frac{1}{2} \; \text{base} \; \times \; \text{the altitude} \text{,}\\
\therefore \; \text{the total area} &= \frac{1}{2} \; (\text{sum of the bases}) \; \times \; \text{the altitude} \\
&= \frac{1}{2} \; \text{arc} \; AB \; \times \; \text{the radius,}
\end{align*}
when the number of $ \triangle $s is infinitely great.
\[ \therefore \; \text{the area of a sector} = \frac{1}{2} \; \text{length of its arc} \; \times \; \text{the radius.} \]
In the case of a complete circle, $ \text{the arc} = \text{the circumference} = 2 \pi r $,
and $ \therefore \; \text{the area of a circle} = \frac{1}{2} \; \times \; 2 \pi r \; \times \; r = \pi r^{2} $
\[ = \frac{\pi d^{2}}{4} = 0.7854 d^{2} \]
\medskip
\begin{wrapfigure}[8]{l}{0.4\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_51_1} \\
\end{center}
\end{wrapfigure}
To find graphically the area of a circular sector.
Let $ OAB $ be a sector. Draw $ BC $ equal in length to the
arc $ AB $. (Prop. 26.)
Join $ OC $. Then $ \triangle OBC = \text{sector} \; OAB $.
\textit{Proof}.\textemdash For area of the sector $ = \frac{1}{2} \; \text{arc} \; \times \; \text{the radius} $, and
the area of the $ \triangle = \frac{1}{2} \; OB \; \times \; BC $,
but $ OB = \text{radius} $, and $ BC = \text{the length of the arc} $. \\
$ \therefore \; \text{area of the} \; \triangle OBC = \frac{1}{2} \; arc \; \times \; \text{the radius} $, \\
and $ \therefore \; \text{area of the sector} \; OAB = \text{area of} \; \triangle OBC $.
\subsection*{PROPOSITION 34.}\label{prop34}
\textbf{To find the area of a circular segment.}
\medskip
Let $ ADB $ be the circular segment.
\begin{wrapfigure}[7]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_51_2} \\
\end{center}
\end{wrapfigure}
Then
\begin{align*}
\text{area} &= \text{area of the sector} \; OADB - \triangle OAB \\
&= \frac{1}{2} \; arc \; \times \; r - \frac{1}{2} \; AB \; \times \; OF.
\end{align*}
\textit{Graphically}.\textemdash By Prop. 33 construct
a $ \triangle OBC = \text{sector} \; OADB $.
Draw $ AE $ parallel to $ OB $, and join $ OE $.
Then $ \triangle OEB = \triangle OAB $ (Prop. 9, Cor. 2),
\begin{align*}
\therefore \; \text{the area of the segment} &= \triangle OBC - \triangle OEB\\
&= \triangle OEC.
\end{align*}
\subsection*{PROPOSITION 35.}\label{prop35}
\textbf{To find areas by a sum-curve.}
\medskip
Let $ OABC $ be an area bounded by the axes $ OX $, $ OY $.
Divide the base into a number of parts (preferably
equal), \textit{viz}., 1, 2, 3, \&c., and draw the vertical ordinates
$ 11' $, $ 22' $, \&c. Also draw the mid-ordinates $ aa' $, $ bb' $, $ cc' $, \&c.,
and project them on the axis $ OY $ at the points $ a'' $, $ b'' $, $ c'' $, \&c.
Take any pole $ Q $ in $ OC $ produced, and join $ Qa'' $, $ Qb'' $,
$ Qc'' $, \&c.
Beginning at $ O $, draw $ O 1'' \parallel Q'' $, $ 1''2'' \parallel Qb'' $, and so on.
\begin{center}
\includegraphics[width=0.75\linewidth]{../images/illus_52_1} \\
\end{center}
\noindent Then the area under the curve $ OABC $ up to any point is
given approximately by the product of the ordinate of the
curve $ O1''2'' $, \&c., up to that point into the polar distance
$ OQ $. This curve is called a \textit{sum-curve}.
\textit{Proof}.\textemdash For consider any $ \triangle 1''2''2''' $ and the $ \triangle QOb'' $.
Since these $ \triangle $s have their sides respectively parallel (by
construction) they are similar,
\[ \text{and} \; \therefore \; \frac{2''2'''}{1''2''} = \frac{Ob''}{QO} \]
and $ \therefore \; 2''2''' \; \times \; QO = Ob'' \; \times \; 1''2'' = bb' \; \times \; 12 = \text{area} \; 11'2'2 $ approximately.
\textit{I.e.}, the area $ 11'2'2 $ which has been added on in going
from 1 to 2 is equal to the increase in the ordinate of the
Sum-Curve $ \; \times \; $ the polar distance, and similarly for the
other elements of area. Wherefore, if we start from $ O $,
the area up to any point is equal to the ordinate up to
that point $ \; \times \; $ the polar distance.
\section*{Solid Geometry.}
42. A \textit{polyhedron} is a figure bounded on all sides by
planes.
43. A \textit{prism} is a polyhedron whose sides are parallelograms,
and whose extremities are equal polygons in
parallel planes.
44. A \textit{parallelopiped} is a polyhedron bounded by three
pairs of parallel planes.
45. A \textit{pyramid} is a polyhedron, one of whose faces is
a polygon, and the others triangles, whose bases are the
sides of the polygon, and having a common vertex.
46. A \textit{tetrahedron} is a pyramid on a triangular base.
47. A \textit{frustum} of a solid is that portion of it contained
between the base and another plane which cuts the solid.
48. A \textit{prismoid} is a solid whose ends are similar
figures, having their sides parallel, and in parallel planes,
or it is a frustum of a pyramid.
\subsection*{PROPOSITION 36.}\label{prop36}
\textbf{The areas of the sections of a pyramid made by
planes parallel to the base, are proportional to the
squares of their distances from the vertex.}
\medskip
\begin{wrapfigure}[7]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_53_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be a pyramid on a $ \triangle $r base $ BCD $, and let
$ EFG $ be a parallel section.
Draw $ APQ $ perpendicular to the
base, meeting the parallel planes in
$ P $ and $ Q $. Join $ EP $, $ BQ $.
Then $ \because \; EF \parallel BC, EG \parallel BD, \text FG \parallel CD $,
the $ \triangle EFG $ is equiangular to the
$ \triangle BCD $, and the $ \triangle AEF $ is similar
to the $ \triangle ABC $, and the $ \triangle AEP $
to the $ \triangle ABQ $,
\[ \therefore \; \frac{\text{area }EFG}{\text{area }BCD} = \frac{\overline{EF^{2}}}{\overline{BC^{2}}} = \frac{\overline{AE^{2}}}{\overline{AB^{2}}} = \frac{\overline{AP^{2}}}{\overline{AQ^{2}}} \quad \text{(Prop. 16 and 13.)} \]
\noindent Also, if the pyramid is on a polygonal base, it can be
decomposed into pyramids on $ \triangle $r bases, and the theorem
proved in the same manner.
\textsc{Cor}.\textemdash If two pyramids are on equal bases, and have
equal altitudes, the sections of them at equal distances
from the base are equal.
\subsection*{PROPOSITION 37.}\label{prop37}
\textbf{The volume of a right prism is equal to the area
of its base multiplied by the height.}
\begin{wrapfigure}[5]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_54_1} \\
\end{center}
\end{wrapfigure}
For, if $ a $, $ b $ be the length and
breadth of the base, there are $ a \; \times \; b $
units of area in the base; and on each
unit of area in the base there are as
many units of volume as there are
units of length in the height $ h $.
$ \therefore \; \text{the whole volume} = abh $.
\subsection*{PROPOSITION 38.}\label{prop38}
\textbf{The volume of an oblique prism is equal to the
area of its right section multiplied by its length.}
\begin{wrapfigure}[7]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_54_2} \\
\end{center}
\end{wrapfigure}
Let $ ABCDEFGHIK $ be an oblique prism, and let
a section at right $ \angle $s to its edges be
made by the plane $ AB'C'D'E' $. Then, if
the wedge-shaped solid $ ABCDEB'C'D'E' $
be placed at the other extremity, $ A $
falling on $ F $, $ B $ on $ G $, and so on, the
volume of the original prism will be
equal to the volume of the right prism
$ AB'C'D'E'FG'H'I'K' = \text{area of its base} \; \times \; \text{height} $ (Prop. 37).
\medskip
\textsc{Cor}. 1.\textemdash The volume of a parallelopiped
= the products of its base
into its altitude.
\medskip
\textsc{Cor}. 2.\textemdash The volume of an oblique $ \triangle $r prism = area of
its base multiplied by the altitude; for it is half the volume
of a parallelepiped.
\medskip
\textsc{Cor}. 3.\textemdash The volume of any oblique prism = area of
its base $\; \times \;$ the altitude, for it may be divided up into
triangular prisms.
\subsection*{PROPOSITION 39.}\label{prop39}
\textbf{Pyramids on equal bases and of equal altitude
are equal in volume.}
\medskip
Let $ ABCD $, $ A'B'C'D' $ be pyramids on equal bases and
having the same altitude.
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_55_1} \\
\end{center}
Then, if each be supposed divided into the same number
of infinitely thin layers, since the corresponding layers in
each are equal in area (Prop. 36, Cor.) and their thickness
is the same, the sum of all those in the first must be equal
to the sum of all those in the second, and the volumes of
the pyramids are therefore equal.
\subsection*{PROPOSITION 40.}\label{prop40}
\textbf{The volume of a pyramid is one-third of the prism
standing on the same base.}
\medskip
\begin{wrapfigure}[6]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_56_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCD $ be a pyramid. Through $ C $, $ D $ draw lines
parallel to $ AB $, and cut them by a plane $ AEF $ parallel to
$ BCD $, forming a prism $ ABCDEF $. Join $ CF $. Then the
prism is divided into three pyramids
$ ABCD $, $ CEAF $, and $ FDCA $.
But $ ABCD = CEAF $, being on
equal bases $ BCD $, $ AEF $, and having
the same altitude (Prop. 39). Also
$ ABCD = FDCA $, being on equal
bases $ ABD $, $ FAD $, and having the
same altitude (Prop. 39).
\[ \therefore \; \text{each pyramid} = \frac{1}{3} \; \text{prism.} \]
\textsc{Cor}. 1.\textemdash The volume of a $ \triangle $r pyramid $ = \frac{1}{3} \; \text{base} \; \times \; \text{altitude} $.
\textsc{Cor}. 2.\textemdash The volume of any pyramid $ = \frac{1}{3} \; \text{base} \; \times \; \text{altitude} $.
\subsection*{PROPOSITION 41.}\label{prop41}
\begin{wrapfigure}[5]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_56_2} \\
\end{center}
\end{wrapfigure}
\textbf{To find the volume of a frustum of a triangular
pyramid between parallel planes, in terms of its
altitude and the areas of its bases.}
\medskip
Let $ ABCDEF $ be a frustum of a pyramid, the plane
$ ABC $ being $ \parallel DEF $.
Let it be cut by planes $ BDF $, $ BAF $
into three pyramids, viz., $ BDEF $,
$ FABC $, $ BADF $.
Let the areas of $ ABC $, $ DEF $ be $ b $, $ B $,
and the altitude of the frustum be $ h $.
Then $ BDEF = \frac{1}{3} \; h \cdot B $ (Prop. 40,
Cor. 1) and $ FABC = \frac{1}{3} \; h \cdot b $ (Prop.
40, Cor. 1).
\begin{align*}
Also \; \frac{BADF}{BACF} = \frac{\text{base} \; ADF}{\text{base} \; ACF} &= \frac{DF}{AC} &\quad \text{(Prop. 11, Cor.)}
&= \frac{\sqrt{B}}{\sqrt{b}} &\text{(Prop. 16.)}
\end{align*}
\[ \text{and} \; \therefore \; BADF = \frac{\sqrt{B}}{\sqrt{b}} \cdot BACF = \frac{\sqrt{B}}{\sqrt{b}} \frac{1}{3} \; bh = \frac{1}{3} \; h \sqrt{Bb} \]
\[ \therefore \; \text{the volume of the frustum} = \frac{1}{3} \; h\{B + \sqrt{Bb} + b\} \]
%% !! Unsure of mathematics here
%% !! This is the sum of the three volumes all with 1/3 h as factors
%% !! But is it \sqrt{Bb} or \sqrt{B}b ?
\noindent that is, is equal to the volume of three pyramids having
the same height as the frustum, and having bases
respectively equal to the parallel faces of the frustum,
and the geometric mean between them.
\textsc{Cor}.\textemdash This property may be extended to the frusta of
all pyramids between parallel planes, by considering
them as made up of the frusta of pyramids on triangular
bases.
\subsection*{Trapezoidal Formula.}
The formula $ \frac{1}{3} \; h \left\{B + \sqrt{Bb} + b \right\} $ may be written
in another form which is convenient in the calculation of
earthwork.
\begin{center}
\includegraphics[width=0.4\linewidth]{../images/illus_57_1} \\
\end{center}
Let $ h_{1} h_{2} h_{3} $ be the distances of the bottom, top and
middle of the frustum from the vertex.
Then if $ M $ be the area of the middle plane,
\[ \frac{M}{b} = \left( \frac{h_{3}}{h_{2}} \right)^{2} \qquad \text{but} \; h_{3} = \frac{h_{1} + h_{2}}{2} \]
\[ \therefore \; M = b \left( \frac{h_{1} + h_{2}}{2h_{2}} \right)^2 = \frac{1}{4} \; b \left( \frac{h_{1}^{2} + h_{2}^{2} + 2h_{1}h_{2}}{h_{2}^{2}} \right) \]
\[ \text{but} \; \frac{h_{1}^{2}}{h_{2}^{2}} = \frac{B}{b} \qquad \therefore \; \frac{h_{1}^{2} + h_{2}^{2}}{h_{2}^{2}} = \frac{B + b}{b} \]
\begin{align*}
\therefore \; M &= \frac{1}{4} \; b \; \left\{ \frac{B + b}{b} + 2 \frac{h_{1}}{h_{2}} \right\} = \frac{1}{4} \; b \; \left\{ \frac{B + b}{b} + 2 \sqrt{ \frac{B}{b}} \right\} \\
&= \frac{B + b}{4} + \frac{1}{2} \sqrt{Bb}
\end{align*}
\[ \therefore \; 4 \;\text{times the middle area} = B + b + 2 \sqrt{Bb} \]
\[ \text{Now,} \; \frac{1}{3} \; h \; \left\{B + \sqrt{Bb} + b \right\} = \]
\[ \frac{h}{6} \; \left\{B + b + B + b + 2 \sqrt{Bb} \right\} \]
= (Sum of the end areas + 4 times the middle area)
multiplied into one-sixth the height of the frustum.
\subsection*{PROPOSITION 42.}\label{prop42}
\textbf{To find the volume of the frustum of a triangular
prism.}
\medskip
\begin{wrapfigure}[5]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_58_1} \\
\end{center}
\end{wrapfigure}
Let $ ABCDEF $ be the frustum.
Let the base $ DEF = b $, and let $ h $, $ h_{1} $, $ h_{2} $ be the altitudes
of $ A $, $ B $, $ C $ above the plane $ DEF $.
First cut the frustum by a plane
$ BDF $.
The volume of the pyramid
$ BDEF = \frac{1}{3} \; bh $. \quad (Prop. 40, Cor. 1.)
Divide the remainder by a plane
$ BAF $ into pyramids $ BADF $, $ ABCF $.
Now since $ BADF $, $ EADF $ stand upon the same base $ ADF $
and have equal altitudes (for $ BE \parallel AD $), $ BADF = EADF= \frac{1}{3} \; bh $ \quad (Prop. 40, Cor. 1);
\noindent also $ ABCF = DBCF = DECF $,
\noindent $ \because $ they stand upon equal bases and have the same altitude,
\noindent $ \therefore \; ABCF = \frac{1}{3} \; bh_{2} $.
$ \therefore \; $ the volume of the whole frustum
\begin{align*}
&= BDEF + BADF + BACF \\
&= \frac{1}{3} \; bh_{1} + \frac{1}{3} \; bh + \frac{1}{3} \; bh_{2}.\\
&= \frac{1}{3} \; b (h_{1} + h_{2} + h).
\end{align*}
\textsc{Cor}. 1.\textemdash In a right prism the volume of the frustum =
the area of its right section multiplied by one−third the
sum of the parallel edges.
\medskip
\textsc{Cor}. 2.\textemdash The volume of the frustum of an oblique
prism = the area of its right section $ \times \; \frac{1}{3} \; $ sum of the
parallel edges.
\begin{wrapfigure}[7]{l}{0.3\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_59_1} \\
\end{center}
\end{wrapfigure}
A useful corollary to Prop. 42 is the case of a prism
bounded by plane surfaces at the
ends, which are parallelograms or
regular polygons.
Let the figure represent such a solid,
where the ends $ AEFB $, $ CDHG $ are
$ \parallelogram $s or regular polygons, and let
$ h_{1} $, $ h_{2} $, $ h_{3} $, $ h_{4} $ be the altitudes of the
points $ A $, $ B $, $ F $, $ E $ above the plane
$ CDHG $, respectively.
Then the volume of the triangular prisms $ ABFHDC $
is $ \frac{1}{3} \; \overline{CDH} (h_{1} + h_{2} + h_{3}) $.
Also $ AEFHGC $ is $ \frac{1}{3} \; \overline{CGH} (h_{1} + h_{4} + h_{3}) $.
But if the ends are $ \parallelogram $s or regular polygons the diagonal
$ CH $ divides the base into equal areas, each equal to
$ \frac{B}{2} $ where $ B $ is the area of the base
$ \therefore \; \text{the volume of the whole solid is} = \frac{1}{3} \; \cdot \frac{B}{2} (2h_{1} + 2h_{3} +
h_{2} + h_{4}) $.
Similarly, if the diagonals $ BE $, $ DG $ be drawn, it may be
shown that the total volume is $ \frac{1}{3} \; \cdot \frac{B}{2} (2h_{4} + 2h_{2} + h_{1} + h_{3}) $
\begin{align*}
\therefore \; 2 \; \text{volume} &= \frac{B}{6} \; (3h_{1} + 3h_{2} + 3h_{3} + 3h_{4}) \\
&= \frac{B}{2} \; (h_{1} + h_{2} + h_{3} + h_{4})\\
\therefore \; \text{the volume} \; V &= \frac{1}{4} \; B (h_{1} + h_{2} + h_{3} + h_{4})
\end{align*}
and similarly, for any other regular polygon of n sides
\[ V = \frac{1}{n} \cdot B (h_{1} + h_{2} + h_{3} + \dots + h_{n}) \]
In calculating the earthwork taken from borrow pits,
the ground is staked out in squares, which must be small
enough that their surfaces may be considered planes,
without sensible error. The height of the horizontal
bottom of the pit being taken with a level, and the heights
of the corners of the square above a known datum being
already determined, the differences give the heights of the
corners, and the volumes can then be determined as above.
In laying out the squares, the tape must be held
horizontally.
\medskip
49. A \textit{wedge} is a polyhedron whose base is a trapezium
and whose edge is parallel to the base.
\subsection*{PROPOSITION 43.}\label{prop43}
\textbf{To find the volume of a wedge.}
\medskip
Let $ ABCDEF $ be a wedge.
\begin{align*}
\text{Then the volume} \; KLCHGD &= \text{base} \; CHL \; \times \; \text{altitude} \; CD. \\
&= \frac{1}{2} \; hl \; \times \; w \text{,}
\end{align*}
where $ l $ is the altitude of the wedge. (Prop. 37.)
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_60_1} \\
\end{center}
Also the volume of the pyramid $ EDGA $
\begin{align*}
&= \frac{1}{3} \; \text{base} \; ADG \; \times \; \text{height} \text{.} \quad \text{(Prop. 40, Cor. 1.)}\\
&= \frac{1}{3} \cdot \frac{1}{2} \cdot AG \cdot hl \text{,} \\ \text{and the volume of the pyramid} \; FCHB \\
&= \frac{1}{3} \; \text{base} \; CHB \; \times \; \text{height} \\
&= \frac{1}{3} \cdot \frac{1}{2} \; \cdot BH \cdot hl \quad \text{(Prop. 40, Cor. 1).}
\end{align*}
Also the volume of the pyramid $ GDKE $
\[ = \frac{1}{3} \; \text{GDK} \; \times \; EK = \frac{1}{3} \; \cdot \frac{1}{3} \; GD \; \times \; l \; \times \; EK\text{,} \]
and the volume of the pyramid HCLF
\[ = \frac{1}{3} \; \text{base} \; HCL \; \times \; LF = \frac{1}{3} \cdot \frac{1}{2} \; \overline{HC} \cdot l \; \times \; \overline{LF} \]
\begin{align*}
\therefore \; \text{the whole volume} &= \frac{1}{2} \; whl + \frac{1}{6} \; \overline{AG} \cdot hl + \frac{1}{6} \; \overline{BH} \cdot hl + \frac{1}{6} \; hl \cdot \overline{EK} + \frac{1}{6} \; hl \cdot \overline{LF}\text{.}\\
&= \frac{lh}{6} \; \left \{3 w + AG + BH + EK + LF \right\}\\
&= \frac{lh}{6} \; \left \{(w + EK + LF) + w + (w + AG + BH) \right\}\\
&= \frac{lh}{6} \; \left(EF + w + AB \right) \text{, that is,}
\end{align*}
Add to the edge of the wedge the sums of the parallel
sides of the base, and multiply the result by one−sixth of
the width of the base multiplied by the altitude of the
wedge.
\section*{VOLUMES BOUNDED BY CURVED SURFACES.}
\section*{The Cylinder.}
50. A \textit{cylinder} is a solid generated by a line which
moves always parallel to itself.
51. A \textit{right circular cylinder} is the solid generated by
the revolution of a rectangle about one of its sides.
\subsection*{PROPOSITION 44.}\label{prop44}
\textbf{To find the lateral surface and volume of a right
circular cylinder.}
\medskip
Since the lateral surface = the area of a rectangle,
whose base is equal to the circumference of the cylinder
$ = \pi d $, and whose height is equal to its height $ h $, the
lateral surface $ = \pi dh $, where $ d $ is the diameter of the
cylinder and $ h $ its height.
Again, since the cylinder may be regarded as a right
prism, whose base is a polygon having an infinite number
of sides, \\
its $ \text{volume} = \text{area of the base} \; \times \; \text{the length} = \pi r^{2}h $, $ r $ being the radius of the base.
\section*{The Cone.}
52. A \textit{cone} is a solid generated by the movement of a
straight line which always passes through a fixed point.
53. A \textit{right circular cone} is the solid generated by the
revolution of a right-angled triangle about one of the sides
containing the right angle.
\subsection*{PROPOSITION 45.}\label{prop45}
\textbf{To find the lateral surface and volume of a right
circular cone.}
\medskip
Let $ ABCDE $ be a right circular cone.
\begin{wrapfigure}[8]{r}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_62_1} \\
\end{center}
\end{wrapfigure}
Inscribe in its base a regular polygon $ BCD\dots $, and let
planes $ ACD $, \&c., be drawn. Thus a polygonal pyramid is
inscribed within the cone, and when
the sides of the polygon are infinite
in number the lateral surface and
volume of the pyramid are equal to
the lateral surface and volume of the
cone.
But the area of a triangle $ ACD = \frac{1}{2} \; CD \; \times $ the perpendicular from $ A $ on
$ CD $, and $ \therefore $ the lateral surface of the
pyramid $ = \frac{1}{2} $ perimeter of the polygon
$ \times $ the perpendicular from $ A $ on one of the sides, and
in the limit, the circumference of the polygon = circumference
of the $\bigcirc$ and the perpendicular from $ A $ on the
side $ = AC $. $ \therefore $ the lateral surface of the cone $ = \frac{1}{2} $ circumference
of the base $ \times $ slant side.
Again, since volume of a pyramid$ = \frac{1}{3} \; \text{base} \; \times \; \text{height} $,
the volume of a cone $ = \frac{1}{3} \; \text{base} \; \times \; \text{height} = \frac{1}{3} \; \pi r^{2}h $.
\subsection*{PROPOSITION 46.}\label{prop46}
\textbf{To find the lateral surface of the frustum of a
cone.}
\medskip
Let $ ABCD $ be the frustum, and let $ EF $ be midway
between $ BC $ and $ AD $. Then the surface may be regarded
as made up of an infinite number
\begin{wrapfigure}[4]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_63_1} \\
\end{center}
\end{wrapfigure}
of trapezia, whose parallel sides are
the sides of regular polygons inscribed
in the circular ends. But
area of a trapezium $ = \frac{1}{2} $ altitude
$\; \times \;$ sum of the parallel sides, and
$ \therefore $ sum of all the trapeziums $ = \frac{1}{2} $ altitude $\; \times $ sum of the circumferences of the parallel
ends.
$ \therefore $ the lateral surface of the frustum slant $ = \frac{1}{2} \; \text{slant side} \; \times $ sum
of the circumferences of the parallel ends $ = \frac{1}{2} \; \text{slant side} \; \times $ circumference of the circle midway between the
ends $ = 2 \pi \cdot AB \cdot EO $.
\section*{The Sphere.}
54. A \textit{sphere} is the solid generated by the revolution of
a semicircle about its diameter.
\subsection*{PROPOSITION 47.}\label{prop47}
\textbf{To find the surface of a sphere.}
\begin{wrapfigure}[6]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_63_2} \\
\end{center}
\end{wrapfigure}
The zone of the sphere generated by a small arc $ AB $ is
ultimately equal to the surface of
the frustum of a cone, whose slant
side is $ AB $ and axis $ HO $, when the
chord $ AD $ becomes infinitely small;
and the surface of the sphere is the
sum of all such zones. Bisect $ AB $ in
$ C $, and draw $ CD $ perpendicular to $ HO $.
The surface of the frustum $ = 2 \pi \cdot CD \cdot AB $ (Prop. 46).
Join $ CO $, and let $ MN $ be the projection of $ AB $ on $ HO $:
then by similar triangles $ \frac{CD}{CO} = \frac{MN}{AB} $
and $ \therefore \; CD \cdot AB = CO \; \times \; MN $
$ \therefore $ the surface of the frustum $ = 2 \pi CO \; \times \; MN $
and $ \therefore $ when the zone is infinitely narrow $ = 2 \pi \cdot r \cdot MN $
Also the sum of all the projections of the arcs, such as
$ MN = 2r $
\[ \therefore \; \text{surface of the sphere} = 2 \; \pi \; r \; \times \; 2 \; r = 4 \pi r^{2} \]
\textsc{Cor}. 1.\textemdash The area of any zone is in proportion to its
height.
\subsection*{PROPOSITION 48.}\label{prop48}
\textbf{To find the volume of a sphere.}
\medskip
Suppose the sphere to be made up of an infinite number
of cones.
Then the volume of each cone $ = \frac{1}{3} \; \text{base} \; \times \; \text{altitude}
= \frac{1}{3} \; \text{base} \; \times \; r $, where $ r $ is the radius of the sphere.
Also the whole volume of the sphere = sum of the
volumes of all the cones
\begin{align*}
&= \frac{1}{3} \; r \; \times \; \text{sum of all the bases} \\
&= \frac{1}{3} \; r \; \times \; 4 \pi r^{2} = \frac{4}{3} \pi r^{3} \quad \text{(Prop. 47).}
\end{align*}
or since $ r^{3} = \frac{d^{3}}{8} $ , the volume of the sphere $ = \frac{\pi d^{3}}{6} $
\section*{PROBLEMS IN PLANE GEOMETRY FOUND USEFUL IN DRAWING.}
\label{p1}
\textsc{Problem 1.}\textemdash \textit{To divide a straight line into two equal
parts.}
Let $ AB $ be the straight line.
\begin{center}
\includegraphics[width=0.25\linewidth]{../images/illus_65_1}
\end{center}
With $ A $ and $ B $ as centres, and with the same radius,
describe arcs intersecting in $ C $ and $ D $. Then a line from
$ C $ to $ D $ bisects $ AB $ in the point $ E $. (Proof by Prop. 5.)
\bigskip
\label{p2}
\textsc{Problem 2.}\textemdash \textit{To divide an angle into two equal parts.}
\begin{wrapfigure}[4]{l}{0.25\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_65_2} \\
\end{center}
\end{wrapfigure}
With $ B $ as centre describe
an arc cutting $ BA $
and $ BC $ in $ D $ and $ E $. With
$ D $ and $ E $ as centres, and
with the same or another
radius describe arcs cutting
in $ F $.
Then $ BF $ bisects the
angle $ ABC $. (Proof by
Prop. 7.)
\medskip
\label{p3}
\textsc{Problem 3.}\textemdash \textit{To divide a line into any number of equal
parts; seven, for example.}
\begin{wrapfigure}[5]{l}{0.5\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{../images/illus_65_3} \\
\end{center}
\end{wrapfigure}
Set off a line $ AC $, and
on it mark off seven
equal parts, starting from
$ A $, and ending at $ D $.
Join $ DB $, and draw
parallels through the points 1, 2, 3, \&c. These will divide
$ AB $ into seven equal parts. (Proof by Prop. 13)
\medskip
\label{p4}
\textsc{Problem 4.}\textemdash \textit{To draw a triangle, whose sides are of
known length.}
\begin{wrapfigure}[4]{r}{0.4\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_66_1} \\
\end{center}
\end{wrapfigure}
Let $ a $, $ b $, $ c $ be the sides of
the triangle. Take any one
of the sides, such as $ c $, and
from its ends, draw arcs
with radii equal to $ a $ and $ b $
intersecting in $ C $. Join $ AC $, $ BC $. (Proof by Prop. 7.)
\medskip
\label{p5}
\textsc{Problem 5.}\textemdash \textit{To inscribe a circle in a given triangle.}
\begin{wrapfigure}[4]{r}{0.4\textwidth}
\vspace{-24pt}
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_66_2} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be the triangle.
Bisect any two of the angles
(Problem 2).
Then the point $ O $, where the
bisecting lines intersect, is the
centre of the circle. (Proof by Prop. 5.)
\medskip
\label{p6}
\textsc{Problem 6.}\textemdash \textit{To circumscribe a circle about a given
triangle.}
\begin{wrapfigure}[4]{r}{0.25\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.25\textwidth]{../images/illus_66_3} \\
\end{center}
\end{wrapfigure}
Let $ ABC $ be the triangle.
Bisect any two of the sides (Problem
1).
Then the point $ O $, where the bisecting
lines intersect, is the centre of the circle.
(Proof by Prop. 6.)
\medskip
\label{p7}
\textsc{Problem 7.}\textemdash \textit{To inscribe a hexagon in a given circle.}
Set off the radius six times round the circle.
\begin{center}
\includegraphics[width=0.25\linewidth]{../images/illus_66_4} \\
\end{center}
Then $ ABCDEF $ will be the hexagon required. (Proof
by Prop. 6, Cor. 2.)
\medskip
\begin{wrapfigure}[7]{l}{0.3\textwidth}
\vspace{-36pt}
\begin{center}
\includegraphics[width=0.3\textwidth]{../images/illus_67_1} \\
\end{center}
\end{wrapfigure}
\label{p8}
\textsc{Problem 8.}\textemdash \textit{To draw a circular
arc through three given points.}
Let $ A $, $ B $, $ C $ be the points.
Bisect $ AB $ and $ BC $ (Problem 1.)
Then the point $ O $, where the
bisecting lines intersect, is the
centre of the arc required.
(Proof by Prop. 20, Cor. 1.)
\medskip
\label{p9}
\textsc{Problem 9.}\textemdash \textit{To inscribe in a given angle a circle of
given radius.}
Let $ ABC $ be the given angle.
Bisect the angle (Problem 2).
\begin{center}
\includegraphics[width=0.4\textwidth]{../images/illus_67_2} \\
\end{center}
Draw $ CD $ at right angles to $ BC $ and equal to the given
radius. Draw $ DO $ parallel to $ BC $. Then $ O $ is the centre
of the circle required. (Proof by Prop. 5.)
\medskip
\label{p10}
\textsc{Problem 10.}\textemdash \textit{To describe a circle of given radius to
touch a given line and a given circle.}
\begin{center}
\includegraphics[width=0.6\linewidth]{../images/illus_67_3} \\
\end{center}
Draw any line $ OE $,
and make $ FE = $ given
radius.
Draw an arc $ EH $ about
$ O $. Set up $ CD = $ radius,
and draw $ CH \parallel $ to $ AB $.
Then $ H $ is the centre of
the circle required.
If it be required to touch on the other side at $ G $, set off
$ GH' = $ given radius, and draw an arc cutting $ CH $ produced
in $ H' $. Then $ H' $ is the centre required.
\medskip
\label{p11}
\textsc{Problem 11.}\textemdash \textit{To describe a circle, whose radius is given,
to touch two given circles.}
Let $ A $ and $ B $ be the centres of the given circles.
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_68_1} \\
\end{center}
Set off $ CD $, $ EF $ equal to the given radius, and draw
arcs to intersect in $ O $. Then $ O $ is the centre of the circle
required.
\medskip
\label{p12}
\textsc{Problem 12.}\textemdash \textit{To describe a circle tangent to a given
line at a given point, and touching a given circle.}
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_68_2} \\
\end{center}
Let $ A $ be the centre of the given circle, $ CD $ the given
line, and $ P $ the given point. Set off $ PE = AF $ and
bisect $ AE $ at right angles. Then $ O $ is centre of circle
required.
\medskip
\label{p13}
\textsc{Problem 13.}\textemdash \textit{To describe a circle tangent to a given line
and touching a given circle
in a given point.}
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_69_1} \\
\end{center}
Let $ A $ be the centre of
the given circle, $ P $ the point,
and $ BC $ the given line.
Join $ AP $ and produce.
Draw $ PD $ perpendicular to
$ AP $. Describe semicircle
$ EPF $. Erect a perpendicular $ FG $. Then $ G $ is centre of
circle required.
\medskip
\label{p14}
\textsc{Problem 14.}\textemdash \textit{To draw a circle to touch three given
straight lines.}
Let $ AB $, $ CD $, $ EF $ be the given lines.
\begin{center}
\includegraphics[width=0.5\linewidth]{../images/illus_69_2} \\
\end{center}
Bisect the angles. The point of intersection of the
bisectors will be the centre of the required circle.
This is equivalent to finding the centre of the escribed
circle of a triangle.
\vskip2cm
\centering{{\small \textbf{BRADBURY, AGNEW, \& CO. LD., PRINTERS, LONDON AND TONBRIDGE.}}}
\begin{center}
[The end of \textbf{Geometry for Technical Students} \\
by E. H. (Ernest Headly) Sprague]
\end{center}
\end{document}