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CHAUVENET'S

TREATISE ON

Elementary Geometry

REVISED AND ABRIDGED

W. E. BYERLY

PROFESSOR OF MATHEMATICS IN HABVABD ONIVERSITY

PHILADELPHIA

J. B. LIPPII^COTT COMPANY

1892.

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3TER£OTYPERSAN0PRI!NT^s]ill'

PREFACE.

In preparing this edition of Chauvenet's Geometry
I have endeavored to compel the student to think and
to reason for himself, and I have tried to emphasize the
fact that he should not merely learn to understand and
demonstrate a few set propositions, hut that he should
acquire the power of grasping and proving any simple
geometrical truth that may he set before him ; and this
power, it must be remembered, can never be gained by
memorizing demonstrations. Systematic practice in de-
vising proofs of new propositions is indispensable.

On this account the demonstrations of the main propo-
sitions, which at first are full and complete, are gradually
more and more condensed, until at last they are some-
times reduced to mere hints, by the aid of which the
full proof is to be developed; and numerous additional
theorems and problems are constantly given as exercises
for practice in original work.

A syllabus, containing the axioms, the postulates, and
the captions of the main theorems and corollaries, has
been added to aid student and teacher in reviews and
examinations, and to make the preparation of new proofs
more easy and definite.

6 ELEMENTS OF GEOMETRY.

In the order of the propositions I have departed con-
siderably from the larger Chauvenet's Geometry, with
the double object of simplifying the demonstrations
and of giving the student, as soon as possible, the few
theorems which are the tools with which he must most
frequently work in geometrical investigation.

Teachers are strongly advised to require as full and
formal proofs of the corollaries and exercises as of the
main propositions, and to lay much stress upon written
demonstrations, which should be arranged as in the
illustrations given at the end of Book I.

In preparing a written exercise, or in passing a written
examination, the student should have the syllabus before
him, and may then conveniently refer to the propositions
by number. In oral recitation, however, he should quote
the full captions of the theorems on which he bases his
proof.

W. E. BYEELY.

Cambridge, Mass., 1887.

CONTENTS.

PA OB

INTKODUCTION 9

PLANE GEOMETEY.

BOOK I.

Rectilinear Figures 12

Exercises on Book 1 49

BOOK II.
The Circle. Ratio. Incommensurables. Doctrine of Limits.

Measure of Angles 55

Exercises on Book II 94

BOOK III.

Theory of Proportion. Proportional Lines. Similar Poly-
gons 101

Exercises on Book III 125

BOOK IV.

Comparison and Measurement of the Surfaces of Recti-
linear Figures 130

Exercises on Book IV 146

BOOK V.

Regular Polygons. Measurement of the Circle 149

Exercises on Book V 169

Miscellaneous Exercises on Plane Geometry 173

Syllabus of Plane Geometry. Postulates, Axioms, and

Theorems 182

8 CONTENTS. ,

GEOMETEY O^ SPACE.
BOOK VI.

PAGK

The Plane. Diedral Angles. Polyedral Angles .... 195
Exercises on Book VI 220

BOOK VII.

POLYEDRONS 224

Exercises on Book VII 254

BOOK VIII.

The Three Bound Bodies. The Cylinder. The Cone.
The Sphere. Spherical Triangles. Spherical Poly-
gons 257

Exercises on Book VIII 287

BOOK IX.

Measurement of the Three Bound Bodies 290

Exercises on Book IX 308

Miscellaneous Exercises on Solid Geometry 309

Syllabus of Propositions in Solid Geometry. Theorems . 311

ELEMENTS OF GEOMETRY.

IKTEODUOTIOK

1. Every person possesses a conception of space indefi-
nitely extended in all directions. Material bodies occupy
finite, or limited, portions of space. The portion of space
which a body occupies can be conceived as abstracted from
the matter of which the body is composed, and is called a
geometrical solid. The material body filling the space is
called a physical solid. A geometrical solid is, therefore,
merely the fornij or figure, of a physical solid. In this work,
since only geometrical solids will be considered, we shall, for
brevity, call them simply solids, and we shall define them
formally, as follows :

Definition. A solid is a limited or bounded portion of space,
and has length, breadth, and thickness.

2, The boundaries of a solid are surface^.

Definition. A surface is that which hae length and breadth,
but no thickness. 1

If a surface is bounded, its boundaries are lir^es.

If two surfaces intersect, their intersection is a line.

Definition. A line is that which has length, but neither
breadth nor thickness.

10 ELEMENTS OF GEOMETRY.

If a line is terminated, it is terminated by points.
If two lines intersect, they intersect in a point.
Definition. A point has position, but neither length, breadth,
nor thickness.

3. If we suppose a point to move in space, its path will
be a line, and it is often convenient to regard a line as the
path, or locus, of a moving point.

If a point starts to move from a given position, it must
move in some definite direction; if it continues to move in
the same direction, its path is a straight line. If the direc-
tion in which the point moves is continually changing, the
path is a curved line.

If a point moves along a line, it is said to describe the
line.

By the direction of a line at any point we mean the direc-
tion in which a point describing the line is moving when it
passes through the point in question.

Definitions. A straight line is a line which -f f

has everywhere the same direction.

A curved line is one no portion of
which, however short, is straight.

A broken line is a line composed of
different successive straight lines.

4. Definitions. A plane surface, or
simply a plane, is a surface such that, if
any two points in it are joined by a
straight line, the line will lie wholly in
the surface.

A curved surface is a surface no portion of which, however
small, is plane.

5. Definitions. A geometrical figure is any combination of
points, lines, surfaces, or solids, formed under given condi-

\ INTRODUCTION. 11

tions. Figures formed by points and lines in a plane are
called plane figures. Those formed by straight lines alone
are called rectilinear^ or right-lined^ figures; a straight line
being often called a right line.

6. Definitions. Geometry may be defined as the science of
extension and position. More specifically, it is the science
which treats of the construction of figures under given con-
ditions, of their measurement and of their properties.

Plane geometry treats of plane figures.
The consideration of all other figures belongs to the geom-
etry of space, also called the geometry of three dimensions.

7. Some terms of frequent use in geometry are here de-
fined.

A theorem is a truth requiring demonstration. A lemma
is an auxiliary theorem employed in the demonstration of
another theorem. A problem is a question proposed for solu^
tion. An axiom is a truth assumed as self-evident. A postu-
late (in geometry) assumes the possibility of the solution of
some problem.

Theorems, problems, axioms, and postulates are all called
propositions.

A corollary is an immediate consequence deduced from one
or more propositions. A scholium is a remark upon one or
more propositions, pointing out their use, their connection,
their limitation, or their extension. An hypothesis is a sup-
position, made either in the enunciation of a proposition or
in the course of a demonstration.

PLANE GEOMETRY.

BOOK I

RECTILINEAR FIGURES.
ANGLES.

1. Definition. A jplane angle^ or simply an angle, is the
amount of divergence of two lines which meet in a point or
which would meet if produced (i.e., prolonged).

The point is called the vertex of the angle,
and the two lines the sides of the angle.

From the definition it is clear that the mag-
nitude of an angle is independent of the length of its sides.

An isolated angle may be designated by the letter at its
vertex, as " the angle O;" but when several angles are formed
at the same point by different lines, as OA, OB,
OC, we designate the angle intended by three 5

letters; namely, by one letter on each of its yC^'^^

sides, together with the one at its vertex, q— -i

which must be written between the other two.

Thus, with these lines there are formed three different angles,

which are distinguished as AOB, BOC, and AOG.

Two angles, such as AOB, BOO, which have the same
vertex 0 and a common side OB between them, are called
adjacent.
12

BOOK I. 13

2. Definitions. Two angles are equal when one can be
superposed upon the other, so that the vertices shall co-
incide and the sides of the first shall fall along the sides of
the second.

Two angles are added by placing them in the same plane
with their vertices together and a side in common, care being
taken that neither of the angles is superposed upon the
other. The angle formed by the exterior sides of the two
angles is their sum.

3. A clear notion of the magnitude of an angle will be
obtained by supposing that one of its sides, as OjB, was at
first coincident with the other side OA^ and

that it has revolved about the point 0 (turning
upon 0 as the leg of a pair of dividers turns
upon its hinge) until it has arrived at the posi-
tion OB. During this revolution the movable side makes
with the fixed side a varying angle, which increases by in-
sensible degrees, that is, continuously ; and the revolving line
is said to describe^ or to generate^ the angle AOB. By con-«
tinning the revolution, an angle of any magnitude may be
generated.

4. Definitions. When one straight line meets another, so as
to make two adjacent angles equal, each of these angles is
called a right angle; and the first line is

said to be perpendicular to the second.

Thus, if ^ 00 and BOC are equal angles,

each is a right angle, and the line CO is

perpendicular to AB.

Intersecting lines not perpendicular are said to be oblique
to each other.

An acute angle is less than a right angle.

An obtuse angle is greater than a right angle.

14 ELEMENTS OF GEOMETRY.

5. Definition. Two straight lines lying in the same plane
and forming no angle with each other — that is, two straight
lines in the same plane which will not meet, however far
produced — are parallel.

TKIANGLES.

6. Definitions. A plane triangle is a portion of a plane
bounded by three intersecting straight lines; as A 50. The
sides of the triangle are the portions of the
bounding lines included between the points of
intersection; viz., AB, BC, CA. The angles
of the triangle are the angles formed by the
sides with each other; viz., CAB, ABC, BCA. The three
angular points. A, B, C, which are the vertices of the angles,
are also called the vertices of the triangle.

If a side of a triangle is produced, the angle
which the prolongation makes with the adja-
cent side is called an exterior angle ; as A CD.

A triangle is called scalene (ABC) when no two of its sides
are equal ; isosceles (DEF) when two of its sides are equal ;
equilateral {GUI) when its three sides are equal.

A right triangle is one which has a right angle ; as MNP,
which is right-angled at ]V. The side MP, opposite to the
right angle, is called the hypotenuse.

The base of a triangle is the side upon which it is supposed

BOOK I. 15

to stand. In general, any side may be assumed as the base ;
but in an isosceles triangle BEF^ whose sides DE and DF
are equal, the third side EF is always called the base.

When any side ^(7 of a triangle has been adopted as the
base, the angle BA G opposite to it is* called
the vertical angle, and its angular point A
the vertex of the triangle. The perpen-
dicular AD let fall from the vertex upon
the base is then called the altitude of the triangle.

7. Definition. Equal figures are figures which can be made
to coincide throughout if one is properly superposed upon
the other.

Eoughly speaking, equal figures are figures of the same
size and of the same shape ; equivalent figures are of the
same size but not of the same shape; and similar figures
are of the same shape but not of the same size.

POSTULATES AND AXIOMS.

8. Postulate I. Through any two given points one straight
line, and only one, can be drawn.

Postulate II. Through a given point one straight line, and
only one, can be drawn having any given direction.

9. Axiom I. A straight line is the shortest line that can
be drawn between two points.

Axiom II. Parallel lines have the same direction.

16 ELEMENTS OF GEOMETEY.

PROPOSITION I.— THEOREM.

10. At a given point in a straight line one perpendicular to
the line can be drawn, and but one.

Let 0 be the given point in the line AB.

Suppose a line OD, constantly passing through 0, to
revolve about 0, starting from the position OA and stopping
at the position OB.

The angle which 01) makes with OA will at first be less
than the angle which it makes with OB, and will eventually
become greater than the angle made with
OB. ^ D

Since the angle DOA increases continu- j /
ously (3), the line OB must pass through ]/_

... BOA

one position m which the angles DOA and

DOB are equal. Let OC he this position. Then 0(7 is

perpendicular to ^jB by (4).*

There can be no other perpendicular to AB at 0, for if
OD is revolved from the position 0(7 by the slightest amount
in either direction, one of the adjacent angles will be in-
creased at the expense of the other, and they will cease to
be equal.

11. Corollary. Through the vertex of any given angle one
line can be drawn bisecting the angle, and but one.

Suggestion. Suppose a line OD to revolve
about O, as in the proof just given.

* An Arabic numeral alone refers to an article in the same Book ; but
in referring to articles in another Book, the number of the Book is also
given.

BOOK I.

PROPOSITION II.— THEOREM.

12. All right angles are equal.

Let AOG and AIQfC be any two right angles.

Superpose A'CfC upon AOC^
placing the point (J upon the
point 0 and making the line C/A'
fall along the line OA; then will
OC coincide with 00, for other- 5 —
wise we should have two perpen-
diculars to the line AB at the same point 0, which, by Propo-
sition I., is impossible.

The angles A'C/C and AOC are then equal by definition (2).

PROPOSITION III.— THEOREM.

13. The two adjacent angles which one straight line makes
with another are together equal to two right angles.

If the two angles are equal, they are right angles by
definition (4), and no proof is necessary.

If they are not equal, as AOD and BOD, still the sum of
AOD and BOB is equal to two right angles.

Let OC be drawn at 0 perpendicular to
AB.

The angle AOD is the sum of the two
angles AOC and COD (2). Adding the
angle BOD, the sum of the two angles
AOD and BOD is the sum of the three angles AOC, COD,
and BOD.

The first of these three is a right angle, and the other two
are together equal to the right angle BOC ; hence the sum
of the angles AOD and BOD is equal to two right angles.

ELEMENTS OF GEOMETRY.

14. Corollary I. The sum of all

angles having a common vertex^ and
formed on one side of a straight line, ^
is two right angles.

15. Corollary II. The sum of all

the angles that can be formed about

a point in a plane is four right
angles.

EXERCISE.

Theorem. — If a line is perpendicular to a second line, then
reciprocally the second line is perpendicular to the first.

16. Definition. When the sum of two angles is equal to a
right angle, each is called the complement
of the other. Thus, J) 0(7 is the comple-
ment of A ODj and A OD is the complement
of DOC.

When the sum of two angles is equal to b
two right angles, each is called the supple-
ment of the other. Thus, BOD is the supplement of AOD,
and AOD is the supplement of BOD.

It is evident that the complements of equal angles are
equal to each other ; and also that the supplements of equal
angles are equal to each other.

BOOK I. 19

PROPOSITION IV.— THEOREM

17. If the sum of two adjacent angles is equal to two right
angles^ their exterior sides are in the same straight line.

Let the sum of the adjacent angles AOD^ BOD, be equal
to two right angles ; then OA and OB are
in the same straight line.

For BOD is the supplement of A 02),

and is therefore identical with the angle

which OD makes with the prolongation of ^ 0 (Proposition

TIL).

Therefore OB and the prolongation of AO are the same
line.

18. Every proposition consists of an hypothesis and a con-
clusion. The converse of a proposition is a second proposition
of which the hypothesis and conclusion are respectively the
conclusion and hypothesis of the first. For example, PropO'
sition III. may be enunciated thus :

Hypothesis — if two adjacent angles have their exterior
sides in the same straight line, then — Conclusion — the sum
of these adjacent angles is equal to two right angles.

And Proposition TV. may be enunciated thus :

Hypothesis — if the sum of two adjacent angles is equal to
two right angles, then — Conclusion — these adjacent angles
have their exterior sides in the same straight line.

Each of these propositions is, therefore, the converse of the
other.

A proposition and its converse are, however, not always
both true.

20 ELEMENTS OF GEOMETRY.

PROPOSITION v.— THEOREM.

19. If two straight lines intersect each other, the opposite (or
vertical) angles are equal.

B 0

Let AB and CD intersect in 0 ; then will
the opposite, or vertical, angles AOG and
BOD be equal.

For each of these angles is, by Proposi-
tion III., the supplement of the same angle BOC, and hence,
they are equal (16).

In like manner it can be proved that the opposite angles
AOD and BOC are equal.

EXERCISES.

1. Theorem. — The line which bisects one of two vertical angles
bisects the other.

2. Theorem. — The straight lines which
bisect a pair of adjacent angles formed
by two intersecting straight lines are per-
pendicular to each other.

Suggestion. Prove EOH = FOH.

PROPOSITION VI.— THEOREM.

20. Two triangles are equal when two sides and the included
angle of the one are respectively equal to two sides and the
included angle of the other.

In the triangles ABC, DEF, let AB be equal to DE, BC to
EF, and the included

' AD D

angle B equal to the
included angle E; then
the triangles are equal.
For, superpose the

.X*

B C E F F E

triangle ABC upon the triangle DEF, placing the point B

BOOK I. 21

upon the point E, and making the side BC fall along the
side EF. Then, since J5(7 is equal to EF by hypothesis, the
point G will fall upon the point F.

Since the angle B is equal to the angle E^ and the side
BG has been made to coincide with the side EF^ BA must
fall along ED^ by definition (2) ; and, as BA is equal to ED^
the point A will fall on the point D.

Since the point G has been proved to coincide with the
point F^ and the point A with the point D, the side GA must
coincide with the side FD^ by Postulate I. (8).

The two triangles have now been proved to coincide
throughout, and are equal, by definition (7).

21. Scholium. When two triangles are equal, the equal
angles are opposite to the equal sides.

PROPOSITION VII.— THEOREM.

22. Two triangles are equal when a side and the two adjacent
angles of the one are respectively equal to a side and the two
adjacent angles of the other.

In the triangles ABG, DEF, let BG be equal to EF, and
let the angles B and G adjacent to BGhe respectively equal
to the angles E and F
adjacent to EF; then
the triangles are equal.

For, superpose the \ \ \ \ ..--''' /

..1.^^ , B C E F F E

triangle ABG upon the

triangle BEF, placing the point B upon the point E, and

making the side BG fall along the side EF.

Since J5(7 is equal to EF, the point G will fall upon the
point F.

Since the angle B is equal to the angle E, by hypothesis,

22 ELEMENTS OF GEOMETRY.

and the side 5(7 has been made to coincide with the side EF^
BA must fall along ED^ and the point A will fall somewhere
on the side ED, or on that side extended.

Since the angle G is equal to the angle P, by hypothesis,
and BC coincides with EF^ CA must fall along FD, and the
point A will fall somewhere on the line FD, or on that line
extended.

Since A has been proved to lie upon ED, and also upon
FD, it must coincide with the only point they have in com-
mon, the point D.

Hence the triangles coincide throughout, and are equal.

PEOPOSITION VIII.— THEOREM.

23. In an isosceles triangle the angles opposite the equal sides
are equal.

Let ^5 and AC he the equal sides of the isosceles triangle
ABC; then the angles B and C are equal.

Through the vertex A draw a line AD, bisect-
ing the angle BAC, and meeting the side BC t
atD. /

In the triangles ABD and ACD the side AB
is equal to the side AChy hypothesis, the side
AD is common, and the included angle BAD is equal to
the included angle CAD by construction. The triangles are
therefore equal, by Proposition YI., and the angle C of the
one is equal to the angle B of the other, by (21).

24. Corollary. The straight line bisecting the vertical angle
of an isosceles triangle bisects the base, and is perpendicular to
the base.

EXERCISE.

Theorem. — An equilateral triangle is also equiangular.

BOOK I. 23

PROPOSITION IX.— THEOREM.

25. Two triangles are equal when the three sides of the one
are respectively equal to the three sides of the other.

In the triangles ABC, DEF, let AB be equal to BE, AG
to DF, and BG to EF ; then
the triangles are equal. ^

For, suppose the triangle ^^ \ ^^

ABG to be placed so that its ^ ^ \^^

base ^(7 coincides with its equal
EF, but so that the vertex A

falls on the opposite side of EF from X), as at G, and join
D and 6^ by a straight line.

The triangle EBG is isosceles, since the side ED is equal
to the side EG by hypothesis; therefore the angles EBG
and EQD are equal, by Proposition YIII.

The triangle FBG is isosceles, since the side FI) is equal
to the side FG by hypothesis; therefore the angles FDG
and FGD are equal, by Proposition YIII.

If to the equal angles EDG and EGB we add the equal
angles FBG and FGB, the sums will be equal, and we have
the whole angle EBF equal to the whole angle EGF.

The two triangles EBF and EGF have now the side EB
equal to the side EG by hypothesis, the side BF equal to
the side FG by hypothesis, and the included angle EBF
proved equal to the included angle EGF. Hence the tri-
angles are equal, by Proposition YI.

EXERCISE.

Theorem. — A line drawn from the vertex of an isosceles tri-
angle to the middle point of the base is perpendicular to the base,
and bisects the vertical angle.

24 ELEMENTS OF GEOMETRY.

PROPOSITION X.— THEOREM.

26. Two right triangles are equal when they have the hypote-
nuse and a side of the one respectively equal to the hypotenuse
and a side of the other.

In the right triangles ABC, A'B'G\ let the hypotenuse AB
be equal to the hypotenuse
A'B', and the side ^C be
equal to the side BC^ ; then
the triangles are equal.

Extend the side BC to D,
making CD equal to BC, and

join A and D ; and extend B'C^ to iX, making C^jy equal to
B'C, and join A' and IT.

The triangle ADC and the triangle ABC having the side
A 0 in common ; the side CD equal to the side CB by con-
struction ; and the included angle A CD equal to the included
angle ACB, since they are adjacent angles and ACB is ^
right angle by hypothesis, are equal, by Proposition YI.

In like manner the triangle A!D^C' may be proved equal
to the triangle MB'C,

The triangles BAD and B'A!D^ having the side AB equal
to the side A!B! by hypothesis; the side BD equal to the
side B'D'^ because they were constructed the doubles oi BG
and B'C'^ which were equal by hypothesis; and the side AD
equal to the side A!D\ since they have been proved to be
equal respectively to the sides AB and A!B' ; are equal to
each other, by Proposition IX., and B is equal to B'.

The triangles ABC and A!B'C' have now been proved to
have two sides and the included angle of the one respectively
equal to two sides and the included angle of the other, and
are equal, by Proposition VI.

BOOK I. 25

PROPOSITION XI.— THEOREM.

27. If two angles of a triangle are equals the sides opposite to
them are equals and the triangle is isosceles.

Let the angles BAG and BCA of the triangle ABC be
equal, then are the sides AB and BG equal.

For, if AB and BG are not equal, one must
be greater than the other. Suppose AB greater
than BG.

Then cut oif from AB a part AD equal to
BG, and join D and G. Compare now the
triangle ADG with the whole triangle ABG, of which it is a
part.

The two triangles have the side A C in common ; the side
AD equal to the side BG hy construction; and the included
angle A equal to the included angle BGA by hypothesis.
Therefore, by Proposition YI., the triangles ADG and ABG
are equal, which is impossible. Consequently, AB could not
have been greater than BG.

In like manner we can prove that BG cannot be greater
than AB.

Therefore, since neither can be greater than the other, AB
and BG are equal.

EXERCISE.

Theorem. — An equiangular triangle is also equilateral.

^ r n (h

26 ELEMENTS OF GEOMETRY.

PKOPOSITION XII.— THEOEEM.

28. If two angles of a triangle are unequal^ the side opposite
the greater angle is greater than the side opposite the less angle.

In the triangle ABC let the angle C be greater than the
angle B ; then AB is greater than A C.

For, suppose the line CD to be drawn, cutting
off from the greater angle a part BCD = B.
Then BDC is an isosceles triangle, by Propo-
sition XI., and DC = DB. But in the triangle
ADC we have ^D + DC > AC, by Axiom I. ; or,
putting DB for its equal DC,AD + DB:>AC;ovAB:>Aa

PROPOSITION XIII.— THEOREM.

29. If two sides of a triangle are unequal, the angle opposite
the greater side is greater than the angle opposite the less side.

In the triangle ABC let the side AB be greater than the
side BC; then will the angle C be greater than
the angle A.

For, if C is not greater than J., it must be
equal to A or less than A.

C cannot be equal to A, for in that case AB ^ ^

and BC would be equal, by Proposition XI.

C cannot be less than J., for in that case AB would be
less than BC, by Proposition XII.

Therefore C is greater than A.

BOOK I. 27

PROPOSITION XIV.— THEOREM.

30. If two triangles have two sides of the one respectively
equal to two sides of the other, and the included angles unequal,
the triangle which has the greater included angle has the greater
third side.

Let ABC and ABB be the two triangles in which the sides
AB, AC are respectively equal to the
sides AB, AD, but the included angle
BAC is greater than the included angle
BAD; then ^Ois greater than BD.

For, suppose the line AE to be drawn,

bisecting the angle CAD and meeting BC

in E; join DE. The triangles AED and AEC are equal, by

Proposition YI., and therefore ED = EC. But in the triangle

BDE we have

BE + ED:> BD, by Axiom I.,

and substituting EC for its equal ED,

BE+ EC> BD, or BC > BD.

PROPOSITION XV.— THEOREM.

31. If two triangles have two sides of the one respectively
equal to two sides of the other, and the third sides unequal, the
triangle which has the greater third side has the greater included
angle.

In the triangles ABC and DEF let AB = DE, AC = DF,
and let 5(7 be greater than EF ;
then will the angle A be greater
than the angle D.

For, if A were equal to D, BC
would be equal to EF, by Proposi-
tion YI. ; and if A were less than
D, BC would be less than EF, by Proposition XIY.

28 ELEMENTS OF GEOMETRY.

PEOPOSITION XVI.— THEOEEM.

32. From a given pointy without a straight line^ one perpen-
dicular can be drawn to the line^ and but one.

Let AB be the given line and P the given point.
Take a second line DJS, and at some point F of DE let
a perpendicular be erected (Proposition I.). Superpose this

second figure upon the

p ^

first, placing the line

DF upon the line AB,
and then move the fig-
ure along, keeping DF ^ ^ J J J
always in coincidence

with AB, until the perpendicular FGr passes through P; we
shall then have a perpendicular to AB drawn through P.
Let PC in the figure below be this perpendicular.

No other perpendicular from P can be drawn to the line AB.
For, suppose that a second perpendicular FD could be drawn.

Extend FG to P', making GF' equal to PC, and join J>
andP'.

The two triangles FGF and F'GF have
the side FG equal to the side F'G by con-
struction; the side GF common; and the
included angle FGF equal to the included
angle F'GD, by Proposition III. There-
fore, by Proposition YI., the triangles are
equal, and the angle FDG is equal to the
angle F'FG. But FDG is a right angle by hypothesis;
therefore F'DG must be a right angle, and FD and DF
must lie in the same straight line, by Proposition lY. ; and
we have two straight lines drawn between P and P', which,
by Postulate I., is impossible.

BOOK I. 29

Since this impossible result follows necessarily from the
assumption that a second perpendicular can be drawn from
P to AB, that assumption must be false.

EXERCISES.

1. Theorem. — A perpendicular let fall from the vertex of an
isosceles triangle upon the base bisects the base and bisects the
vertical angle.

2. Theorem. — Two right triangles are equal when they have
the hypotenuse and an adjacent angle of the one respectively
equal to the hypotenuse and an adjacent angle of the other.

Suggestion. Superpose the second triangle upon the first,
making the given equal angles coincide.

PROPOSITION XVII.— THEOREM.

33. The perpendicular is the shortest line that can be drawn
from a point to a straight line.

Let PC be the perpendicular and PD
any oblique line from the point ^P to the
line AB. Then PC is shorter than PD.

Extend PC to P', making CP' equal to
PC, and join D and P'.

The triangles PCD and P'CD are equal,
by Proposition YI. Therefore P'B = PD.

PP' <^PP + DP', by Axiom I.

Therefore PC, the half of PP', is less than PD, the half
of PJ)P\

EXERCISES.

1. Theorem. — Two oblique lines drawn from a -point to a line,
and meeting the line at equal distances from the foot of the per-
pendicular from the given point, are equal.

2. Theorem. — Two equal oblique lines drawn from a point to
a line meet it at equal distances from the foot of the perpen-
dicular.

ELEMENTS OF GEOMETEY.

PROPOSITION XVIII.— THEOREM.

34. If a perpendicular is erected at the middle of a straight
linCj then,

1st. Uvery point in the perpendicular is equally distant from
the extremities of the line ;

2d. Every point without the perpendicular is unequally distant
from the extremities of the line.

Let AB be a finite straight line and
CD a perpendicular at its middle point.

1st. Then is any point P on CD equi-
distant from A and B. For, join P and
A and P and J5.

The triangles FCA and FOB are equal,
by Proposition YI. ; therefore FA and FB are equal.

2d. Any point Q without the perpen-
dicular is unequally distant from A and
B. For, Q being on one side or the other
of the perpendicular, one of the lines QA,
QB must cut the perpendicular; let it
be QA and let it cut in P; join FB. The straight line QB
is less than the broken line QFB^ by Axiom I. ; that is,
C^ < §P + FB. But FB = FA ; therefore QB <: QF +
FA, or QB < QA.

35. Definition. A geometric locus is the geometric figure
containing all the points which possess a common property,
and no others.

In this definition, points are understood to have a common
property when they satisfy the same geometrical conditions.

Thus, since all the points in the perpendicular erected at
the middle of a line possess the common property of being
equally distant from the extremities of the line (that is,

BOOK I. 31

satisfy the condition that they shall be equally distant from
those extremities), and no other points possess this property^
the perpendicular is the locus of these points; so that the
preceding proposition is fully covered by the following brief
statement :

The perpendicular erected at the middle of a straight line is
the locus of the points which are equally distant from the extrem-
ities of that line,

PROPOSITION XIX.— THEOREM.

36. Uvery point in the bisector of an angle is equally distant
from the sides of the angle ; and every point not in the bisector,
but within the angle, is unequally distant from the sides of the
angle ; that is, the bisector of an angle is the locus of the points
within the angle and equally distant from its sides.

1st. Let AD be the bisector of the angle JBAC, P any
point in it, and PE, PF, the perpendicular
distances of P from AB and AC ; then
PE = PF.

For, the right triangles APE, APF,
having the angles PAE and PAF equal,
and AP common, are equal (32, Exercise
2) ; therefore PE = PF.

2d. Let Q be any point not in the bisector, but within
the angle; then the perpendicular distances QE and QH
are unequal.

For, suppose that one of these distances, as QE, cuts the
bisector in some point P; from P let PF be drawn perpen-
dicular to AC, and join QF. We have QJff <C QF ; also^
§P < §P + PF, or QF<iQP + PE, or QF < QE ^ there-
fore qe:< QE.

ELEMENTS OF GEOMETRY.

When the angle BAG is obtuse, the point Q, not in the
bisector, may be so situated that the perpendicular on one
of the sides, as AB, will fall at the
vertex A ; the perpendicular QS is
then less than the oblique line QA.
Or, a point §' may be so situated
that the perpendicular §'^', let fall
on one of the sides, as AB, will meet

that side produced through the vertex A ; this perpendicular
must cut the side AC in. some point, K^ and we then have

EXEBCISE.

Theorem, — The locus of the points equally distant from two
intersecting straight lines is the pair of lines which bisect all
the angles formed by the given lines.

(v. 19, Exercise 2.)

37. Definition. A broken line, as ABCBE^ is called convex
when no one of its component straight

lines, if produced, can enter the space ^

enclosed by the broken line, and the
straight line joining its extremities.

PROPOSITION XX.— THEOREM.

38. A convex broken line is less than any other line which
envelops it and has the same extremities.

Let the convex broken line AFQE have the same extrem-
ities A, E^ as the line ABODE, and be
enveloped by it; that is, wholly in-
cluded within the space bounded by
ABCDE and the straight line AE.
Then AFGE < ABCDE.

For, produce AF and FG to meet the enveloping line in

BOOK I. 33

H and K. Imagine ABODE to be the path of a point
moving from A to E. If the straight line AH be substituted
for ABCH, the path AHDE will be shorter than the path
ABODE, the portion HDE being common to both. If,
further, the straight line FK be substituted for FHDK, the
path AFKE will be a still shorter path from A to E. And
if, finally, GE be substituted for GKE, AFGE will be a still
shorter path. Therefore AFGE is less than any enveloping
line.

39. Scholium. The preceding demonstration applies when
the enveloping line is a curve, or any species of line whatever.

/ PROPOSITION XXI.— THEOREM.

40. If two oblique lines drawn from a point to a line meet the
line at unequal distances from the foot of the perpendicular, the
more remote is the greater.

Ist. If the lines lie on the same side of the perpendicular.

Let PO be the perpendicular and FD and FE the two
oblique lines, EO being greater than
DO; then is FE greater than FD.

For, produce FO to P', making OP'
equal to FO, and join P' with D and
with E.

The triangles FDO and F'DO and
the triangles FEO and F^EO are equal,
by Proposition YI. ILence FD = F'D,
and FE = F'E.

FDF' is less than FEF\ by Proposition XX. ; therefore
FE, the half of FEF\ is greater than FD, the half of
FDF\

ELEMENTS OF GEOMETEY.

2d. If the lines lie on opposite sides of the perpendicular.

Let PC be the perpendicular and FD and PE the oblique
lines, EG being greater than CD.

Lay off CD' equal to CD, and
join P and D'. Then the triangles
PDC and PD'C are equal, by Prop-
osition YI., and PD' =: PD.

But PD' is less than PE, by the
proof given above. Hence its equal PD is less than PE,

PROPOSITION XXII.— THEOREM.

41. Two straight lines perpendicular to the same straight line
are parallel.

Let AB and CD be two lines per-
pendicular to the same line EF;

then are they parallel. For, if AB

and CD are not parallel, they must

meet if produced ; but this is impos-
sible, for in that case we should
have two perpendiculars from their

point of meeting to the same straight line EF, which is
contrary to Proposition XYI.

PROPOSITION XXIII.— THEOREM.

42. Through a given point one line, and only one, can be
drawn parallel to a given line.

Let A be the given point and —

BC the given line.

From A draw AD perpendicular ^

to BC, and through A draw AE

perpendicular to AD. AE and DC, being perpendicular to

BOOK I. 35

the same line AI), are parallel, by Proposition XXII. No
other line can be drawn through A parallel to BC^ for, by
Axiom II., it would have the same direction as AE^ and
therefore, by Postulate II., it would coincide with AE,

EXERCISES.

1. Theorem. — Lines having the same direction are parallel.
Suggestion. Suppose them to meet ; v. Postulate II.

2. Theorem. — Lines parallel to the same line are parallel to
each other.

43. Definitions. When two straight lines AB, CD, are cut
by a third EF, the eight angles formed
at their points of intersection are
named as follows :

The four angles, 1, 2, 3, 4, without
the two lines, are called exterior angles.

The four angles, 5, 6, 7, 8, within
the two lines, are called interior angles.

Two exterior angles on opposite sides of the secant line
and not adjacent — as 1, 3 — or 2, 4 — are called alternate-exterior
angles.

Two interior angles on opposite sides of the secant line
and not adjacent — as 5, 7 — or 6, 8 — are called alternate-interior
angles.

Two angles similarly situated with respect both to the
secant and to the line intersected by it are caUed correspond-
ing angles ; as 1, 5 — 2, 6 — 3, 7 — 4, 8.

36 ELEMENTS OF GEOMETRY.

PROPOSITION XXIV.— THEOREM.

44. When two straight lines are cut by a third, if the alter-
nate-interior angles are equal, the two straight lines are parallel.

Let the line AB cut the lines CD and UF, making the
alternate-interior angles CAB and ABF equal ; then are CD
and FF parallel.

Through Gr, the middle
point of AB, draw GS per-
pendicular to CD, and cut-
ting FF in I.

Then the triangles AGH
and BGI, having the side AG equal to the side GB by-
construction, the angle GAH equal to the angle GBI^y
hypothesis, and the angle AGS equal to the angle IGB by-
Proposition Y., are equal, by Proposition YII. Therefore the
angle GIB is equal to the angle GHA. But GHA is a right
angle by construction ; hence GIB is a right angle, and CD
and FF are perpendicular to the same line HI, and are
therefore parallel, by Proposition XXII.

If FBA and BAD are the given equal alternate angles,
their supplements CAB and ABF are equal, and the proof
just given is valid.

If the given alternate-interior angles are right angles, the
lines are parallel, by Proposition XXII.

45. Corollary I. When two straight lines are cut by a third,
if a pair of corresponding angles are equal, the lines are parallel.

Suggestion. Show that in that case a pair of alternate-
interior angles are equal.

46. Corollary II. When two straight lines are cut by a
third, if the sum of two interior angles on the same side of the
secant line is equal to two right angles, the two lines are parallel.

BOOK I. 37

Suggestion. Show that a pair of alternate-interior angles
are equal.

PEOPOSITION XXV.— THEOKEM.

47. If two parallel lines are cut by a third straight line, the
alternate-interior angles are equal.

Let the parallel lines CI> and EF be cut by the line AB ;
then will the angles CAB and ABF be equal.
For, if they are not equal,

draw through A sl line AG,
making the angles GAB and
ABF equal ; then, by Propo-
sition XXIV., GA and EF
are parallel, and we have

two parallels to the same line EF drawn through the same
point A, which is contrary to Proposition XXIII., and there-
fore impossible. Hence the angles CAB and ABF are equal.

48. Corollary I. If two parallel lines are cut by a third
straight line, any two corresponding angles are equal.

49. Corollary II. If two parallel lines are cut by a third
straight line, the sum of the two interior angles on the same side
of the secant line is equal to two right angles.

EXERCISE.

Theorem. — A line perpendicular to one of two parallel lines is
perpendicular to the other.

/f^^ Off rn^.'^^^

/^ . ay

J^IFOUll^

ELEMENTS OF GEOMETRY.

PK0P08ITI0N XXVL— THEOREM.

50. The sum of the three angles of any triangle is equal to
two right angles.

Let ABC he any triangle ; then the sum of its three angles
is equal to two right angles.

Produce BGj and through G draw C£J parallel to BA»
Since the line AG meets the parallel

lines AB and EG, the alternate-interior
angles AGJ3 and BAG are equal, by
Proposition XXV.

Since the line BD cuts the parallel
lines AB and EG, the corresponding angles EGD and ABG
are equal, by Proposition XXY., Corollary I.

Therefore the sum of the angles of the triangle is equal
to BGA-\- AGE + EGD, which is two right angles, by Propo-
sition III., Corollary I.

51. C0R01.LARY. If a side of a triangle is eoctended, the exterior
angle is equal to the sum of the two interior opposite angles.

EXERCISE.

Theorem. — If the sides
of an angle are respectively
perpendicular to the sides
of a second angle, the an-
gles are equal, or supple-
mentary.

BOOK I. 39

POLYGONS.

52. Definitions. A polygon is a portion of a plane bounded
by straight lines ; as ABODE. The bounding

lines are the sides ; their sum is the perimeter
of the polygon. The angles which the adja-
cent sides make with each other are the angles
of the polygon; and the vertices of these
angles are called the vertices of the polygon.

Any line joining two vertices not consecutive is called a
diagonal; as AG.

53. Definitions. Polygons are classed according to the num-
ber of their sides :

A triangle is a polygon of three sides.

A quadrilateral is a polygon of four sides.

A pentagon has five sides ; a hexagon, six ; a heptagon, seven ;
an octagon, eight ; an enneagon, nine j a decagon^ ten ; a dodec-
agon, twelve ; etc.

An equilateral polygon is one all of whose sides are equal ;
an equiangular polygon, one all of whose angles are equal.

54. Definition. A convex polygon is one no side of which,
when produced, can enter within the space enclosed by the
perimeter; as ABODE in (52). Each of the angles of such
a polygon is less than two right angles.

It is also evident from the definition that the perimeter
of a convex polygon cannot be intersected by a straight line
in more than two points.

A concave polygon is one of which two
or more sides, when produced, will enter
the space enclosed by the perimeter; as
MNOPQ, of which OP and QP, when ^
produced, will enter within the polygon.
The angle OPQ, formed by two adjacent re-entrant sides,

ELEMENTS OF GEOMETRY.

is called a re-entrant angle ; and hence a concave polygon is
^OTCLQiimQ^ cdXlQ^ 2i re-entrant polygon.

All the polygons hereafter considered will be understood
to be convex.

PROPOSITION XXVII.— THEOREM.

55, The sum of all the angles of any polygon is equal to twice
as many right angles, less four, as the figure has sides.

Join any point 0 within the polygon to each of the ver-
tices, thus dividing the polygon into as many triangles as it
has sides.

The sum of the angles of these tri-
angles will, by Proposition XXYI.,
be twice as many right angles as the
figure has sides. But the angles of
the triangles form the angles of the
polygon plus the angles at 0, which
are equal to four right angles, by
Proposition III., Corollary II.

EXERCISE.

1. Theorem. — If each side of a polygon is extended, the sum
of the exterior angles is four right angles.

Suggestion. The sum of all the an-
gles, exterior and interior, is obvi-
ously twice as many right angles as
the figure has sides.

BOOK I. 41

QUADRILATERALS.

56. Definitions. Quadrilaterals are divided into classes, as
follows :

Ist. The trapezium (J.), which has no two
of its sides parallel.

2d. The trapezoid (jB), which has two
sides parallel. The parallel sides are called

the hases^ and the perpendicular distance /^ ^ \ \
between them the altitude of the trapezoid.

3d. The parallelogram ( (7), which is bounded v r\

by two pairs of parallel sides. \ j \

The side upon which a parallelogram is
supposed to stand and the opposite side are called its lower
and upper bases. The perpendicular distance between the
bases is the altitude.

57. Definitions. Parallelograms are divided into species, as
follows :

1st. The rhomboid (a), whose adjacent 's*

sides are not equal and whose angles are \
not right angles.

2d. The rhombus^ or lozenge (&), whose sides
are all equal.

3d. The rectangle (c), whose angles are all
equal, and therefore right angles.

4th. The square (d), whose sides are all equal and
whose angles are all equal.

The square is at once a rhombus and a rectangle.

42 ELEMENTS OF GEOMETRY.

PROPOSITION XXVIII.— THEOREM.

58. Two parallelograms are equal when two adjacent sides
and the included angle of the one are equal to two adjacent sides
and the included angle of the other.

AD == A'jy, and the angle BAD I j ^/ j

= ^'J.'i/; then these parallelo- ^ ^ ^' ^

grams are equal.

For they may evidently be applied the one to the other, so
as to coincide throughout, (v. Proposition XXIII.)

59. Corollary. Two rectangles
are equal when they have equal bases
and equal altitudes.

PROPOSITION XXIX.— THEOREM.

60. The opposite sides of a parallelogram are equal and the
opposite angles are equal.

Suggestion. Draw a diagonal AC. ACB J^ ^

and CAD are equal, by Proposition XXY.

CAB and AGD are equal, by Proposition
XXY.

Hence the triangles ABC and ADC are equal, by Propo-
sition YII.

EXERCISES.

1. Theorem. — If one angle of a parallelogram is a right angle,
all the angles are right angles, and the figure is a rectangle.

2. Theorem. — If two angles
have the sides of one respectively
parallel to the sides of the other,
they are equal, or supplementary.

3. Theorem. — Two parallel
lines are everywhere equidistant.

BOOK I. 43

PROPOSITION XXX.— THEOREM.

61. If two opposite sides of a quadrilateral are equal and
parallel, the figure is a parallelogram.

Suggestion. Let AD be equal and parallel i^ ^

to BC. Draw a diagonal AC. / \

The triangles ABC and ABO are equal, £
by Proposition YI. Therefore the angles
BAG and ACD are equal, and AB and CD are parallel, by
Proposition XXIY.

PROPOSITION XXXI.— THEOREM.

62. If the opposite sides of a quadrilateral are equals the
figure is a parallelogram.

Suggestion. Draw a diagonal, and prove the two triangles
equal.

PROPOSITION XXXII.— THEOREM.

63. The diagonals of a parallelogram bisect each other.

A D

Suggestion. The triangles AED and BEC
are equal, by Proposition YII.

EXERCISES.

1. Theorem. — The diagonals of a rectangle are equal.

2. Theorem. — The diagonals of a rhombus are perpendicular
to each other.

3. Theorem. — If the diagonals of a quadrilateral bisect each
other, the figure is a parallelogram.

4. Theorem. — If the diagonals of a parallelogram are equal,
the figure is a rectangle.

5. Theorem. — If the diagonals of a parallelogram are perpen-
dicular to each other, the figure is a rhombus.

ELEMENTS OF GEOMETRY.

ARRANGEMENT OF WRITTEN EXERCISES.

64. In writing out a demonstration, brevity of statement
and clearness of arrangement should be carefully studied,
and symbols and abbreviations may bo used with profit.
The following list is recommended :

SYMBOLS AND

ABBREVIATIONS.

. • . therefore.

Bef.

definition.

= equal to.

Fost

postulate.

O equivalent to.

Ax.

axiom.

> greater than.

Prop.

proposition.

<; less than.

Cor.

corollary.

II parallel to.

Hyp.

hypothesis.

J_ perpendicular to.

Cons.

construction.

/_ angle.

Adj.

adjacent.

^ angles.

Inc.

included.

rt. /_ right angle.

Alt.-int.

alternate-interior.

l\ triangle.

Sup.

supplementary.

l^ triangles.

Comp.

complementary.

rt. I\ right triangle.

Q.E.B,

, quod erat demonstran-

/ / parallelogram.

dum (= which was

/ * / parallelograms.

to be proved).

O circle.

0 circles.

BOOK I. 45

65. In arranging a written demonstration, it is well to
begin each statement on a separate line, giving the reason
for the statement at the end of the line, if it can be written
briefly, or in parenthesis immediately below the line, if it
cannot be written briefly. The following examples of demon-
strations prepared as written exercises, or for a written
examination, will serve as illustrations.

(1) PROPOSITION XII.— THEOREM.

If two angles of a triangle are unequal^ the side opposite the
greater angle is greater than the side opposite the less angle.

In A ABC, let it be given that Z_ACBy /^B,
we are to prove AB > J. (7.

Draw CDj cutting off from /^ACB si part
/ BCD = /_B.
Then in /\ BCD we have

/ BCD = l^B. Cons.

BD = CD, Prop. XL

and BD + DA = CD ^ DA.

But AC<:CD-^DA. Ax. I.

AC<iBD ^DA,
^^^ AC<:^AB, Q. E. D.

46 ELEMENTS OF GEOMETRY.

(2) PROPOSITION XXIV.— THEOREM.

When two straight lines are cut by a third, if the alternate*
interior angles are equal, the two straight lines are parallel.

Let AB cut CD and EF in the points A and B, making
/^BAG= /^ABF,
we are to prove CD \\ to EF.

Through G, the middle point of AB, draw HI ]_ to CD.

Then in the A ^^^ and BGI
we have

and

But

Aa^BG,

Cons.

/_GAH= /_ GBI,

Hyp.

/_AGH= /_^BGI.

Prop. Y.

/\AGII=: ^BGI,

Prop. YII.

/_ GIB = /_ GHA

(homologous angles of =

A).

/ G^^isar^. /.

Cons.

/_GIBiBSirt. /_,

Him ]_to EF.

J?7is_|_to CD.

Cons.

CDand j^i^are||.

Prop. XXII.

E.D.

and
But

If the given equal ^ are ABE and BAD,
we have / ABE = /_ BAD, Hyp.

/ ABE + / ABF =2rt. /^, Prop. III.
/ BAD -{- Z,S^0=2rt. ^. Prop. III.
/^ABF= /_BAC,
and the proof given above applies.

BOOK I. 47

If the given equal alt.-int. /^ are rt. ^ the two given lines
are I to the third line and are ||, by Proposition XXII.

(3) PEOPOSITION XXVI. COEOLLARY.

If one side of a triangle is extended^ the exterior angle is equal
to the sum of the two interior opposite angles.

In the /^ ABC let the side AC he extended,
we are to prove ^ BCD ^ /_A -\- /_B.
We have the sum of the adj. ^

BCD -{-BGA = 2rt. ^. Prop. III.

But /_A + /^B+ /^BCA = 2rt. ^. Prop. XXYL

/^BCD== /_A-\- /^B. Q. E. B.

(4) EXERCISE 3, PAGE 43.— THEOREM.

If the diagonals of a quadrilateral bisect each other j the figure
is a parallelogram.

..--'.B**.

In the quadrilateral ABCD^ let the diagonals AD and BG
bisect each other.
i We are to prove ABCD a / "],

48 ELEMENTS OF GEOMETRY.

In the ^ AEB and CED
we have GE = EB, Hyp.

ED = AE, Hyp.

/ CED = /_ AEB. Prop. Y.

A ^^^ = A C'^A I'rop. YI.

and OD = A^

(homologous sides of equal ^^
and /_EDC= Z_EAB

(homologous /» of equal /^).
But JE^DCand jEAJ5are alt.-int. ^.

CD is II to AB, Prop. XXIY.

and since CD = AB, Proved above.

^^CD is a O- ^^^P- ^^^'

Q. E. D.

EXERCISES ON BOOK I.

1. The straight line AJEJ which bisects the angle
exterior to the vertical angle of an isosceles tri-
angle ABCis parallel to the base BC,

2. If from a variable point in the base of an isos-

1/ celes triangle parallels to the sides are drawn, a

parallelogram is formed whose perimeter is constant.

3. The sum of the four lines drawn to the vertices of a quadri-
f^ lateral from any point except the intersection of the diagonals, is

greater than the sum of the diagonals.
^ 4. The lines drawn from the extremities of the base of an isos-
celes triangle to the middle points of the opposite sides are equal.
J 6. The perpendiculars from the extremities of the base of an
•^ isosceles triangle upon the opposite sides are equal.

6. The bisectors of the base angles of an isosceles triangle are
equal.

7. A perpendicular let fall from one end of the base of an isos-
celes triangle upon the opposite side makes with the base an
angle equal to one-half the vertical angle.

8. If the vertical angle of an isosceles triangle is one-half as
great as an angle at the base, a line bisecting a base angle will
divide the given triangle into two isosceles triangles.

9. If two isosceles triangles have the sides of one equal to the
sides of the other, and the base of one double the altitude of the

*/ other, the two triangles are of the same size.

c d 5 49

ELEMENTS OP GEOMETRY.

10. If from a variable point P in the base of an
isosceles triangle ABC^ perpendiculars, PM^ PN^
to the sides are drawn, the sum of PM and PN
is constant, and equal to the perpendicular from
C upon AB.

Suggestion, The triangles PNC and PEC are
equal, by Proposition VII.

11. The line joining the feet of perpendiculars let fall from the
extremities of the base of an isosceles triangle upon the opposite
sides is parallel to the base.

12. If BE bisects the angle ^ of a triangle
ABC^ and CE bisects the exterior angle ACD^
the angle E is equal to one-half the angle A,

13. The medial line to any side of a triangle is less
than the half sum of the other two sides.

Definition. A line joining a vertex of a triangle
with the middle point of the opposite side is called
a medial line.

14. If from two points, A and -B, on the
same side of a straight line JfiV, straight
lines, APj BPy are drawn to a point P in
that line, making with it equal angles APM
and BPN^ the sum of the lines AP and BP
is less than the sum of any other two lines,
AQ and BQ^ drawn from A and B to any
other point Q in MN.

/ 16. If the medial line from the vertex of a triangle to the base
^ is equal to one-half the base, the vertical angle is a right angle.

^ 16. The altitude of a triangle divides the vertical angle into
two parts, whose difference is equal to the difference of the base
angles of the triangle.

BOOK I.

17. The perpendicular erected at the middle point
^ of one side of a triangle meets the longer of the
other two sides.

18. Lines drawn from a point within
a triangle to the extremities of the base
include an angle greater than the verti-
cal angle of the triangle, (v. Proposi-

tion XXVI., Corollary.)

19. The sum of the angles at the vertices of
a five-pointed star (pentagram) is equal to
two right angles.

20. The three perpendiculars erected at the
middle points of the sides of a triangle meet
in the same point.

Suggestion. The point of intersection of
EH and FK is equidistant from the three
vertices, and therefore must lie on DQ
(Proposition XVIII.).

21. The three bisectors of the three an-
gles of a triangle meet in the same point.

Suggestion. The point of intersection of
BE and CF is equidistant from the three
sides, and therefore must lie on AD
(Proposition XIX.).

22. The bisectors of two external angles of a triangle and the
bisector of the remaining internal angle meet in a point.

23. If from the diagonal BD of a square ABCD,
BE is cut off equal to BC, and EF is drawn per-
pendicular to BDy then DE = EF= FG,

ELEMENTS OF GEOMETRY.

24. In a trapezoid, the straight line joining A

the middle points of the non-parallel sides is /

parallel to the bases, and is equal to one-half y"^

their sum. L

Suggestion. Draw HO parallel to AB^ and -B
extend ^Z). DOF= Offi^CProposition VII.),
and EFHB is a parallelogram, by Proposition XXX.

D Q

H 0

25. If the sides of a trapezoid which are
"^ not parallel are equal, the base angles are
equal and the diagonals are equal.

A E

26. If through the four vertices of a quadrilateral lines are
^ drawn parallel to the diagonals, they will form a parallelogram
twice as large as the quadrilateral.

27. The three perpendiculars from
^ the vertices of a triangle to the op-

posite sides meet in the same point.
Suggestion. Draw through the
three vertices lines parallel to the
opposite sides of the triangle. By
the aid of the three parallelograms
ABCB', ABA'C, and ACBG', prove
that the sides of A'B^C are bisected
by -4, -B, and C, See now Exercise 20.

28. If a straight line drawn parallel to the base
^ of a triangle bisects one of the sides, it also bisects

the other side ; and the portion of it intercepted
between the two sides is equal to one-half the base.
Suggestion. Draw DF parallel to A C. See now
Proposition VII. and Proposition XXIX.

y 29. The straight line joining the middle points of two sides of
a triangle is parallel to the third side. {v. Exercise 28.)
*
vT 30. The three straight lines joining the middle points of the
sides of a triangle divide the triangle into four equal triangles.

^ 31. In any right triangle, the straight line ^

, drawn from the vertex of the right angle to the
"^ middle of the hypotenuse is equal to one-half
the hypotenuse, (v. Exercise 28.) b

BOOK I.

/^ 32. The straight lines joining the middle
points of the adjacent sides of any quadrilat-
eral form a parallelogram whose perimeter is
equal to the sum of the diagonals of the quad-
rilateral (Exercise 29).

33. If E and -Pare the middle points of the
/ opposite sides, AD, BC, of a parallelogram
ABCD, the straight lines BE, DF, trisect the
diagonal AC (Exercise 28).

/^ 34. The four bisectors of the angles
of a quadrilateral form a second quad-
rilateral, the opposite angles of which
are supplementary.

If the first quadrilateral is a paral-
lelogram, the second is a rectangle.
If the first is a rectangle, the second
is a square.

36. The point of intersection of the diagonals of a parallelogram
bisects every straight line drawn through it and terminated by
the sides of the parallelogram.

36. If from each vertex of a parallelogram
the same given distance is laid ofi" on a side
of the parallelogram, care being taken that
no two distances are laid off on the same
side, the points thus obtained will be the
vertices of a new parallelogram.

37. If from two opposite vertices of a par-
allelogram equal distances are laid off on
the sides adjacent to those vertices, the
points thus obtained will be the vertices
of a parallelogram.

A A'
D

£■:

■:--)

A A'

38. The three medial lines of a triangle
m.eet in the same point.

Suggestion. Let O be the point of inter-
section of AD and BE, and H and O the
middle points of OB and OA. Hence prove
OD = ^AD and OE = ^BE. In like man-
ner the point of intersection of AD and CE ^
can be shown to cut off one-third of AD.

ELEMENTS OF GEOMETRY.

89. The intersection of the straight
lines which join the middle points of
opposite sides of any quadrilateral is
the middle point of the straight line
which joins the middle points of the
diagonals (Exercise 29).

he
he

a«i

SYLLABUS TO BOOK L

POSTULATES, AXIOMS, AND THEOKEMS.

POSTULATE I.

Through any two given points one straight line, and only one, can

be drawn.

POSTULATE II.

Through a given point one straight line, and only one, can be drawn

having any given direction.

AXIOM I.

A straight line is the shortest line that can be drawn between two
points.

AXIOM IL

Parallel lines have the same direction.

PROPOSITION I.

At a given point in a straight line one perpendicular to the line can
be drawn, and but one.

Corollary. Through the vertex of any given angle one straight line
can be drawn bisecting the angle, and but one.

PROPOSITION IL

All right angles are equal.

PROPOSITION III. ^

The two adjacent angles which one straight line makes with another
are together equal to two right angles.

Corollary I. The sum of all the angles having a common vertex, and
formed on one side of a straight line, is two right angles.

Corollary IL The sum of all the angles that can be formed about a
point in a plane is four right angles.

PROPOSITION IV.

If the sum of two adjacent angles is two right angles, their exterior
sides are in the same straight line.

ELEMENTS OF GEOMETRY.

PKOPOSITION V.

If two straight lines intersect each other, the opposite (or . *■
angles are equal.

PKOPOSITION VI.

Two triangles are equal when two sides and the included ang^v w; ihtt
one are respectively equal to two sides and the included angk of the
other.

PROPOSITION VII

Two triangles are equal when a side and the two adjacent angles «)f the
one are respectively equal to a side and the two adjacent angh " tlin
other.

~-^. PROPOSITION VIII.

In an isosceles triangle the angles opposite the equal sides are equal.
Corollary. The straight line bisecting the vertical angle of an isosceles
triangle bisects the base, and is perpendicular to the base.

PROPOSITION IX.

Two triangles are equal when the three sides of the one are respectively
equal to the three sides of the other.

PROPOSITION X.

Two right triangles are equal when they have the hypotenus : •.

side of the one respectively equal to the hypotenuse and a sid(
other.

PROPOSITION XI.

If two angles of a triangle are equal, the sides opposite to tt
equal, and the triangle is isosceles.

PROPOSITION XII.

If two angles of a triangle are unequal, the side opposite the
angle is greater than the side opposite the less angle.

PROPOSITION XIII.

If two sides of a triangle are unequal, the angle opposite the
side is greater than the angle opposite the less side.

PROPOSITION XIV.

If two triangles have two sides of the one respectively equal ' wo
sides of the other, and the included angles unequal, the triangle
has the greater included angle has the greater third side.

ELEMENTS OF GEOMETRY. HI

PROPOSITION XV. "^

If two triangles have two sides of the one respectively equal to two
sides of the other, and the third sides unequal, the triangle which has the
greater third side has the greater included angle.

PROPOSITION XVI.

From a given point, without a straight line, one perpendicular can be
3ut one.

PROPOSITION XVII.

The perpendicular is the shortest line that can be drawn from a point
to a straight line.

PROPOSITION XVIII.

If a perpendicular is erected at the middle of a straight line, then
ever}^ point on the perpendicular is equally distant from the extremities
of the line ; and every point not on the perpendicular is unequally
distant from the extremities of the line.

PROPOSITION XIX.

Every point in the bisector of an angle is equally distant from the
sides of the angle ; and every point not in the bisector, but within the
angle, is unequally distant from the sides of the angle; that is, the
bisector of an angle is the locus of the points within the angle and
equally distant from its sides.

PROPOSITION XX.

A convex broken line is less than any other line which envelops it
and has the same extremities.

PROPOSITION XXI.

If two oblique lines drawn from a point to a line meet the line at
unequal distances from the foot of the perpendicular, the more remote
is the greater.

PROPOSITION XXII.

Two straight lines perpendicular to the same straight line are parallel.

PROPOSITION XXIII.

Through a given point one line, and only one, can be drawn parallel
to a given line.

BOOK 11.

THE CIRCLE.

1. Definitions, A circle is a portion of a plane bounded t)y
a curve, all the points of which are equally distant from a
point within it called the centre.

The curve which bounds the circle is
called its circumference.

Any straight line drawn from the cen-
tre to the circumference is called a
radius.

Any straight line drawn through the
centre and terminated each way by the circumference is
called a diameter.

In the figure, 0 is the centre, and the curve ABCEA is
the circumference of the circle; the circle is the space in-
cluded within the circumference; OA, OBy OC, are radii;
-4.0(7 is a diameter.

By the definition of a circle, all its radii are equal; also
all its diameters are equal, each being double the radius.

If one extremity, 0, of a line OA is fixed, while the line
revolves in a plane, the other extremity, J., will describe a
circumference, whose radii are all equal to OA.

2. Definitions. An arc of a circle is any portion of its cir-
cumference ; as DEF.

A chord is any straight line joining two points of the cir-
cumference; as DF. The arc DFF is said to be subtended
by its chord DF.

ELEMENTS OF GEOMETRY. ill

PROPOSITION XV. "^

If two triangles have two sides of the one respectively equal to two
sides of the other, and the third sides unequal, the triangle which has the
greater third side has the greater included angle.

PROPOSITION XVI.

From a given point, without a straight line, one perpendicular can be
drawn to the line, and but one.

PROPOSITION XVII.

The perpendicular is the shortest line that can be drawn from a point
to a straight line.

PROPOSITION XVIII.

If a perpendicular is erected at the middle of a straight line, then
ever}'- point on the perpendicular is equally distant from the extremities
of the line ; and every point not on the perpendicular is unequally
distant from the extremities of the line.

PROPOSITION XIX.

PROPOSITION XX.

A convex broken line is less than any other line which envelops it
and has the same extremities.

PROPOSITION XXI.

If two oblique lines drawn from a point to a line meet the line at
unequal distances from the foot of the perpendicular, the more remote
is the greater.

PROPOSITION XXII.

Two straight lines perpendicular to the same straight line are parallel.

PROPOSITION XXIII.

Through a given point one line, and only one, can be drawn parallel
to a given line.

(

BOOK 11.

THE CIRCLE.

1. Definitions. A circle is a portion of a plane bounded Dy
a curve, all the points of which are equally distant from a
point within it called the centre.

The curve which bounds the circle is
called its circumference.

Any straight line drawn from the cen-
tre to the circumference is called a
radius.

Any straight line drawn through the
centre and terminated each way by the circumference is
called a diameter.

In the figure, 0 is the centre, and the curve ABCEA is
the circumference of the circle; the circle is the space in-
cluded within the circumference; OA^ OB, OC, are radii;
J. 0(7 is a diameter.

By the definition of a circle, all its radii are equal; also
all its diameters are equal, each being double the radius.

If one extremity, 0, of a line OA is fixed, while the line
revolves in a plane, the other extremity. A, will describe a
circumference, whose radii are all equal to OA.

2. Definitions. An arc of a circle is any portion of its cir-
cumference ; as DEF.

A chord is any straight line joining two points of the cir-
cumference; as DF. The arc DEF is said to be subtended
by its chord DF.

66 ELEMENTS OF GEOMETRY.

Every chord subtends two arcs, which together make up
the whole circumference. Thus, DF subtends both the arc
DBF and the arc DGBAF. When an
arc and its chord are spoken of, the arc
less than a semi-circumference, as DFF,
is always understood, unless otherwise
stated.

A segment is a portion of the circle
included between an arc and its chord;
thus, by the segment DFF is meant the space included
between the arc BF and its chord.

A sector is the space included between an arc and the two
radii drawn to its extremities ; as A OB.

3. From the definition of a circle it follows that every
point within the circle is at a distance from the centre which
is less than the radius ; and every point without the circle is
at a distance from the centre which is greater than the
radius. Hence the locus of all the points in a plane which
are at a given distance from a given point is the circumference
of a circle described with the given point as a centre and with
the given distance as a radius.

4. Postulate. A circumference may be described with any
point as centre and any distance as radius.

PEOPOSITION I.— THEOREM.

5. Two circles are equal when the radius of the one is equal
to the radius of the other. \

Let the second circle be superposed upon the first, so that
its centre falls upon the centre of the first ; then will the two
circumferences coincide throughout.

BOOK II. 57

For, if any point of either circumference falls outside of
the other circle, the line joining it with the common centre
must cross the circumference of that circle.

The whole line will be a radius of one circle, the portion
of it within the other circle will be a radius of that other
circle, and we shall have two unequal radii, which is con-
trary to our hypothesis.

PROPOSITION II.— THEOREM.

6. Every diameter bisects the circle and its circumference.

Let AMBN be a circle whose centre is
0 ; then any diameter AOB bisects the
circle and its circumference.

For, if the figure ANB be turned
about AB as an axis and superposed
upon the figure AMB^ the curve ANB
will coincide with the curve AMB, since
all the points of both are equally distant from the centre.
(v. Proof of Proposition I.) The two figures then coincide
throughout, and are therefore equal in all respects. There-
fore AB divides both the circle and its circumference into
equal parts.

7. Definitions. A segment equal to one-half the circle, as the
segment AMB, is called a semicircle. An arc equal to half
a circumference, as the arc AMB, is called a semi-circum-
ference.

ELEMENTS OF GEOMETRY.

PROPOSITION III.— THEOREM.

8. In equal circles^ or in the same circle^ equal angles at the
centre intercept equal arcs on the circumference.

Let 0, 0\ be the centres of equal circles, and AOB, A'O'B',
equal angles at these centres ; then
the intercepted arcs AB^ A'B', are
equal.

For the angle 0' may be super-
posed upon, and made to coincide
with, its equal 0. The extrem-
ities of the arc A'B' will then fall on the extremities of the
arc ABj and the arcs must coincide throughout and be equal,
since the radii are equal, (y. Proof of Proposition I.)

9. Corollary. Conversely, in the same circle, or in equal
circles, equal arcs subtend equal angles at the centre.

10. Definition. A fourth part of a cir-
cumference is called a quadrant. It is
evident from the preceding theorem that
a right angle at the centre intercepts a
quadrant on the circumference.

Thus, two perpendicular diameters,
AOC, BCD, divide the circumference
into four quadrants, AB, BC, CD, DA.

PROPOSITION IV.— THEOREM.

11. In equal circles, or in the same circle, equal arcs are sub-
tended by equal chords.

Let 0, 0', be the centres of equal circles, and AB, A!B\
equal arcs ; then the chords AB, A!B', are equal.

BOOK II.

For, drawing the radii to the extremities of the arcs, the
angles 0 and 0' are equal (Propo- ^ ^
sition III., Corollary), and conse- /^\~/^
quently the triangles AOB, A'0'B\ [ V
are equal (Proposition YI., Book
I.). Therefore AB = A'B',

If the arcs are in the same circle the demonstration is
similar.

12. Corollary. Conversely, in equal circles, or in the same
circle, equal chords subtend equal arcs,

EXERCISES.

1. Theorem. — A diameter is greater than
any other chord.

2. Theorem. — The shortest line that can be
drawn from a point within a circle to the cir-
cumference is a portion of the diameter drawn
through the point.

PROPOSITION v.— THEOREM.

13. In equal circles, or in the same circle, the greater of twfr
unequal arcs is subtended by the greater chord, the arcs being
each less than a semi-circumference.

Let the arc AC he greater than the arc AB ; then the chord
AC iQ greater than the chord AB.

Superpose the arc AB upon the
arc AC, placing centre upon cen-
tre and A upon A; B must fall
between A and C, since AB is less

ELEMENTS OF GEOMETRY.

than AC. Draw now the radii OA, 0J5, 0(7. In the triangles

AOG and AOB the angle AOG is

obviously greater than the angle

AOB; therefore, by Proposition

XIV., Book I., the chord A a is

greater than the chord AB.

14. Corollary. Conversely, in

equal circles, or in the same circle, the greater of two unequal
chords subtends the greater arc.

Suggestion, v. Proposition XY., Book I.

PROPOSITION VI.— THEOREM.

15. The diameter perpendicular to a chord bisects the chord
and the arcs subtended by it.

The triangles ACQ, BCO, are equal, by Proposition X.,
Book I. Therefore AC=CB.

The triangles AOD, BOB, are equal, by
Proposition VI., Book I. Therefore AD =
BD, and hence, by Proposition IV., Corol-
lary, the arc AD is equal to the arc BD.

In the same way we can prove the arc
AD' equal to the arc BD'. ^

16. Corollary I. The perpendicular

erected at the middle point of a chord passes through the centre
of the circle, (v. Proposition XVIII., Book I.)

17. Corollary II. When two circumferences intersect, the
straight line joining their centres bi-
sects their common chord at right
angles.

Suggestion. Erect a perpendicular
at the middle point of the common
chord, (v. Corollary I.)

/ /

1
1 /

^*''J

BOOK II. 61

EXERCISE.

Theorem. — The locus of the middle points of a set of parallel
chords is the diameter perpendicular to the chords.

PROPOSITION VII.— THEOREM.

18. In the same circle, or in equal circles, equal chords are
equally distant from the centre ; and of two unequal chords, the
less is at the greater distance from the centre.

1st. Let AB, CD, be equal chords; OE, OF, the perpen-
diculars which measure their distances
from the centre 0 ; then OE = OF. ^ ^^

For, since the perpendiculars bisect f ^L^-'^^'^x
the chords, ^^ = (7i^; hence (Propo- ^LrCT^T^. I

sition X., Book I.) the right triangles I ...--"'''/V /

AOE and COF are equal, and OE z= ^V-v^ y^

OF. M>%^

2d. Let CG, AB, be unequal chords ;
OE, OH, their distances from the centre ; and let CG be less
than ^5; then OlfyOE.

For, since chord AB > chord CG, we have arc AB > arc
CG ; so that if from C we draw the chord CD = AB, its
subtended arc CD, being equal to the arc AB, will be greater
than the arc CG. Therefore the perpendicular OS will
intersect the chord CD in some point I. Drawing the per-
pendicular OF to CD, we have, by the first part of the
demonstration, OF = OE. But OH > 01, and 01 > OF
(Proposition XYIL, Book I.) ; still more, then, is OS > OF,
or OS^OE.

If the chords be taken in two equal circles, the demonstra-
tion is the same.

ELEMENTS OF GEOMETRY.

19. Corollary. Conversely^ in the same circle^ or in equal
circles^ chords equally distant from the centre are equal ; and
of two chords unequally distant from the centre, that is the
greater whose distance from the centre is the less.

EXERCISE.

Theorem. — The least chord that can be
drawn in a circle through a given point is
the chord perpendicular to the diameter
through the point

Suggestion, v. Proposition XYII., Book I.

TANGENTS AND SECANTS.

20. Definitions. A tangent is an indefinite straight line
which has but one point in common

with the circumference; as ACB.
The common point, C, is called the
point of contact, or the point of tan-
gency. The circumference is also
said to be tangent to the line AB
at the point C,

A secant is a straight line which
meets the circumference in two points ; as JEF.

Two circumferences are tangent to each other when they
are both tangent to the same straight line
at the same point.

21. Definition. A rectilinear figure is said
to be circumscribed about a circle when all
its sides are tangents to the circumference.

In the same case, the circle is said to be
inscribed in the figure.

BOOK II. 63

PROPOSITION VIII.— THEOREM.

22. A straight line cannot intersect a circumference in more
than two points.

For, if the line could intersect the circumference in three
points, the radii drawn to these points would meet the line
at unequal distances from the perpendicular let fall from the
centre of the circle upon the line, and would be unequal, by
Proposition XXL, Book I.

PROPOSITION IX.— THEOREM.

23. A straight line tangent to a circle is perpendicular to the
radius drawn to the point of contact.

For any other point of the tangent, 2)^/
as D, must lie outside of the circle, and ^y\\ X
therefore the line OD^ joining it with a^^ \1 \

the centre, must be greater than the I q ]

radius 0(7, drawn to the point of con- V /

tact. ^

OCiSj then, the shortest line that can
be drawn from 0 to the tangent AB, and is therefore perpen-
dicular to AB, by Proposition XYII., Book I.

24. Corollary I. A perpendicular to a tangent line drawn
through the point of contact must pass through the centre of the
circle.

25. Corollary II. If two circumferences are tangent to each
other, their centres and their point of contact lie in the same
straight line.

Suggestion. Through their point of contact draw a line
perpendicular to the tangent at that point, (v. Corollary I.)

ELEMENTS OF GEOMETRY.

PROPOSITION X.— THEOREM.

26. When two tangents to the same circle intersect, the dis-
tances from their j)oint of intersection to their points of contact
are equal.

Eor the right triangles OAP and
OBP (Proposition IX.) are equal,
by Proposition X., Book I.

EXERCISES.

1. Theorem. — In any circumscribed quadrilateral, the sum of
two opposite sides is equal to the sum of the other two opposite sides,

2. Theorem. — If two circumferences are tangent, and from any
point, P, of the tangent at their point of contact, tangents are
drawn to the two circles, the points of contact of these tangents
are equally distant from P.

PROPOSITION XI.— THEOREM.

27. Two parallels intercept equal arcs on a circumference.

We may have three cases :

1st. When the parallels AB, CI), are
both secants, then the intercepted arcs
A G and BB are equal. For, let OM be
the radius drawn perpendicular to the
parallels. By Proposition YI., the point
M is at once the middle of the arc AMB
and of the arc CMD, and hence we have

AM = BM and CM = DM,

whence, by subtraction,

AM— CM= BM — DM,

AG = BD.

that is,

G/^

<^ \ — V

( ^ \

BOOK II. 65

2d. When one of the parallels is a secant, as AB, and the
other is a tangent, as EF at Jf, then the intercepted arcs
AM and BM are equal. For the radius OM drawn to the
point of contact is perpendicular to the tangent (Proposition
IX.), and consequently perpendicular also to its parallel AB ;
therefore, by Proposition YI., AM = BM.

3d. When both the parallels are tangents, as JSF at Mj
and GH at iV, then the intercepted arcs MAN" and MBN
are equal. For, drawing any secant AB parallel to the tan-
gents, we have, by the second case,

AM = BM and AN = BM,

whence, by addition,

AM+A]Sr=BM+BW,
that is,

MAN=MBJSr;

and each of the intercepted arcs in this case is a semi-circum-
ference.

MEASURE OF ANGLES.

As the measurement of magnitude is one of the principal
objects of geometry, it will be proper to premise here some
principles in regard to the measurement of quantity in
general.

28. Definition. To measure sl quantity of any kind is to find
how many times it contains another quantity of the same
kind, called the unit.

Thus, to measure a line is to find the number expressing
how many times it contains another line, called the unit of
length, or the linear unit.

The number which expresses how many times a quantity
contains the unit is called the numerical measure of that
quantity.

66 ELEMENTS OF GEOMETRY.

29. Definition. The ratio of two quantities is the quotient
arising from dividing one by the oth-er : thus, the ratio of A

to B is ~.

To find the ratio of one quantity to another is, then, to
find how many times the first contains the second; there-
fore it is the same thing as to measure the first by the
second taken as the unit (28). It is implied in the defini-
tion of ratio that the quantities compared are of the same
kind.

Hence, also, instead of the definition (28), we may say that
to measure a quantity is to find its ratio to the unit.

The ratio of two quantities is the same as the ratio of their
numerical measures. Thus, if P denotes the unit, and if P
is contained m times in A and n times in B^ then

A mP m

B~~nP~ n

30. Definition. Two quantities are commensurable when there
is some third quantity of the same kind which is contained
a whole number of times in each. This third quantity is
called the common measure of the proposed quantities.

Thus, the two lines A and B are commensurable if there
is some line, (7, which is contained a

whole number of times in each, as, ^' ' ^ ' ' ' ' '
for example, 7 times in J., and 4 times £< — • — « — « — «
in B. 0' — «

The ratio of two commensurable
quantities can, therefore, be exactly expressed by a number,

whole or fractional (as in the preceding example by -), and

is called a commensurable ratio.

BOOK II. 67

31. Definition. Two quantities are incommensurable when
they have no common measure. The ratio of two such
quantities is called an incommensurable ratio.

If A and B are two incommensurable quantities, their ratio

is still expressed by - .

32. Problem. To find the greatest common measure of two
quantities. The well-known arithmetical process may be ex-
tended to quantities of all kinds. Thus, suppose AB and CD
are two straight lines whose common measure is required.
Their greatest common measure can-
not be greater than the less line CD. ^' ' ' J"^

Therefore let CD be applied to AB as Cr—T-pD

many times as possible, suppose three

times, with a remainder EB less than CD. Any common
measure of AB and CD must also be a common measure of
CD and EB ; for it will be contained a whole number of
times in CD, and in AE^ which is a multiple of OD, and
therefore to measure AB it must also measure the part
EB. Hence the greatest common measure of AB and CD
must also be the greatest common measure of CD and EB,
This greatest common measure of CD and EB cannot be
greater than the less line EB ; therefore let EB be applied
as many times as possible to CD, suppose twice, with a
remainder FD. Then, by the same reasoning, the greatest
common measure of CD and EB^ and consequently also
that of AB and CD, is the greatest common measure of
EB and FD. Therefore let FD be applied to EB as many
times as possible: suppose it is contained exactly twice in
EB without remainder; the process is then completed, and
we have found FD as the required greatest common meas-
ure.

68 ELEMENTS OF GEOMETRY.

The measure of each lin-e, referred to FD as the unit, will
then be as follows : we have

EB = 2FD,

CD = 2EB +FD = 4FD + FD = 6FD,

AB = SCD + EB= WFD + 2FD = 17FD.

The proposed lines are therefore numerically expressed, in

terms of the unit FD, by the numbers 17 and 5 j and their

.. . 17
ratio IS -— .
5

33. When the preceding process is applied to two quanti-
ties and no remainder can be found which is exactly con-
tained in a preceding remainder, however far the process bo
contmued, the two quantities have no common measure ; that
is, they are incommensurable, and their ratio cannot be exactly
expressed by any number, whole or fractional.

34. As the student often has difficulty in realizing the
possibility of an incommensurable ratio, and imagines that if
two lines are given it must be possible to take a divisor so
small that it will go exactly into each of them, it seems
worth while to consider at some length an important exam-
ple,— namely, the ratio of the diagonal of a square to one of
the sides.

Let the method of (32) be applied
to finding the common measure of the
diagonal and a side of the square
ABCD.

AO 18 clearly less than twice AB ;
i.e., than AB -^ BC. Lay off on J. (7
AB\ equal to AB. Our problem is now
reduced to finding a common measure of B'C and AB, or its
equal, CB.

BOOK II. 69

Erect at B' a perpendicular B'A' to AC. A'B\ B'C, and
A'B are all equal (y. Exercise 23, Book I.). If, then, we
lay off CB" equal to CB\ B'G goes into BC twice, with a
remainder B"A'^ by which we must proceed to divide B'C.
But A' B'C is half a square, precisely similar to ABC^ and in
performing the division of B'C, or its equal, A'B'^ by A'B"^
we are merely repeating, on a smaller scale, the process just
performed in dividing BC loy B'C This will lead us to
another repetition, on a still smaller scale, and so on indefi-
nitely, and we shall never reach an exact division. The
diagonal and the side of a square have then no common
divisor, and are absolutely incommensurable.

35. Although an incommensurable ratio cannot be exactly
expressed by a number, a number can be found by the fol-
lowing method that will approximately express it, and the
approximation may be made as close as we choose.

Suppose that - denotes the ratio of two incommensurable

quantities, A and B. Let B be divided into n equal parts,
n being some number taken at pleasure ; and then let A be
divided by one of these parts. Suppose A is found to con-
tain this divisor m times, with a remainder, which, of course,

is less than the divisor; then - is an approximation to the

value of - , and an approximation that may be made as close

as we please by taking a sufficiently great value of n.

For, if X is the magnitude of one of the parts into which
B is divided, we have

B = nXj while A^ mx and < (m -|- l)x.
Hence

^>^and<(^^+^)^:
B nx nx

70 ELEMENTS OF GEOMETEY.

that is,

^ Hes between ^ and ^ + 1 ;
B n n n

and by increasing n we may make -, which is the difference

between two numbers, one less and one greater than -, as

small as we choose, and may thus make the less number -

A
as close an approximation to the value of - as we please.

As a numerical example, take the ratio of the diagonal of
a square to one of the sides (34). If the side is divided into
three equal parts, the diagonal will contain one of these parts

four times, with a remainder less than the divisor. - is then

an approximation, though a very rough one, to the value of

4 5

the ratio in question, which must lie between - and -.

o o

If the side is divided into five equal parts, the diagonal

7
will contain seven of them, and - is a closer approximation.

141 1414

and are still closer approximations.

100 1000 ^^

36. Definition. A proportion is an equation between two
itios. ']
equation

A A!

ratios. Thus, if the ratio - is equal to the ratio — , the

4= ^
B B'

is a proportion. It may be read, " Eatio of ^ to jB equals
ratio of A' to 5'," or, "J. is to B as A' is to B'r

BOOK II. 71

A proportion is often written as follows :

where the notation A : B is equivalent to A -i- B. When
thus written, A and B' are called the extremes^ B and A' the
means, and B' is called a fourth proportional to A, B and A' ;
the first terms, A and A\ of the ratios are called the ante-
cedents ; the second terms, B and B\ the consequents.
When the means are equal, as in the proportion
A:B = B:C,

the middle term B is called a mean proportional between A
and Cj and C is called a 3f/^^r^ proportional to ^ and j5.

37. In cases where it is necessary to prove the equality of
incommensurable ratios, it is usually best to employ what is
called the method of limits.
. 38. Definitions. A variable quantity, or simply a variable, is
a quantity whose value is supposed to change.

A constant quantity, or simply a constant, is a quantity
whose value is fixed.

The value of a variable may be changed at pleasure, in
which case it is called an independent variable ; or it may be
changed by changing at pleasure the value of some other
variable or variables on which it depends, and in this case it
is called a dependent variable.

39. Definition. If, by changing in some specified way the
variable on which it depends, we can make a dependent vari-
able approach as near as we please to some given constant, but
can never make the values of the variable and the constant
exactly coincide ; or, in other words, if we can make the dif-
ference between the variable and the constant as small as we
please, but cannot make it absolutely zero, the constant is called
the limit of the variable under the circumstances specified.

72 ELEMENTS OF GEOMETRY.

40. For example, consider the fraction -, where n is sup-

n
posed to be an independent variable, — Le.^ one whose value

may be changed arbitrarily and to any extent, — the fraction
- is then a dependent variable. By increasing n at pleasure, _

may be made to approach as near as we please to the value
zero, but can never be made exactly equal to zero.

We say, therefore, that zero is the limit of -, as n is indefi-

nitely increased.

Again, the numerical approximation to the value of an
incommensurable ratio (v. 35) is a dependent variable^ depend-
ing upon the arbitrarily chosen number, n, of equal parts
into which the denominator of the ratio is divided, and it
has been shown to differ from the actual value of the ratio

by an amount less than -. By increasing n at pleasure we

can make this difference as small as we please, but can never
make it absolutely zero, for in that case we should have
found a common measure of the incommensurable numerator
and denominator of the given ratio.

The actual value of an incommensurable ratio is, then, the
limit approached by the approximation described in (35), as
n is indefinitely increased.

41. The usefulness of the method of limits flows entirely
from the following fundamental theorem, the truth of which
is almost axiomatic.

Theorem. — If two variables dependent upon the same variable
are so related that they are always equals no matter what value
is given to the variable on which they depend^ and if as the inde-
pendent variable is changed in some specified way, each of them
approaches a limit, the two limits must be absolutely equal.

For, in considering two variables that are and that always

BOOK II. 73

remain equal to each other, we are dealing with a single vary-
ing value, — i.e., their common value, — and it is clear that a
single variable cannot be made to approach as near as we
please to two diiferent constant values at the same time, as
if it is once brought between the two values in question,
afterward, in approaching nearer to one, it must inevitably
recede from the other.

The student should study this demonstration in connection
with that of Proposition XII., which follows.

PROPOSITION XII.— THEOREM.

42. In the same circle, or in equal circles, two angles at the
centre are in the same ratio as their intercepted arcs.

Let AOB and AGO be two angles at the centre of the same

circle, or at the centres of

equal circles; AB and AC,

their intercepted arcs ; then

AOB ^AB

AOC AC' o

1st. Suppose the arcs to have a common measure, x, which
is contained m times in AB and n times in A C Then AB =t

mx and AC = nx, and

AB rnx m

AC nx n
Apply the measure x to the arcs AB and AC, and draw radii
to the points of division. The angle AOB is thus divided
into m parts, and the angle AOC into n parts, all of which
are equal, by Proposition III., Corollary. Call any one of these
smaller angles y ; then JLO^ = my and AOC == ny, and

AOB rny m

AOC ny n'

Therefore A^ = ^,

AOC AC'

or (v. 36) AOB : AOC =z AB : AC.

74 ELEMENTS OF GEOMETRY.

2d. If the arcs are incommensurable, suppose the arc AG
to be divided into any arbitrarily chosen number, w, of equal
parts, and let one of the parts
be applied as many times as
possible to the arc AB ; let B'
be the last point of division,
and draw the radius OB'.

By construction, the arcs AB'
and AG are commensurable. Therefore, by the proof above,

AOB' ^ A^

AOG AG'

If, now, we change n the number of parts into which A G
is divided, AB' and AOB' will change, and consequently
AOB' , AB' .|| , AOB' . AB' ..

AOG ""^ AG ^^^^ '^"^^'- AOG ^^^ AG "'' '^'^ ^""^-
bles depending upon the same variable, n.

By increasing n at pleasure we can make each of the
equal parts into which AC is divided as small as we please,
and consequently the remainder B'B, which is necessarily
less than one of these parts, can be made as small as we
please. It cannot, however, be made zero, for the arcs AB
and AG are incommensurable, by hypothesis.

It is clear, then, that if n is indefinitely increased, AB' will
have AB for its limitj and A OB' will have A OB for its limit
Hence

-j^^ IS the limit of -j^,

and

41 is the limit of 41?',
AG AG

as n is indefinitely increased.

BOOK II. 75

A OB' AB'
As the two variables and — — , both depending upon

-4.C/0 Aiy

n, are always equal, no matter what the value of w, and each
approaches a limit as n is indefinitely increased, the two
limits in question are absolutely equal (41). Hence

AOB ^AB

AOG AG'

PROPOSITION XIII.— THEOREM.

43. The numerical measure of an angle at the centre of a
circle is the same as the numerical measure of its intercepted arc^
if the adopted unit of angle is the angle at the centre which
intercepts the adopted unit of arc.

Let AOB be an angle at the centre
0, and AB its intercepted arc. Let
-10(7 be the angle which is adopted as
the unit of angle, and let its intercepted
arc AC he the arc which is adopted as
the unit of arc. By Proposition XII., we have

AOB ^AB

AOO AC'

But the first of these ratios is the measure (28) of the angle
AOB referred to the unit AOC; and the second ratio is the
measure of the arc J.J5 referred to the unit AC. Therefore,
with the adopted units, the numerical measure of the angle
AOB is the same as that of the arc AB.

44. Scholium I. This theorem, being of frequent applica-
tion, is usually more briefly, though less accurately, expressed
t>y saying that an angle at the centre is measured by its inter-

76 ELEMENTS OF GEOMETRY.

cepted arc. In this conventional statement of the theorem,
the condition that the adopted units of angle and arc cor-
respond to each other is understood ; and the expression " is
measured by" is used for " has the same numerical measure
as."

45. Scholium II. The right angle is, "by its nature, the most
simple unit of angle ; nevertheless custom has sanctioned a
different unit.

The unit of angle generally adopted is an angle equal to
^ part of a right angle, called a degree, and denoted by the
symbol °. The corresponding unit of arc is ^^^ part of a
quadrant (10), and is also called a degree.

A right angle and a quadrant are therefore both expressed
by 90°. Two right angles and a semi-circumference are both
expressed by 180°. Four right angles and a whole circum-
ference are both expressed by 360°.

The degree (either of angle or arc) is subdivided into
minutes and seconds, denoted by the symbols ' and " : a minute
being -^ part of a degree, and a second being -^ part of a
minute. Fractional parts of a degree less than one second
are expressed by decimal parts of a second.

An angle, or an arc, of any magnitude is, then, numeri-
cally expressed by the unit degree and its subdivisions.
Thus, for example, an angle equal to ^ of a right angle, as
well as its intercepted arc, will be expressed by 12° 51'
25''.714

46. Definition. When the sum of two arcs is a quadrant
(that is, 90°), each is called the complement of the other.

When the sum of two arcs is a semi-circumference (that
is, 180°), each is called the supplement of the other. See (I.,
16).

BOOK II. 77

47. Definitions. An inscribed angle is one whose vertex is
on the circumference and whose sides are
chords ; as BA C.

In general, any rectilinear figure, as
ABCj is said to be inscribed in a circle
when its angular points are on the circum*
ference; and the circle is then said to be
circumscribed about the figure.

An angle is said to be inscribed in a segment when its vertex
is in the arc of the segment, and its sides pass through the
extremities of the subtending chord. Thus, the angle BAG
is inscribed in the segment BAG.

PROPOSITION XIV.— THEOREM.

48. An inscribed angle is measured by one-half its intercepted
arc.

There may be three cases :

1st. Let one of the sides AB of the inscribed angle BAG
be a diameter; then the measure of the
angle BAG is one-half the arc BG.

For, draw the radius OG. Then, AOG
being an isosceles triangle, the angles OAG
and OGA are equal (I., Proposition YIII.).
The angle BOG, an exterior angle of the
triangle AOG, is equal to the sum of the
interior angles OAG and OGA (I., Proposition XXYI., Cor-
ollary), and therefore double either of them. But the an-
gle BOG, at the centre, is measured by the arc BG (44) ;
therefore the angle OAG is measured by one-half the arc
BG.

ELEMENTS OF GEOMETRY.

2d. Let the centre of the circle fall within the inscribed
angle BAG ; then the measure of the angle BAC is one-half
of the arc BC.

For, draw the diameter AD. The meas-
ure of the angle BAB is, by the first case,
one-half the arc BD ; and the measure of
the angle GAD is one-half the arc GD ;
therefore the measure of the sum of the
angles BAD and GAD is one-half the sum
of the arcs BD and GD ; that is, the measure of the angle
BAG is one-half the arc BG.

3d. Let the centre of the circle fall without the inscribed
angle BAG ; then the measure of the angle
BAG is one-half the arc BG.

For, draw the diameter AD. The meas-
ure of the angle BAD is, by the first case,
one-half the arc BD ; and the measure of
the angle GAD is one-half the arc GD ;
therefore the measure of the difference of
the angles BAD and GAD is one-half the
difference of the arcs BD and GD ; that is, the measure of
the angle BAG is one-half the arc BG.

49. Corollary. An angle inscribed in a
semicircle is a right angle.

EXERCISE.

Theorem. — The opposite angles of an inscribed quadrilateral
are supplements of each other.

c D

BOOK II.

PROPOSITION XV.— THEOREM.

50. An angle formed hy a tangent and a chord is measured
by one-half the intercepted arc.

Let the angle BAC be formed by
the tangent AB and the chord AC;
then it is measured by one-half the
intercepted arc AMC.

For, draw the diameter AD. The
angle BAD, being a right angle (Prop-
osition IX.), is measured by one-half

the semi-circumference AMJD ; and the angle CAD is meas-
ured by one-half the arc CD; therefore the angle BAC,
which is the difference of the angles BAD and CAD, is
measured by one-half the difference of AMD and CD ; that
is, by one-half the arc AMC.

Also, the angle B'AC is measured by one-half the inter-
cepted arc ANC. For, it is the sum of the right angle B'AD
and the angle CAD, and is measured by one-half the sum
of the semi-circumference AND and the arc CD ; that is, by
one-half the arc ANC.

EXERCISE.

Prove Proposition XY. by the aid of
this figure, OE being a radius perpendic-
ular to A C.

Suggestion. Complements of the same
angle are equal.

ELEMENTS OF GEOMETRY.

PROPOSITION XVI.— THEOREM.

51. An angle formed by two chords^ intersecting within the
circumference^ is measured by one-half the sum of the arcs inter-
cepted between its sides and between the sides of its vertical angle.

Let the angle AEC be formed by the chords AB, CD, in-
tersecting within the circumference ; then
will it be measured by one-half the sum
of the arcs AG and BD^ intercepted be-
tween the sides of AEG and the sides of
its vertical angle BED,

For, join AD. The angle AEG is equal
to the sum of the angles EDA and EAD, and these angles
are measured by one-half of AG and one-half of BD, re-
spectively ; therefore the angle AEG is measured by one-half
the sum of the arcs AG and BD.

EXERCISE.

Prove Proposition XYI. by the aid of this
figure, DF being draT\jn parallel to AB. (v.
Proposition XI.)

PROPOSITION XVII.— THEOREM.

52. An angle formed by two secants, intersecting without the
circumference, is measured by one-half the difference of the inter-
cepted arcs.

Let the angle BA G be formed by the se-
cants AB and AG; then will it be measured
by one-half the difference of the arcs BG
and DE.

For, join GD. The angle BDG is equal
to the sum of the angles DAG and AGD ;
therefore the angle A is equal to the differ-

BOOK II.

ence of the angles BDC and ACD. But these angles are
measured by, one-half of BG and one-half of JDE respec-
tively; hence. the angle A is measured by one-half the differ-
ence of BG aad DB,

EXERCISE.

Prove Proposition XYII. by the aid of Proposition XI.,
drawing a suitable figure.

PROPOSITION XVIII.-THEOEEM.

53. An angle formed by a tangent and a secant is measured
by half the difference of the intercepted arcs.

For the angle A is equal to BDG minus
ABD, by I., Proposition XXYI., Corollary.

54. Corollary. An angle formed by two
tangents is measured by half the difference of
the intercepted arcs.

EXERCISE.

1. Prove Proposition XYIII. and its Corollary by the aid
of Proposition XI.

2. Theorem. — If through the point of contact of two tangent
circles, two secants are drawn, the

chords joining the points where the
secants cut the circles are parallel.
Suggestion. FED = CEa,

. • . DBE = GAE.

Consider, also, the case where
the given circles are internally
tangent.

82 ELEMENTS OF GEOMETRY.

PROBLEMS OF CONSTRUCTION.

Heretofore our figures have been assumed to be constructed
under certain conditions, although methods of constructing
them have not been given. Indeed, the precise construction
of the figures was not necessary, inasmuch as they were only
required as aids in following the demonstration of principles.
We now proceed, first, to apply these principles in the solu-
tion of the simple problems necessary for the construction
of the plane figures already treated of, and then to apply
these simple problems in the solution of more complex ones.

All the constructions of elementary geometry are effected
solely by the straight line and the circumference, these being
the only lines treated of in the elements ; and these lines are
practically drawrij or described, by the aid of the ruler and
compasses, with the use of which the student is supposed to
be familiar.

PROPOSITION XIX.— PROBLEM.

65. To bisect a given straight line.

Let AB be the given straight Kne.

"With the points A and B as centres, and with a radius
greater than the half of AB, describe arcs

intersecting in the two points D and E.

Through these points draw the straight line

DJS, which bisects AB at the point G. For,

J) and B being equally distant from A and

B, the straight line DE is perpendicular to

AB at its middle point (I., Proposition ^

XYIII.).

-iB

"jk"

BOOK II. 83

PROPOSITION XX.— PROBLEM.

56. At a given point in a given straight line, to erect a perpen-
dicular to that line.

Let AB be the given line and G the given .^ ..^

point.

Take two points, D and B, in the line
and at equal distances from C. With D J"d o Fb
and E as centres, and a radius greater than
DCov GEj describe two arcs intersecting in F. Then CF is
the required perpendicular (I., Proposition XYIII.).

57. Another solution. Take any point
O, without the given line, as a centre.

from O to C, describe a circumference

and with a radius equal to the distance ^y

intersecting J.jB in G and in a second **• -''

point D. Draw the diameter DOE^

and join EG. Then EG will be the required perpendicular ;
for the angle EGD^ inscribed in a semicircle, is a right angle
(Proposition XIY., Corollary).

This construction is often preferable to the preceding, es-
pecially when the given point G is at, or near, one extremity
of the given line, and it is not convenient to produce the line
through that extremity. The point O must evidently be so
chosen as not to lie in the required perpendicular.

,t7l?I?B

ELEMENTS OF GEOMETRY.

PROPOSITION XXI.— PROBLEM.

58. From a given point without a given straight line, to lei

fall a perpendicular to that line.

o
Let AB be the given line and G the

given point.

With C as a centre, and with a radius a — "-^^ — I — r^rr- — b
sufficiently great, describe an arc in-
tersecting AB in D and E. With D
and E as centres, and a radius greater
than the half of DE^ describe two arcs intersecting in F.
The line CF is the required perpendicular (I., Proposition
XYIII.).

59. Another solution. With any point 0 in the line AB as a
centre, and with the radius 0(7, describe

an arc GDE intersecting AB in D.

With i) as a centre, and a radius equal

to the distance DC, describe an arc in- ^

tersecting the arc CDE in E. The line /%

CE is the required perpendicular. For,

the point I) is the middle of the arc CJDE^ and the radius

OD drawn to this point is perpendicular to the chord CE

(Proposition YI.).

li)

PROPOSITION XXII.-PROBLEM.

60. To bisect a given arc or a given angle.

Ist. Let AB be a given arc.

Bisect its chord AB by a perpendicular, as
in (55). This perpendicular also bisects the
arc (Proposition YI.).

BOOK II.

2d. Let BAC be a given angle. With
^ as a centre, and with any radius, de-
scribe an arc intersecting the sides of the
angle in D and E, With D and E as
centres, and with equal radii, describe
arcs intersecting in F. The straight line
AF bisects the arc DE^ and consequently also the angle
BAG,

61. Scholium. By the same construction, each of the halves
of an arc, or an angle, may be bisected ; and thus, by succes-
sive bisections, an arc, or an angle, may be divided into 4, 8,
16, 32, etc., equal parts.

PROPOSITION XXIII.— PROBLEM.

62. At a given point in a given straight line, to construct an
angle equal to a given angle.

Let J. be the given point in the straight
line AB, and 0 the given angle.

With 0 as a centre, and with any radius,
describe an arc MJS^ terminated by the sides
of the angle. With A as a centre, and with
the same radius, OM, describe an indefi-
nite arc BC. With ^ as a centre, and with
a radius equal to the chord of MJ^, de-
scribe an arc intersecting the indefinite arc BC in D. Join
AI). Then the angle BAD is equal to the angle 0. For
the chords of the arcs MN and BD are equal; therefore
these arcs are equal, and consequently also the angles 0
and A.

86 ELEMENTS OF GEOMETRY.

PROPOSITION XXIV.— PROBLEM.

63. Through a given point, to draw a parallel to a given
straight line.

Let A be the given point, and JBC ^

the given line. ,\

From any point B in BO draw the a/. — \ jp

straight line BAD through A. At the

point A, by the preceding problem, b ' ^

construct the angle DAE equal to the

angle ABC. Then AB is parallel to BC (I., Proposition

XXIY., Corollary L).

64, Scholium. This problem is, in practice, more accurately
solved by the aid of a triangle, con-
structed of wood or metal. This
triangle has one right angle, and its
acute angles are usually made equal
to 30° and 60°.

Let A be the given point, and
BG the given line. Place the tri-
angle, BJFD, with one of its sides
in coincidence with the given line

BO. Then place the straight edge of a ruler, MN, against
the side EF of the triangle. Now, keeping the ruler firmly
fixed, slide the triangle along its edge until the side ED
passes through the given point A. Trace the line EAD along
the edge ED of the triangle ; then it is evident that this line
will be parallel to BG.

EXERCISE.

Prohlem. Two angles of a triangle being given, to find the
third, (v. I., Proposition XXYI., and I., Proposition III., Cor-
ollary I.)

BOOK II. ^7

PROPOSITION XXV.— PROBLEM.

65. Two sides of a triangle and their included angle being
given, to construct the triangle.

y &

Let 6 and c be the given sides, and Xj^

A their included angle.

Draw an indefinite line AE, and
construct the angle EAF=^A. On
AE take AC = b, and on AF take
AB = c; join BC. Then ABC is
the triangle required ; for it is formed with the data.

With the data, two sides and the included angle, only one
triangle can be constructed ; that is, all triangles constructed
with these data are equal, and thus only repetitions of the
same triangle (I., Proposition YL).

66. Scholium. It is evident that one triangle is always pos-
sible, whatever may be the magnitude of the proposed sides
and their included angle.

PROPOSITION XXVL— PROBLEM.

67. One side and two angles of a triangle being given, to
construct the triangle.

Two angles of the triangle being given, ^^ ^^^^

the third angle can be found; and we c

shall therefore always have given the
two angles adjacent to the given side.
Let, then, c be the given side, A and B
the angles adjacent to it.

Draw a line AB = c ; at J. make an angle BAD = A, and
at B an angle ABE = B. The lines AD and BE intersecting
in (7, we have ABC as the required triangle.

With these data but one triangle can be constructed (I.,
Proposition YIL).

88 ELEMENTS OF GEOMETRY.

68. Scholium. If the two given angles are together equal
to or greater than two right angles, the problem is impos-
sible ; that is, no triangle can be constructed with the data ;
for the lines AD and BG will not intersect on that side of
AB on which the angles have been constructed.

PROPOSITION XXVII.— PROBLEM.

69. The three sides of a triangle being given, to construct the
triangle,

Let a, bj and c be the three given sides. 6. .

Draw BC = a; with (7 as a centre and *
a radius equal to b describe an arc ; with
.B as a centre and a radius equal to c de-
scribe a second arc intersecting the first
in A. Then ABC is the required triangle.

With these data but one triangle can be constructed (I.,
Proposition IX.).

70. Scholium. The problem is impossible when one of the
given sides is equal to or greater than the sum of the other
two (I., Axiom I.).

PROPOSITION XXVIII.— PROBLEM.

7L Two sides of a triangle and the angle opposite to one of
them being given, to construct the triangle.

i a

1st. When the given angle A is

acute, and the given side a, oppo- r/

site to it in the triangle, is less than VwiXo

the other sjiven side c. X 7 I \

Construct an angle DAE = A. ^ ^,' '^z, ^

In one of its sides, as AD, take

AB = c; with J5 as a centre and a radius equal to a, describe

BOOK II. 89

an arc which (since a <^ c) will intersect AE in two points, C"
and C", on the same side of A. Join BC and BG", Then
either ABC or ABC" is the required triangle, since each is
formed with the data ; and the problem has two solutions.

There will, however, be but one solution, even with these
data, when the side a is so much less than the side c as to be
just equal to the perpendicular from B upon AE. For then
the arc described from -S as a centre, and with the radius a,
will touch AE m2i single point, (7, and the required triangle
will be ABG^ right angled at C.

2d. When the given angle A is
either acute, right, or obtuse, and
the side a opposite to it is greater
than the other given side c.

The same construction being "^(""a' ^^^

made as in the first case, the arc

described with j5 as a centre, and with a radius equal to a,
will intersect AE in only one point, C, on the same side of A.
Then ABC will be the triangle required, and will be the only-
possible triangle with the data.

The second point of intersection, C", will fall in EA pro-
duced, and the triangle ABC thus formed will not contain
the given angle.

72. Scholium. The problem is impossible when the given
angle A is acute and the proposed side opposite to it is less
than the perpendicular from B upon AE; for then the arc
described from B will not intersect AE.

The problem is also impossible when the given angle is
right, or obtuse, if the given side opposite to the angle is less
than the other given side ; for either the arc described from
B would not intersect AEj or it would intersect it only when
produced through A.

90 ELEMENTS OF GEOMETRY.

EXERCISE.

Problem. — The adjacent sides of a parallelogram and their
included angle being given, to construct the parallelogram.

PROPOSITION XXIX.— PROBLEM.

73. To find the centre of a given circumference, or of a given
arc.

Take any three points, A, B, and (7, in
the given circumference or arc, and join
them by chords AB, BC. The perpen-
diculars erected at the middle points of
these chords will intersect in the required
centre (Proposition YI., Corollary I.).

74. Scholium I. Only one solution is possible ; for, since the
centre is equidistant from B and (7, it must lie in the perpen-
dicular erected at the middle point of BC (I., Proposition
XYIII.), and since it is equidistant from A and B, it must lie
in the perpendicular erected at the middle point of AB ; and
these perpendiculars can have but one point in common.

75. Scholium IL The same construction serves to describe
a circumference which shall pass through three given points,
Aj B, C; or to circumscribe a circle about a given triangle,
ABC; that is, to describe a circumference in which the given
triangle shall be inscribed (47).

76. Scholium III. It follows from Scholium 1. that three
points not in the same straight line will determine a circum-
ference,— i.e, through three points not in the same straight
line one circumference, and only one, can be drawn.

Hence two circumferences cannot intersect in more than
two points; for if they had three points in common they
would coincide throughout.

BOOK II.

PROPOSITION XXX.— PROBLEM.

77, At a given point in a given circumference^ to draw a tan-
gent to the circumference.

Let A be the given point in the given
circumference. Draw the radius OA^ and
at A draw BAC perpendicular to OA ;
BG will be the required tangent (Propo-
sition IX.).

If the centre of the circumference is
not given, it may first be found by the
preceding problem, or we may proceed
more directly as follows : take two points,
D and E^ equidistant from A ; draw the
chord DE^ and through A draw BAC
parallel to BE. Since A is the middle
point of the arc BE^ the radius drawn
to A will be perpendicular to BE (Proposition YI.), and con-
sequently also to BG ; therefore jBC is a tangent at A,

PROPOSITION XXXI.— PROBLEM.

78. Through a given point without a given circle^ to draw a
tangent to the circle.

Let 0 be the centre of the given circle
and P the given point.

Upon OP, as a diameter, describe a cir-
cumference intersecting the circumference
of the given circle in two points, A and A'.
Draw PA and PA\ both of which will be
tangent to the given circle. For, drawing
the radii OA and OA', the angles OAP and OA'P are right

ELEMENTS OP GEOMETRY.

angles (Proposition XIY., Corollary) ; therefore PA and PA^
are tangents (Proposition IX.).

In practice, this problem is accurately-
solved by placing the straight edge of a
ruler through the given point and tangent
to the given circumference, and then tracing
the tangent by the straight edge. The pre-
cise point of tangency is then determined
by drawing a perpendicular to the tangent
from the centre.

79. Scholium, This problem always admits of two solutions.

EXERCISE.

Problem. — To draw a
common tangent to two
given circles.

Suggestion. For an ex-
terior common tangent, in
the larger circle draw a
concentric circle whose
radius is the difference of the
radii of the given circles. For
an interior common tangent,
about one of the circles draw
a concentric circle whose ra-
dius is the sum of the radii of
the given circles.

BOOK II.

PROPOSITION XXXII.— PROBLEM.

80. To inscribe a circle in a given triangle.

Let ABC be the given triangle. Bisect any two of its
angles, as B and (7, by straight lines meet-
ing in 0. From the point 0 let fall per-
pendiculars OX), OE^ OF, upon the three
sides of the triangle ; these perpendiculars
will be equal to each other (I., Proposi-
tion XIX.). Hence the circumference of
a circle, described with the centre 0, and
a radius = OD, will pass through the three points D, E, F,
will be tangent to the three sides of the triangle at these
points (Proposition IX.), and will therefore be inscribed in
the triangle.

EXERCISE.

Problem. — Upon a given straight line, to describe a segment
which shall contain a given angle.

Suggestion. Through one end of the given
line AB draw a line BC, making with it
the given angle. The two lines will be
one a chord and the other a tangent.
Hence the centre of the circle can be
found.

EXERCISES ON BOOK II.

THEOREMS.

1. If two circumferences are tangent in-
ternally, and the radius of the larger is the
diameter of the smaller, then any chord of
the larger drawn from the point of contact
is bisected by the circumference of the
smaller {v. Proposition XIV., Corollary, and
Proposition VI.).

2. If two equal chords intersect within a circle, the segments
of one are respectively equal to the segments of the other. What
is the corresponding theorem for the case where the chords meet
when produced?

3. A circumference described on the hypotenuse of a right tri-
angle as a diameter passes through the vertex of the right angle.
(v. Proposition XIV., Corollary.)

4. The circles described on two sides of a triangle as diameters
intersect on the third side.

Suggestion, Drop a perpendicular from the opposite vertex upon
the third side.

5. The perpendiculars from the angles upon the opposite sides
of a triangle are the bisectors of the angles of the triangle formed
by joining the feet of the perpendiculars.

Suggestion. On the three sides of the given triangle as diam-
eters describe circumferences, {v. Exercise 3, Proposition XIV.,
and I., Proposition XXVI.).

6. If a circle is circumscribed about an equilateral triangle, the
perpendicular from its centre upon a side of the triangle is equal
to one-half of the radius.

BOOK II.

7. The portions of any straight line which
are intercepted between the circumferences
of two concentric circles are equal.

8. Two circles are tangent internally
at P, and a chord AB of the larger circle
touches the smaller at C; prove that PC
bisects the angle APB.

Suggestion. CPQ = BCP, BPQ =
BAP, BCP—BAP = APC (I., Propo-
sition XXVI., Corollary).

9. If a triangle ABC is formed by the intersection of three
tangents to a circumference, two of
which, A3f and AN, are fixed, while
the third, BC, touches the circumfer-
ence at a variable point P, prove that
the perimeter of the triangle ^^C is
constant, and equal to AM -\- AN, or
2 AN (Proposition X.).

Also, prove that the angle P 0(7 is
constant.

10. If through one of the points of intersection of two circum-
ferences a diameter of each circle is drawn, the straight line
which joins the extremities of these diameters passes through the
other point of intersection, and is parallel to the line joining the
centres.

Suggestion. Draw the common chord and the line joining the
centres, (v. Proposition VI., Corollary II., and Exercise 29,
Book I.)

11. The difference between the hy-
potenuse of a right triangle and the
sum of the other two sides is equal to
the diameter 'of the inscribed circle.

ELEMENTS OF GEOMETRY.

12. A circle can be entirely sur-
rounded by six circles having the
same radius with it.

13. The bisectors of the vertical angles of all triangles having
the same base and equal vertical angles have a point in common.

Suggestion, The triangles may all be inscribed in the same
circle.

14. If the hypotenuse of a right triangle is double one of the
sides, the acute angles of the triangle are 30° and 60° respectively.

15. If, from a point whose distance from the centre of a given
circle is equal to a diameter, tangents are drawn to the circle,
they will make with each other an angle of 60°.

LOCI.

16. Find the locus of the centre of a circumference which passes
through two given points, (v. I., Proposition XVIII.)

17. Find the locus of the centre of a circumference which is
tangent to two given straight lines, {v. I., Proposition XIX.)

18. Find the locus of the centre of a circumference which is
tangent to a given straight line at a given point of that line, or
to a given circumference at a given point of that circumference.

19. Find the locus of the centre of a circumference passing
through a given point and having a given radius.

20. Find the locus of the centre of a circumference tangent to
a given straight line and having a given radius.

21. Find the locus of the centre of a circumference of given
radius, tangent externally or internally to a given circumference.

BOOK II.

22. A straight line MN, of given length,
is placed with its extremities on two
given perpendicular lines ^^, CD; find
the locus of its middle point P (Exercise
31, Book I.).

23. A straight line of given length is inscribed in a given circle;
find the locus of its middle point, {v. Proposition VII.)

24. A straight line is drawn through a
given point A^ intersecting a given circum-
ference in B and C; find the locus of the
middle point, P, of the intercepted chord
BC.

Note the special case in which the point
A is on the given circumference.

25. From any point ^ in a given circumference, a straight line
-4P of fixed length is drawn parallel to a given line MN ; find
the locus of the extremity P, {v. I., Proposition XXX.)

26. From one extremity -4 of a fixed
diameter ABy any chord AC is drawn,
and at (7 a tangent CD. From P, a per-
pendicular BD to the tangent is drawn,
meeting AC in P, Find the locus of P.

Suggestion. (Draw radius OC. v. I.,
Exercise 28.)

27. The base BC of a triangle is given, and
the medial line BE, from jB, is of a given
length. Find the locus of the vertex A.

Suggestion. Draw AO parallel to EB. Since
BO=BC, O is a fixed point; and since AO
= 2BEy OA is a constant distance.

98 ELEMENTS OF GEOMETRY.

PROBLEMS.

The most useful general precept that can be given, to aid the
student in his search for the solution of a problem, is the follow-
ing : Suppose the problem solved, and construct a figure accord-
ingly ; study the properties of this figure, drawing auxiliary lines
when necessary, and endeavor to discover the dependence of the
problem upon previously solved problems. This is an analysis
of the problem. The reverse process, or synthesis^ then furnishes
a construction of the problem. In the analysis, the student's in-
genuity will be exercised especially in drawing useful auxiliary
lines ; in the synthesis, he will often find room for invention in
combining in the most simple form the several steps suggested by
the analysis.

The analysis frequently leads to the solution of a problem by
the intersection of loci. The solution may turn upon the deter-
mination of the position of a particular point. By one condition
of the problem it may appear that this required point is neces-
sarily one of the points of a certain line ; this line is a locus of
the point satisfying that condition. A second condition of the
problem may furnish a second locus of the point ; and the point
is then fully determined, being the intersection of the two loci.

Some of the following problems are accompanied by an analysis
to illustrate the process.

28. To determine a point whose distances from two given inter-
secting straight lines, AB^
A^B\ are given. •.... c

Analysis. The locus of
all the points which are at
a given distance from AB
consists of two parallels to
ABy CE, and DF, each at
the given distance from
AB, The locus of all the

points at a given distance from A^B^ consists of two parallels,
C^E^ and D^F^^ each at the given distance from A^B\ The re-
quired point must be in both loci, and therefore in their inter-
section. There are in this case four intersections of the loci, and
the problem has four solutions.

Construction. At any point of AB, as A, erect a perpendicular
CD, and make AC = AD = the given distance from AB ; through
O and 2) draw parallels to AB. In the same manner, draw par-
allels to A^B^ at the given distance A^C = A'D\ The intersec-

E 0

\ /

\»*

^^ \

/'o

'\ \

y*.

BOOK II. 99

tion of the four parallels determines the four points P^, Pj? P^^ ^4»
each of which satisfies the conditions.

29. Given two perpendiculars, AB and CD,
intersecting in O, to construct a square, one
of whose angles shall coincide with one of
the right angles at O, and the vertex of the
opposite angle of the square shall lie on a
given straight line EF, (Two solutions.)

30. In a given straight line, to find a point equally distant from
two given points without the line.

31. To construct a square, given its diagonal.

32. Through a given point P within a given angle, to draw a
straight line, terminated by the sides of the anglfe, which shall
be bisected at P. {v. Exercise 28, Book I.)

33. Given two straight lines which caimot be produced to their
intersection, to draw a third which would pass through their
intersection and bisect their contained angle.

Suggestion. Find two points equidistant from the two lines.
(v. I., Proposition XIX.)

34. Given the middle point of a chord in a given circle, to draw
the chord.

36. To draw a tangent to a given circle which shall be parallel
to a given straight line.

36. To draw a tangent to a given circle, such that its segment
intercepted between the point of contact and a given straight
line shall have a given length.

Suggestion. The tangent, the radius drawn to the point of con-
tact, and a line drawn from the centre to the end of the tangent
form a right triangle, two of whose sides are known. A simple
construction gives the hypotenuse.

In general there are four solutions. Show when there will be
but two ; also, when no solution is possible.

37. Through a given point within or without a given circle, to
draw a straight line, intersecting the circumference, so that the
intercepted chord shall have a given length. (Two solutions.)
{v. Exercise 23 and Section 78.)

38. Construct an angle of 60°, one of 120°, one of 30°, one of 150°,
one of 45°, and one of 135°.

100 ELEMENTS OF GEOMETRY.

39. Construct a triangle, given the base, the angle opposite to
the base, and the altitude.

Analysis. Suppose BAC to be the re- ^ s^^

quired triangle. The side BC being fixed ^ 7^^. ^\\ ^

in position and magnitude, the vertex A is / / ^X; 1 1

to be determined. One locus of A is an k.^^ '^-vV

arc of a segment, described upon AB, con- ^ ^

taining the given angle. Another locus

of J. is a straight line MN drawn parallel to ^(7, at a distance
from it equal to the given altitude. Hence the position of A will
be found by the intersection of these two loci, both of which are
readily constructed.

Limitation. If the given altitude were greater than the perpen-
dicular distance from the middle of ^C to the arc BAC, the arc
would not intersect the line MN, and there would be no solution
possible.

The limits of the data within which the solution of any prob-
lem is possible should always be determined.

40. Construct a triangle, given the base, the medial line to the
base, and the angle opposite to the base.

41. With a given radius, describe a circumference, 1st, tangent
to two given straight lines ; 2d, tangent to a given straight line
and to a given circumference ; 3d, tangent to two given circum-
ferences ; 4th, passing through a given point and tangent to a
given straight line ; 5th, passing through a given point and tan-
gent to a given circumference ; 6th, having its centre on a given
straight line, or a given circumference, and tangent to a given
straight line, or to a given circumference. (Exercises 19, 20, 21.)

42. Describe a circumference, 1st, tangent to two given straight
lines, and touching one of them at a given point (Exercises 17,
18) ; 2d, tangent to a given circumference at a given point and
tangent to a given straight line ; 3d, tangent to a given straight
line at a given point and tangent to a given circumference (Exer-
cise 18) ; 4th, passing through a given point and tangent to a
given straight line at a given point ; 5th, passing through a given
point and tangent to a given circumference at a given point.

43. Draw a straight line equally distant from three given points.
When will there be but three solutions, and when an indefinite

number of solutions ?

44. Inscribe a straight line of given length between two given
circumferences, and parallel to a given straight line. {v. Exer-
cise 25.)

BOOK III.

PROPORTIONAL LINES. SIMILAR FIGURES.
THEORY OF PROPORTION.

1. Definition. One quantity is said to be proportional to
another when the ratio of any two values, A and JB, of the
first, is equal to the ratio of the two corresponding values
A' and 5', of the second ; so that the four values form the
proportion (II., 36)

A:B=A':B%
or

B B''

This definition presupposes two quantities, each of which
can have various values, so related to each other that each
value of one corresponds to a value of the other. An exam-
ple occurs in the case of an angle at the centre of a circle
and its intercepted arc. The angle may vary, and with it
also the arc ; but to each value of the angle there corresponds
a certain value of the arc. It has been proved (II., Proposi-
tion XII.) that the ratio of any two values of the angle is
equal to the ratio of the two corresponding values of the
arc ; and, in accordance with the definition just given, this
proposition would be briefly expressed as follows : " The angle
at the centre of a circle is proportional to its intercepted arc."

2. Definition. One quantity is said to be reciprocally propor-
tional to another when the ratio of two values, A and B^ of
the first, is equal to the reciprocal of the ratio of the two

«* 101

102 ELEMENTS OF GEOMETRY.

corresponding values, A' and B\ of the second, so that the
four values form the proportion

A\B=B':A\
or

A^B^ ^. _^ ^
B A' ' B''

For example, if the product p of two numbers, x and y, is
given, so that we have

xy=p, ^

then X and y may each have an indefinite number of values,
but as X increases y diminishes. If, now, A and B are two
values of rr, while A' and B' are the two corresponding values

of ?/, we must have

A XA' = p,

BX B' = p,
whence, by dividing one of these equations by the other,

B^ B'~ '
and therefore

A _ J_ ^ ^' .
B A^ A"
B'

that is, two numbers whose product is constant are reciprocally
proportional.

3. Let the quantities in each of the couplets of the pro-
portion

j = §,ovA:B = A':B', [1]

be measured by a unit of their own kind, and thus expressed

BOOK III. 103

by numbers (II., 28) ; let a and h denote the numerical meas-
ures of A and B, a' and h' those of A' and B' ; then (II., 29)

A^a A!. =9l

B 6' B' b"

and the proportion [1] may oe replaced by the numerical pro-
portion

0 0

4. Conversely, if the numerical measures a, 6, a', h\ of four
quantities, A^ J5, A\ B', are in proportion, these quantities
themselves are in proportion, provided that A and B are
quantities of the same kind, and J.' and B' are quantities of
the same kind (though not necessarily of the same kind as A
and B) ; that is, if we have

a\h = a' :h\

we may, under these conditions, infer the proportion

A:B = A' :B'.

5. Let us now consider the numerical proportion

a\h = a' :h\
Writing it in the form

and multiplying both members of this equality by hh\ we
obtain

ah' = a'b,

whence the theorem : the product of the extremes of a (numer-
icaT) proportion is equal to the product of the means.

4ii»tt.

104 ELEMENTS OP GEOMETRY.

Corollary. If the means are equal, as in the proportion
a : b = b : Cj we have b^ = ac, whence b = j/ac; that is, a
mean proportional (II., 36) between two numbers is equal to the
square root of their product.

6. Conversely, if the product of two numbers is equal to the
product of two others, either two may be made the extremes, and
the other two the means, of a proportion. For, if we have given

ab' = a^bj
then, dividing by bb', we obtain

^ = ^'ora:b = a':b\
b b

Corollary. The terms of a proportion may be written in
any order which will make the product of the extremes
equal to the product of the means. Thus, any one of the
following proportions may be inferred from the given equal-
ity ab' = a'b :

:b =a'

:b\

:a'=b

:6',

:a=b'

:a'.

:b' = a

:a',

U '. a' =b '.a, etc.
Also, any one of these proportions may be inferred from any
other.

7. Definitions. When we have given the proportion

a-.b = a' :b',
and infer the proportion

a: a' =b :¥,
the second proportion is said to be deduced by alternation.
When we infer the proportion

b : a = b' '. a',
this proportion is said to be deduced by inversion.

BOOK III. 105

8. It is important to observe that when we speak of the
products of the extremes and means of a proportion, it is
implied that at least two of the terms are numbers. If, for
example, the terms of the proportion

A:B=A': B'

are all lines^ no meaning can be directly attached to the
products ^ X ^'j ^ X A\ since in a product the multiplier
at least must be a number.

But if we have a proportion such as

A \ B = m : n^

in which m and n are numbers, while A and B are any two
quantities of the same kind, then we may infer the equality
nA = mB.

Nevertheless, we shall, for the sake of brevity, often speak
of the product of two lineSj meaning thereby the product of the
numbers which represent those lines when they are measured hy a
common unit

9. If A and B are any two quantities of the same kind,
and m any number whole or fractional, we have, identically,

mA A ,

mB~ B'

that is, equimultiples of two quantities are in the same ratio as
the quantities themselves.

Similarly, if we have the proportion

A:B = A':B%

and if m and n are any two numbers, we can infer the pro-
portions

mA : mB = nA' : nB\

mA : nB = mA' : nB\

106 ELEMENTS OF GEOMETRY.

10. Composition and Division. If we have given the pro-

A A'
portion - = — , we have, by alternation,
B B'

A' B''
Let r be the common value of these two ratios ; then

4, = r,and|=r,

and

A = rA', and B == rB\

Adding the second equation to the first, we have

A + B^KA'+B'),
or

A_±B_ ^ A^B_

A' + B' A' B''

The proportions ^^±|-, = A, and ^,i|, = | are said to

be formed from the given proportion

A:B^=A':B\hj composition.

If we subtract the equation B = rB' from A = rA!^ we have

A—B = r(^Al — B'),
whence, as above,

A —B ^A

A' — B' A"
and

A —B ^B
A' — B' B"

two proportions which are said to be formed from the given

proportion

A : B = A' : B^hj division.

11. Definition. A continued proportion is a series of equal
ratios, as

A:B=zA' :B'=:A" : B" = A'" : B"' = etc.

BOOK III. 107

12. Let r denote the common value of the ratio in the
continued proportion of the preceding article ; that is, let

B B' B" B"'
then we have

A=::Br, A'=B'r, A" = B'^r, A"' = B'^'r, etc.,
and, adding these equations,

A-{.A' + A''+ A'" + etc. = (B-{-B' + B" + B'" + etc.) r,
whence

^ + ^^+^"+^"^+etc. ^^^A^A^^^^^,
BJ^B'^B^'^ B"' + etc. B B'

that is, the sum of any number of the antecedents of a continued
proportion is to the sum of the corresponding consequents as any
antecedent is to its consequent.

In this theorem the quantities J., 5, (7, etc., must all be
quantities of the same kind.

If, instead of a continued proportion, we have an ordinary
proportion, the theorem just proved obviously holds good.

13. If we have any number of proportions, as

a '. h =^ c '. dj
a'-.h' = c' : d\
a" : 6" = c" : d'\ etc. ;
then, writing them in the form

a c a! c^ a" c"

h~ d' h'~d" W~dr' '

and multiplying these equations together, we have

aa' a" . . . c d d' . . .

hh'h'' ...~ dd'd'\.:
or

aaW ...'.hh'h" ... = cdd'...:dd'd"...]

that is, if the corresponding terms of two or more proportions
are multiplied together, the products are in proportion.

108 ELEMENTS OF GEOMETRY.

If the corresponding terms of the several proportions are
equal, that is, ii a z= a' = a", b = b' = V'^ etc., then the mul-
tiplication of two or more proportions gives

that is, if four numbers are in jprojportion^ like powers of these
numbers are in proportion.

PEOPOETIONAL LI:N^ES.
PROPOSITION I.— THEOREM.

14. A parallel to the base of a triangle divides the other two

sides proportionally.

Let DE be a parallel to the base, BC, of the triangle ABC;

then

AB:AD=AC :AE. ^

1st. Suppose the lines J.J5, AD, to have # i^

a common measure which is contained m
times in AB and n times in AD. Then / [S

I ^^u'

AB_m L A

AD n ^ ^o

Apply this measure to AB, and through
the points of division draw lines parallel to the base BC o^
the triangle ; then through the points of intersection of these
lines with A G draw lines parallel to AB. The small triangles
thus formed are all equal, by Propositions XXIX. and YII.,
Book I. Hence the m parts into which ACiQ divided are all
equal, and, as AE contains n of these parts,

AC ^m
AE n'

Therefore

AB ^ AC

AD AE

BOOK III. 109

2d. If AB and AD are incommensurable, let AD be divided
into any arbitrarily chosen number n of
equal parts, and let AB be divided by one ^

of these parts. Let B' be the last point of A

division, 5 '5 beinff of course less than the di- / \

visor. Through B' draw B' C parallel to DE, / \ ~

Since AB' and AD are commensurable, / \

A W A C I \

= — — -, and this holds true no matter i//-— -\c"

AD AE t \

what value may be given to n. By taking

a sufficiently great value for /i, we can make B' come as near

''is we please to B ; but we cannot make B' and B coincide,

since no divisor of AD can divide AB without remainder.

AB' AC

AB' and AC, and consequently and , are then

^ ^ AD AE'

variables dependent upon the same variable, n; and, as we

have seen above, they are equal, no matter what value is

given to n. If n is indefinitely increased,

and

Therefore, by the fundamental theorem in the Doctrine of
Limits (41, Book II.), these limits are equal, and therefore

AB ^ AC

AD~ AE' \

Compare this reasoning with that in II., 42. \

EXERCISE. \

Show that in Proposition I. AD : DB = AE : EC Q>. 10),

and also that

AB ^AD ^ DB

AC AE EC'

AB'
AD

has the limit

AB
AD'

AC
AE

has the limit

AC
AE

110

ELEMENTS OF GEOMETRY.

PKOPOSITION II.— THEOREM.

15. Conversely, if a straight line divides two sides of a tri-
angle proportionally, it is parallel to the third side.

Let DE divide the sides AB, AC, of the tri-
angle ABC, proportionally; then DE is par-
allel to BC

For, if DE is not parallel to BC, let some
other line DE', drawn through D, be parallel
to BC. Then, by Proposition I.,

AB:AD=AC:AE'.

But, by hypothesis, we have

AB:AD = AC:AE.

Hence

AC^
AE'

AG
AJEf

whence it follows that AE' = AE, which is impossible unless
DE' coincides with DE. Therefore DE is parallel to BC.

EXERCISE.

1. Theorem. — The line bisecting the vertical angle of a triangle
divides the base into segments proportional to the adjacent sides
of the triangle.

Suggestion. Through B draw
a line parallel to the bisector
and extend the side CA to meet
it. The triangle EAB is isos-
celes. .'.AE = AB. CE and

CB are divided proportionally (Proposition I.). Hence CD :
DB = CA : AB.

2. Prove the converse of Exercise I. (v. Proposition II.)

BOOK III. Ill

SIMILAR POLYGONS.

16. Definitions. Two polygons are similar when they are
mutually equiangular and have their homologous sides pro-
portional.

In similar polygons, any points, angles, or lines, similarly
situated in each, are called homologous.

The ratio of a side of one polygon to its homologous side
in the other is called the ratio of similitude of the polygons.

PROPOSITION III.— THEOREM.

17. Two triangles are similar when they are mutually equi-
angular.

Let ABC, A'B'C, be mutually ^

equiangular triangles, in which y^i -4'

A=A', B = B\ C= C; then / / ^f

these triangles are similar. / / / /

For, superpose the triangle b c b! c

A'B'C upon the triangle ABC,

making the angle A' coincide with its equal, the angle A, and

let B' fall at h and C at e. Since the angle Abe is equal to

B, he is parallel to BC (I., Proposition XXIY., Corollary I.),

and we have (Proposition I.)

AB:Ah=AC '.Ac,
or

AB\A'B'=^AC'.A'C'.

If, now, we superpose A'B'C upon ABC, making B' co-
incide with B, we may prove, in the same manner, that

AB'.A'B'=BC'.B'C';
and, combining these proportions,

AB ^ AC_ ^ BC_ PJ-,

A'B' A'C B'C '- ^

Therefore the homologous sides are proportional, and the tri-
angles are similar (16).

112 ELEMENTS OF GEOMETRY.

18. Scholium I. The homologous sides lie opposite to equal
angles.

19. Scholium II. The ratio of similitude (16) of the two
similar triangles is any one of the equal ratios in the con-
tinued proportion [1].

EXERCISE.

Theorem. — The altitudes of two similar triangles are to each
other in the ratio of similitude of the triangles.

PROPOSITION IV.— THEOREM.

20. Two triangles are similar when an angle of the one is
equal to an angle of the other, and the sides including these
angles are proportional.

In the triangles ABC, A'B'C, ^ ^'

let A = A', and yf /j

AB AC , .X / X /

A'B' A'C' / Bf G'

then these triangles are similar. ^ ^

For, place the angle A' upon its

equal angle A ; let B' fall at h, and C at c. Then, by the

hypothesis,
^ AB^AC

Ah Ac

Therefore he is parallel to BC (Proposition II.), and the tri-
angle Ahc is similar to ABC (Proposition III.). But Ahc is
equal to A'B'C; therefore A'B'C is also similar to ABC.

BOOK III. 113

PROPOSITION v.— THEOREM.

21. Two triangles are similar when their homologous sides are
propoj'tional.

In the triangles ABC, A'B'C, let

AB_^AG_^BG_, ry.

A'B A'C B'G" •- ■'

then these triangles are similar. ^ ^'

For, take Ah == A!B', and Ac ' /I /I

= ^'(7', and join .6 and c. /^ / / /

Abe is similar to ABC, by Prop- y^' J n c

osition lY. Therefore b o

AB BO AB BC

-, or

Ab be ' A'B' be

But, by hypothesis,

AB ^ BC

A'B' B'C'
Hence

The triangle A!B'G' is then equal to Abe, by I., Proposition
IX., and is consequently similar to ABC.

PROPOSITION VI.— THEOREM.

22. If two polygons are composed of the same number of tri-
angles similar each to each and similarly placed, the polygons
are similar.

Let the polygon ABCD, /"^^^•^^^^ /t^^^

etc., be composed of the tri- ^/:;.'.. \^ A'k::y.. \n

angles J.5 (7, A CD, etc.; and \ " '"^^^ V---^'^'

let the polygon A'B'C'D\ V-^-^""''^

etc., be composed of the

triangles A'B'C\ A'C'D\ etc., similar to ABC, ACD, etc.,

2^ 10*

114

ELEMENTS OF GEOMETEY.

respectively, and similarly placed; then the polygons are
similar.

1st. The polygons are
mutually equiangular. For,
the homologous angles of
the similar triangles are
equal ; and any two corre-
sponding angles of the poly-
gons are either homologous angles of two similar triangles,
or sums of homologous angles of two or more similar trian-
gles. Thus, B = B\; BCD = BCA -f ACD = B'C'A' +
A'C'D'=B'C'D' ; etc.

2d. Their homologous sides are proportional. For, from
the similar triangles, we have

AB ^ BG ^ AC ^ CD ^ AD ^ DE ^
A'B' B'C A'C CD' A'D' D'E' ^ ^'
Therefore the polygons fulfil the two conditions of similarity
(16). \y—

PEOPOSITION VII.— THEOREM.

23. Conversely, two similar polygons may be decomposed into
the same number of triangles similar each to each and similarly
placed.

Let ABCD, etc., A'B'C'D', etc., be two similar polygons.
From two homologous ver-
tices, A and A', let diag-
onals be drawn in each
polygon; then the poly-
gons will be decomposed
as required.

For, 1st. We have, by the definition of similar polygons,

AB BC

Angle B = B\ and

A'B' B'C

BOOK III. 115

therefore the triangles J.5(7and A^B'C are similar (Proposi-
tion lY.).

2d. Since ABC and A!B'G' are similar, the angles BGA and
B'C'A! are equal; subtracting these equals from the equals
BCD and B'C'I)\ respectively, there remain the equals ACJ)
and A' CD', Also, from the similarity of the triangles ABC
and A'B'C, and from that of the polygons, we have

AC BC CD

A'C B'C CD''

therefore the triangles ACD and A' CD' are similar (Propo-
sition lY.).

Thus, successively, each triangle of one polygon may be
shown to be similar to the triangle similarly situated in the
other.

PKOPOSITION VIII.— THEOREM.

24. The 'perimeters of two similar polygons are in the same
ratio as any two homologous sides.

For we have (see preceding figures)

^^ ^^ ^^ etc.;

A'B' B'C CD'
whence (12)

AB -{- BC + CD -{- etc. ^ AB ^ BC
A'B' + B'C+ CD' + etc. A'B' B'C

etc.

.1^^

116 ELEMENTS OF GEOMETRY.

PROPOSITION IX.— THEOREM.

25. If a perpendicular is drawn from the vertex of the right
angle to the hypotenuse of a right triangle :

1st. The two triangles thus formed are similar to each other
and to the whole triangle ;

2d. The perpendicular is a mean proportional between the seg-
ments of the hypotenuse ;

3d. Each side about the right angle is a mean proportional
between the hypotenuse and the adjacent segment.

Let G be the right angle of the triangle
ABC, and CD the perpendicular to the
hypotenuse; then,

1st. The triangles ACD and CBD are
similar to each other and to ABC. For the triangles ACD
and ABC have the angle A common, and the right angles
ADCj ACBj equal; therefore they are mutually equiangular,
and consequently similar (Proposition III.). For a like
reason CBD is similar to ABCj and consequently also to
ACD.

2d. The perpendicular CD is a mean proportional between

the segments AD and DB. For the similar triangles ACD,

CBD, give

AD:CD = CD : BD.

3d. The side J. (7 is a mean proportional between the hy-
potenuse AB and the adjacent segment AD. For the similar
triangles, ACD, ABC, give

AB:AC = AC:AD.

In the same way the triangles CBD and ABC give

AB:BC = BC :BD.

BOOK III. 117

26. Scholium. If the lengths of the lines in the figure
above are expressed in terms of the same unit, the results
just obtained may be written (5, Corollary)

Vlf =ABX I>B,
TO' = AB X AD,
'mJ' = AB X BD.

27. Corollary. If from any point in the circumference of a
circle a perpendicular is let fall upon a diameter, the perpen-
dicular is a mean proportional between the segments of the diam-
eter.

Suggestion. Draw the chords A C and CB,
(v. 11., Proposition XIY., Corollary.)

'< ' PROPOSITION X.— THEOREM.

28. The square of the length of the hypotenuse of a right
triangle is the sum of the squares of the lengths of the other two
sides, the three lengths being expressed in terms of the same unit.

Let ABC be right angled at C; then

Xg' = TO' + BU'.
For, by Proposition IX., we have

TU' = AB X AD, and 277' = AB X BD,
the sum of which is

■AJ]' + 'mj'=.AB X {AD ^ BD) = AB X AB = A^.

29. Scholium I. By this theorem, if the numerical measures
of two sides of a right triangle are given, that of the third
is found. For example, if ^(7= 3, BO = 4; then AB =
1/ [3^ + 4^ = 5.

If the hypotenuse, AB, and one side, AC, are given, we
have BC^ = TB'' — X0\' thus, if there are given AB = b,
AC = 3, then we find BO = ^^[5^ — 3'] = 4.

118

ELEMENTS OF GEOMETRY.

30. Scholium II. If J. 0 is the diagonal of a square ABCD,
we have, by the preceding theorem,

whence

and, extracting the square root,

-j£ = \/^= 1.41421 + ad inf.

Since the square root of 2 is an incommensurable number, it
follows that the diagonal of a square is incommensurable with
its side. (v. II., 34.)

31. Definition. The projection of a point
A upon an indefinite straight line XY is
the foot P of the perpendicular let fall
from the point upon the line.

The projection of a finite straight line AB upon the line XY
is the distance FQ between the projections of the extremities
oi AB.

If one extremity B of the line AB is in

the line- XY^ the distance from B to P (the
projection of A) is the projection of AB on
XY; for the point B is in this case its own
projectioik

PM,

EXERCISES.

1. Theorem. — In any triangle, the square of the side opposite
to an acute angle is equal to the sum of the squares of the other
two sides diminished by twice the product of one of these sides
^dnd the projection of the other upon that side.

BOOK III.

119

PCy, Fig. 1, or

YiQ. 2.

Suggestion. Let (7 be the acute angle in question in Fig. 1
or Fig 2.

A Fig. 1.

== XF' + (5(7
-^'^ XF' + (PC

== 2T^ + TO' + :BT' - 2BC X Pc;
= Xa' + IB^C — 2BC X PC,

2. Theorem. — J/i an obtuse angled triangle,
the square of the side opposite to the obtuse
angle is equal to the sum of the squares of the
other two sides, increased by twice the product
of one of these sides and the projection of the
other upon that side.

PROPOSITION XI.— THEOREM.

32. If two chords intersect within a circle, their segments are
reciprocally proportional.

For the triangles APB' and A'PB are mu-
tually equiangular (II., Proposition XIY.),
and therefore similar (Proposition III.).

Hence

AP : A'P = PB' : PB.

33. Scholium. If the lengths of the lines in

question are expressed in terms of the same unit, the result
above can be written

APXPB= A'P X PB',

and the proposition may be stated : if through a fixed point
within a circle any chord is drawn, the product of the lengths of
its segments is the same whatever its direction.

120

ELEMENTS OF GEOMETRY.

EXERCISE.

Theorem. — Either segment of the least chord
that can he drawn through a fixed point is a
mean proportional between the segments of any
other chord drawn through that point, (v.
II., 19, Exercise.)

PROPOSITION XII.— THEOREM.

34. If two secants intersect without a circle, the whole secants
and their external segments are reciprocally proportional.

For the triangles PAB^ and PA'B are mu-
tually equiangular (II., Proposition XIY.),
and therefore similar. Consequently

PB : PB' = PA' : PA.

35. Corollary. If a tangent and a secant
intersect, the tangent is a mean proportional
between the whole secant and its external seg-
ment.

Suggestion. Show that the triangles PA T
and PTB are similar.

36. Scholium. If the lengths of the lines
are expressed in terms of the same unit,
the result of (34) can be written PB X PA

= PB' X PA\ and Proposition XII. can be stated : if through
a fixed point without a circle a secant is drawn, the product of
the lengths of the whole secant and its external segment has the
same value in whatever direction the secant is drawn.

BOOK III.

121

EXERCISE.

Theorem. — If from any point on the common chord of two
intersecting circles^ produced^ tangents are drawn to the two
circles^ the lengths of these tangents are equal.

PEOBLEMS OF CONSTEUCTIOK
PROPOSITION XIII.— PROBLEM.

37. To divide a given straight line into any given number of
equal parts.

Let AB be the given line. Through A draw
an indefinite line AX^ upon which lay off the
given number of equal distances, each distance
being of any convenient length ; through M the
last point of division on AX draw MB^ and
through the other points of division of AX
draw parallels to MB^ which will divide AB
into the required number of equal parts. This
follows from the first part of the proof of Proposition I.

PROPOSITION XIV.— PROBLEM.

38. To divide a given straight line into parts proportional to
two given straight lines.

Let it be required to divide AB
into parts proportional to M and N,
From A draw the indefinite line AX^
upon which lay off AG = M and CD
= N. Join BB^ and through G draw
GE parallel to DB. Then we shall
have AE :JEB=AC:GD, by Propo-
sition I.

122

ELEMENTS OF GEOMETRY.

EXERCISE.

Problem. — To divide a given straight line
into parts proportional to given straight
lines.

...•'5

M N

PROPOSITION XV.— PROBLEM.

39. To find a fourth proportional to three given straight lines.

Let it be required to find a fourth pro-
portional to M, JV, and P. Draw the in-
definite lines AJT, AY, making an angle
with each other. Upon AJC lay off AB
= M, AD = N; and upon ^ Ylay off ^(7
= P; join BG, and draw BE parallel to
BC ; then AEi^ the required fourth pro-
portional.

For we have (Proposition I.)

AB : AD=AG: AE, or M : W= P : AE.

EXERCISE.

Prohlem.-
lines.

To find a third proportional to tvx) given straight

PROPOSITION XVI.— PROBLEM.

40. To find a mean proportional between two given straight
lines.

Let it be required to find a mean pro-
portional between M and N. Upon an
indefinite line lay ofl AB=M,BG = JSF;
upon AG describe a semi-circumference,
and at B erect a perpendicular, BD, to
AG. Then BD is the required mean proportional (Proposi-
tion IX., Corollary).

BOOK III.

123

41. Definition. When a given straight line is divided into
two segments such that one of the segments is a mean pro-
portional between the given line and the other segment, it is
said to be divided in extreme and mean ratio.

Thus, AB is divided in extreme and mean ratio ^

H — I

at C,ii AB\AC=:AG : CB.

PROPOSITION XVII.— PROBLEM.

42. To divide a given straight line in extreme and mean ratio.

Let AB be the given straight line. At B erect the per-
pendicular BO equal to one-half of AB. With the centre 0
and radius OB^ describe a circumference,
and through A and 0 draw AO cutting
the circumference first in D and a second
time in D'. Upon AB lay ofl AG == AD.
Then AB is divided at C in extreme and
mean ratio.

For we have (Proposition XIL, Corollary) ^

^J • ^^^
AD' :AB=AB:AD or AC, [1]

whence, by division (10),

AD' — AB:AB = AB — AC:Aa,

or, since DD' =20B = AB, and therefore AD' — AB = AD^
— DD' = AD=AC,

AO:AB = CB:AC,
and, by inversion (7),

AB:AC = AC:CB;
that is, AB is divided at C in extreme and mean ratia

124

ELEMENTS OF GEOMETRY.

PROPOSITION XVIII.— PROBLEM.

43. On a given straight line^ to construct a polygon similar to
a given polygon.

Let it be required to construct upon A'B' a polygon similar
to ABCDEF,

Divide AB CDEF mio tri-
angles by diagonals drawn
from A. Make the angles
B'A'C and A'B'C equal
to BAG and ABC respec-
tively; then the triangle

A'B'C will be similar to ABC (Proposition III.).
same manner construct the triangle A'D'C similar to ADC,
A' WD' similar to AED, and A'E'F' similar to AEF. Then
AIB'C'D'E'F' is the required polygon (Proposition YL).

In the

EXERCISES ON BOOK III.

THEOREMS.

1. If two straight lines are intersected by any number of par-
allel lines, the corresponding segments of the two lines are pro-
portional, {v. Proposition I.)

2. The diagonals of a trapezoid divide each other into segments
which are proportional.

3. In a triangle any two sides are reciprocally proportional to
the perpendiculars let fall upon them from the opposite vertices.

4. The perpendiculars from two vertices of a triangle upon the
opposite sides divide each other into segments which are recipro*
cally proportional.

6. If the three sides of a triangle are respectively perpendicular
to the three sides of a second triangle, the triangles are similar.

6. If ABC and A^BO are two triangles Laving
a common base BC and their vertices in a line
AA^ parallel to the base, and if any parallel to
the base cuts the sides AB and ^ C in Z> and jE7,
and the sides A^B and A^C in D^ and E^^ then
DE=^J)'E' (Proposition III.).
""7. If two sides of a triangle are divided propor-
tionally, the straight lines drawn from correspond-
ing points of section to the opposite angles intersect
on the line joining the vertex of the third angle
and the middle of the third side.

Suggestion. Draw the line ADE through the in-
tersection of B^C and BC^. B'E'D and CED are

similar ;

similar

BC ^
B'C

Hence

EC
B'E'

DC
B'D

DC
B'D
BC

B'DC^ and BDC are

^ AB^C^ and ABC are similar

AB''

B'C

AB'E' and ABE are similar ;
BE

AB
AB'

BE
B'E'

B'E' B'E'

and BE = EC,

11* 125

126

ELEMENTS OF GEOMETRY.

8. The difference of the squares of two sides of any triangle is
equal to the difference of the squares of the projections of these
j--^ sides on the third side (Proposition X.).

^ ^ 9. J[f from any point in the plane df a polj[gon\peitoendiculars

^"^^ -^alfe^rawn to all the sides, the two siiml of\the^qu\res of the
alternate segments of the sides are equkl. ^

\J 10. If through a point P in the circum-
ference of a circle two chords are drawn,
the chords and the segments cut from them
by a line parallel to the tangent at P are re-
ciprocally proportional.
Suggestion, Prove PAB and Pba similar.

-^ -41

11. If three circles int^fSect, their three
common chords pass through the same
point, {v. Proposition XI.)

12. If two tangents are drawn to a circle
at the extremities of a diameter, the por-
tion of any third tangent intercepted be-
tween them is divided at its point of con-
tact into segments w^hose product is equal
to the square of the radius.

Suggestion. Prove AOB a right triangle.

13. The perpendicular from any point of a cir-
cumference upon a chord is a mean proportional
between the perpendiculars from the same point
upon the tangents drawn at the extremities of
the chord.

Suggestion. PBD and PAE are similar ; . • .
PB^PD
PA PE

PCE and PAD are similar

PA
PC

PD
PE'

Hence ^ =
PA

PA
PO'

BOOK III.

127

14. If two circles touch each other, secants drawn through their
point of contact and terminating in the two circumferences are
divided proportionally at the point of contact, {v. II., 54, Exer-
cise 2.) ^ - ' ^

15. If two circles are tangent exter-
nally, the portion of their common
tangent included between the points
of contact is a mean proportional be-
tween the diameters of the circles.

Suggestion. Show that OBO^ is a right
triangle.

16. If a fixed circumference is cut by any circumference which
passes through two fixed points, the com-
mon chord passes through a fixed point.

Suggestion. PA . PB = PC. PD = PT^
by Proposition XII. and Corollary. Join P
with (7^, and show that PC^ will cut both
circles at the same distance from P, and
will be their common chord. "^ ''-::.-.-.-...-=''£ ^

^f :z X

LOCI.

17. From a fixed point O, a straight line OA is
drawn to any point in a given straight line JJfJV,
and divided at P in a given ratio m : n (ie, so
that OP:PA='m:n); find the locus of P. (v.
Proposition II.)

18. From a fixed point O, a straight line
OA is drawn to any i:)oint in a given cir-
cumference, and divided at P in a given
ratio ; find the locus of P.

Suggestion. PC is a fixed length.

19. Find the locus of a point whose distances from two given
straight lines are in a given ratio. (The locus consists of two
straight lines.)

^^r

l^i^r-*^ ELEMENTS OF GEOMETRY. /

yy- 20. Find the locus of the points which divide the various chords
of a given circle into segments whose product is equal to a given
constant, k'^ (33, Exercise).

21. Find the locus of a point the sum of whose distances from
two given straight lines is equal to a given constant, k. {v, I.,
Exercise 10.)

22. Find the locus of a point the difference of whose distances
from two given straight lines is equal to a given constant, k.

Suggestion, Reduce it to I., Proposition XIX., by drawing a
third line parallel to the more distant of the given lines at a
distance from it equal to k,

^^ PROBLEMS.

23. To divide a given straight line into three segments. A, B,
and C, such that A and B shall be in the ratio of two given
straight lines M and iV, and B and C shall be in the ratio of two
other given straight lines P and Q.

■24. Through a given point, to draw a straight line so that the
portion of it intercepted between two given straight lines shall
be divided at the point in a given ratio.

Suggestion. Through the point draw a line parallel to one of
the given lines, {v. II., Exercise 32.)

25. Through a given point, to draw a straight line so that the
distances from two other given points to this line shall be in a
given ratio.
Suggestion. Divide the line joining the two other given pointe

26. To determine a point whose distances from three given in-
deiinite straight lines shall be proportional to three given straight
lines. (Exercise 19.)

4LJ' A

'h^sq

27. In a given triangle ABC, to inscribe a
square DEFQ. (Exercises 6 and II., 29.)

i:r.~

BOOK m.

-Hr

129

^ 28. Griven two circumferences inter-

secmng in -4, to draw through A a
i l^ant, JSACj such that AB shall be

TO AO in Si given ratio.

Suggestion. Divide 00^ in the given

ratio, (v. Exercise 1.)

29. To aescribe a circumference passing through two given
points A and B and tangent to a
given circumference O.

Analysis. Suppose ATB is the re-
guired circumference tangent to the
given circumference at T, and ACDB
any circumference passing through A
and B and cutting the given circum-
ference in C and D. The common
chords AB and CD^ and the common
tangent at T^ all pass through a com-
mon point P (Exercise 16) ; from

which a simple construction may be inferred. There are two
solutions given by the two tangents that can be drawn from P.

<| 30. To describe a circumference passing through two given
^ points and tangent to a given straight line. (Two solutions.)
{v. Proposition XII., Corollary.)

31. To describe a circumference passing through a given point
and tangent to two given straight lines, {v. Exercise 13.)

-'lY,'//.

<^>^xjeJ?i

>^ir

BOO

COMPARISON AND MEASUREMENT OP THE
SURFACES OP RECTILINEAR PIGURES.

1. Definition. The area of a surface is its numerical meas-
ure, referred to some other surface as the unit ; in other words,
it is the ratio of the surface to the unit of surface (II., 29).

The unit of surface is called the superficial unit. The most
convenient superficial unit is the square whose side is the
linear unit.

2. Definition. Equivalent figures are those whose areas are

equal.

PROPOSITION I.— THEOREM.

3. Parallelograms having equal bases and equal altitudes are
equivalent.

Let ABCD and AJECF be two
parallelograms having equal bases
and equal altitudes. ^ ^

Superpose the second upon the
first, making the equal bases coincide. Since the altitudes
are equal, the upper bases will lie in the same straight line.
The triangles ABE and CDF are equal (I., Proposition YL).
If the triangle CDF is taken from the whole figure, ABFC,
the first parallelogram ABCD is left; if the equal triangle
ABE is taken from the same figure, the second parallelogram
AECF is left. The magnitudes of the two parallelograms
are therefore equal, and the parallelograms are equivalent.

4. Corollary. Any parallelogram is equivalent to a rectangle
having the same base and the same altitude.

130

BOOK IV.

131

PROPOSITION II.— THEOREM.
5. Two rectangles having equal altitudes are to each other as
their bases.

D O D F

Let ABCD and AEFD be two
rectangles having equal altitudes ;
then are they to each other as
AB : AE.

1. Suppose the bases have a common measure which is
contained m times in AB and n times in AK Then we have

AB
AE

Apply this measure to the two bases, and through the points
of division draw perpendiculars to the bases. The two rec-
tangles are thus divided into smaller rectangles, all of which
are equal, by I., Proposition XXYIIL, Corollary, and of
which ABCD contains m and AEFD contains n. Then

ABOD m

and consequently

AEFD

ABCD ^
AEFD~

AB
AE'

c c

2. If the bases are incommensurable, divide AE in any
arbitrarily chosen number n of
equal parts, and apply one of
these parts to AB. Let B' be the
last point of division, B'B being
of course less than the divisor.

Since AB' and AE are commensurable, we have =

AEFD

AB'

£f B

, and this holds, no matter what the value of n. If, now,

132

ELEMENTS OF GEOMETRY.

C O

n is increased at pleasure, we can make J5'J5, and consequently
B'BGG'j as small as we please, but
cannot make them absolutely zero.
Hence, as n is indefinitely in-
creased, AB' has AB for its limit,
AB'C'D has ABGD for its Hmit,

Bf B

has . for its limit.

and

AEFD
AB'

AEFD

, „ has — — for its limit.
AE AE

Therefore, by II., Theorem^ Doctrine of Limits,

ABCD AB

AEFD

^. (^v. IL, 42, and III., 14.)

6. Corollary. Two rectangles having equal bases are to each
other as their altitudes.

Note. In these propositions, by " rectangle" is to be under-
stood " surface of the rectangle."

v^Vfiv^jfi^^-^

V PBOPOSITION III.— THEOKEM.

7. Any two rectangles are to each other as the products of their
bases by their altitudes.

Let B and i2' be two rec-
tangles, k and k their bases,
h and h' their altitudes ; then

kXh
J<fXh''

For, let >S^ be a third rectan-
gle, having the same base k

BOOK IV.

133

as the rectangle -B, and the same altitude h! as the rectangle
R' ; then we have, by Proposition II., Corollary, and Propo-
sition II.,

S h" R' k"
and multiplying these ratios, we find

Ji'X h''

8. Scholium. It must be remembered that by the product
of two lines is to be understood the product of the numbers
which represent them when they are measured by the linear
unit (III., 8).

PEOPOSITION IV.— THEOREM.

9. The area of a rectangle is equal to the product of its base
and altitude.

Let R be any rectangle, k its base,
and h its altitude numerically ex-
pressed in terms of the linear unit ;
and let Q be the square whose side
is the linear unit ; then, by Proposition III.,

R _k X h

Q IX 1

= k X h.

But since Q is the unit of surface, - = the numerical meas-
ure, or area, of the rectangle, R, (1) ; therefore

Area of R = k X h.

134 ELEMENTS OF GEOMETRY.

10. Scholium I. When the base and altitude are exactly
divisible by the linear unit, this proposition is rendered evi-
dent by dividing the rectangle into squares

each equal to the superficial unit. Thus, if
the base contains 7 linear units and the alti-
tude 5, the rectangle can obviously be divided
into 35 squares each equal to the superficial
unit ; that is, its area = 5x7. The propo-
sition, as above demonstrated, is, however, more general,
and includes also the cases in which either the base or the
altitude, or both, are incommensurable with the unit of
length.

11. Scholium II. The area of a square, being the product
of two equal sides, is the second power of a side. Hence it is
that in arithmetic and algebra the expression " square of a
number" has been adopted to signify "second power of a
number."

We may also here observe that many writers employ the
expression "rectangle of two lines" in the sense of "product
of two lines," because the rectangle constructed upon two
lines is measured by the product of the numerical measures
of the lines.

PROPOSITION v.— THEOREM.

12. The area of a parallelogram is equal to the product of its
base and altitude.

For, by Proposition I., the parallelogram is equivalent to a
rectangle having the same base and the same altitude.

BOOK IV. { I '^ < / ' 135
PROPOSITION VI.— THEOREM.

13. The area of a triangle is equal to half the product of its
base and altitude.

Let ABC be a triangle, k the numerical f- --^

measure of its base BC^h that of its alti- •'' ^^^^"^^^ / •*
tude AB^ and 8 its area \ then B^~h o""d

S=ikXh.

For, through A draw AE parallel to CB, and through B draw
B£J parallel to CA. The triangle ABC is one-half the paral-
lelogram AEBC (I., Proposition IX.); but the area of the
parallelogram = k y^ h ; therefore, for the triangle, we have
S=lk X h.
/ 14. Corollary I. A triangle is equivalent to one-half of any

i^parallelogram having the same base and the same altitude.

J 15, Corollary II. Triangles having equal bases and equal

\laltitudes are equivalent.

16. CoROLLA'feY III. Triangles having equal altitudes are to
each other as their bases ; and triangles having equal bases are
to each other as their altitudes.

X PROPOSITION VII.— THEOREM.

17. The area of a trapezoid is equal to the product of its alti^

tude by half the sum of its parallel bases.

Let ABCD be a trapezoid ; MN= h, its
altitude ; AD = a, BC = b, its parallel
bases ; and let S denote its area ; then

S=i(^a + b)X h.

136

ELEMENTS OF GEOMETRY.

A M

For, draw the diagonal AC. The altitude of each of the
triangles ABC and ABC is equal to h, and
their bases are respectively a and b ; the
area of the first is ^ a y^ h^ that of the
second is ^ 6 X ^ ; and the trapezoid being
the sum of the two triangles, we have b n o

S=iaXh-]-ibXh = i(a + b)Xh.

18. Scholium. The area of any polygon may be found by
finding the areas of the several triangles into which it may
be decomposed by drawing diagonals from any vertex.

The following method, however, is usually preferred, espe-
cially in surveying. Draw the longest
diagonal AD of the proposed polygon
ABCDEF; and upon AD let fall the
perpendiculars BM^ CW, UP, FQ.
The polygon is thus decomposed into
right triangles and right trapezoids,
and by measuring the lengths of the

perpendiculars and also of the distances AM, MN^ ND, AQ,
QP, PD, the bases and altitudes of these triangles and trape-
zoids are known. Hence their areas can be computed by the
preceding theorems, and the sum of these areas will be the
area of the polygon.

i PROPOSITION VIII.— THEOREM.

■' 19. Similar triangles are to each other as the squares of their

homologous sides.

Let ABC, A'B'C, be similar tri-
angles; then

ABC ^ ZB^

BOOK IV.
Let AD and A'D' be the altitudes ; then

137

ABC _ ^AD X BC _ 4^ BC

A'B'C ^A'D' X B'C A'D' '^ B'C

But the triangles ABB and A'D'B' are similar (III., Propo
sition III.) ] therefore

AB AB

A'D' A'B"

and from the similarity of ABC and A'B'G\

BC AB

hence

and we have

B'C A'B"

AB BC ^ TB''
A'B' ^ B'G' A^''

ABC ^ X5'

EXERCISE.

\ I

^ Theorem. — Ti^o triangles having an angle of the one equal to
an angle of the other are to each other as the products of the ^
sides including the equal angles. L^

Suggestion. Let ABE and ABC be the two
triangles. Draw BE^ and compare the two
triangles with AEB. (v. Proposition YI.,
Corollary III.)

12»

138 ELEMENTS OF GEOMETRY.

PROPOSITION IX.— THEOREM.

20. Similar polygons are to each other as the squares of their
homologous sides.

Let ABCDEF, A' B' C D' E' F' , be two similar polygons,
and denote their surfaces
by /Sand >S"; then b ^ ^ c' •

For, let the polygons be p

decomposed into homolo-
gous similar triangles (III., Proposition YII.). The ratio of
the surfaces of any pair of homologous triangles, as ABC and
A'B'C, ACD and A' CD', etc., will be the square of the ratio
of two homologous sides of the polygons (Proposition YIII.) ;
therefore we shall have

ABO ACD ADE AEF Tff

A'B'C A' CD' A'D'E' A'E'F' Z^'

Therefore, by addition of antecedents and consequents (III.,
12).

ABC + ACD + ADE + AEF ^S^ Z^
A'B'C + A' CD' + A'D'E' + A'E'F' S' ATB"^'

PROPOSITION X.— THEOREM.

21. The square described upon the hypotenuse of a right tri-
angle is equivalent to the sum of the squares described on the
other two sides.

BOOK IV.

139

Let the triangle ABC be right angled at C; then the
square AIT, described upon the
hypotenuse, is equal in area to
the sum of the squares AF and
£D, described on the other two
sides.

For, from C draw CP perpen-
dicular to AB and produce it to
meet KH in Z. Join CK, BG.
Since ACF and ACB are right
angles, CF and CB are in the
same straight line (I., Proposition
lY.) ; and for a similar reason AG and CD are in the same
straight line.

In the triangles OAK, GAB, we have AK equal to AB,
being sides of the same square; AC equal to AG, for the
same reason; and the angles CAK, GAB, equal, being each
equal to the sum of the angle GAB and a right angle; there-
fore these triangles are equal (L, Proposition VI.).

The triangle CAK and the rectangle AL have the same
base AK; and since the vertex C is upon LP produced, they
also have the same altitude; therefore the triangle CAK is
equivalent to one-half the rectangle AL (Proposition VI
Corollary I.). '

The triangle GAB and the square AF have the same base
AG; and smce the vertex B is upon FG produced, they also
have the same altitude ; therefore the triangle GAB is equiva-
lent to one-half the square AF (Proposition VI., Corollary I )

But the triangles CAK, GAB, have been shown to be equal •
therefore the rectangle AL is equivalent to the square AF

In the same way it is proved that the rectangle BL is
equivalent to the square BD.

140

ELEMENTS OF GEOMETRY.

Therefore the square AH^ which is the sum of the rec-
tangles AL and BL, is equivalent
to the sum of the squares AF and
BD.

22. Scholium. This theorem is
ascribed to Pythagoras (born
about 600 B.C.), and is commonly
called the Pythagorean Theorem.
The preceding demonstration of
it is that which was given by
Euclid, in his Elements (about
300 B.C.).

It is important to observe that we may deduce the same
result from the numerical relation XS^ ==:AC^ -\- ^FU^ already
established in III., Proposition X. For, since the measure
of the area of a square is the second power of the number
which represents its side, it follows directly from this numer-
ical relation that the area of which AB^ is the measure is
equal to the sum of the areas of which AC and BTf are the
measures.

EXERCISES.

1. Theorem. — If the three sides of a right triangle be taken as
the homologous sides of three similar polygons. constructed upon
them, then the polygon constructed upon the hypotenuse is equiva-
lent to the sum of the polygons constructed upon the other two
sides, (v. Proposition IX.)

2. Theorem. — The squares on the sides of a right triangle are
proportional to the segments into which the hypotenuse is divided
by a perpendicular let fall from the vertex of the right angle,
(v. Figure, Proposition X.)

BOOK IV. 141

PEOBLEMS OF C0:N^STEUCTI0K.
PROPOSITION XI.— PROBLEM.

23. To construct a triangle equivalent to 4: given polygon.

Let ABCDEF be the given polygon.
' Take any three consecutive vertices,
as J., 5, (7, and draw the diagonal AC'
Through B draw BP parallel to AO
Wk meeting DC produced in P; join AP.

The triangles APG, ABC, have the
same base AG ; and since their vertices,

P and 5, lie on the same straight line BP parallel to AG^
they also have the same altitude ; therefore they are equiva-
lent. Therefore the pentagon APDEF is equivalent to the
hexagon ABGDEF. Now, taking any three consecutive ver-
tices of this pentagon, we shall, by a precisely similar con-
struction, find a quadrilateral of the same area ; and, finally,
by a similar operation upon the quadrilateral we shall find a
triangle of the same area.

Thus, whatever the number of the sides of the given poly-
gon, a series of successive steps, each step reducing the num-
ber of sides by one, will give a series of polygons of equal
areas, terminating in a triangle.

PROPOSITION XII.— PROBLEM.

24. To construct a square equivalent to a given parallelogram
or to a given triangle. ^ ^

1st. Let J. (7 be a given parallelogram, k its \ ^i \

base, and h its altitude.

Find a mean proportional x between h and j y

k, by III., 40. The square constructed upon

X will be equivalent to the parallelogram, since x^ = h y^ k.

142

ELEMENTS OF GEOMETRY.

2d. Let ABC be a given triangle, a its base, and h its
altitude.

Find a mean proportional x between a
and ih; the square constructed upon x
will be equivalent to the triangle, since ^ a u

= a X i h = i ah. I 1

25. Scholium. By means of this problem
and the preceding, a square can be found equivalent to any-
given polygon.

Pi-

PEOPOSITION XIII.-PROBLEM.

26. To construct a square equivalent to the sum of two or more
given squares, or to the difference of two given squares.

1st. Let m, w, ^, q, be the sides of given
squares.

Draw AB = m, and BC = n, perpendic-
ular to each other at B; join AC. Then
(Proposition X.) ZU^ ^ m^ -\- n\

Draw CD = p perpendicular to A (7, and
join AD. Then AJf = TU' + p' = m' +
n^ + _p^

Draw DB = q perpendicular to AD, and
join AK Then TE^ = Jlf -\- q' = m,' +
n^ _|_ p^ _|_ ^2 Therefore the square con-
structed upon AE will be equivalent to the sum of the
squares constructed upon m, 7i, p, q.

In this manner may the areas of any number of given
squares be added.

2d. Construct a right angle ABC, and lay
off BA = n. With the centre A and a radius
= m, describe an arc cutting BCinC. Then

BOOK IV.

143

:ro' = zu^ - zs'

therefore the square con-

structed upon BC will be equivalent to the difference of the
squares constructed upon m and n.

EXERCISE.

Prohlem.^Upon a given straight line, to construct a rectangle
equivalent to a given rectangle.

\ PBOPOSITION XIV.-PROBLEM.

27. To construct a rectangle, having given its area and the
sum of two adjacent sides.

Let MN be equal to the given sum of

the adjacent sides of the required rec- ^ .^

tangle ; and let the given area be that of i,,..^.^::C:r\9
the square whose side is AB. '' ^ — ^^;--—

Upon MJSr as a diameter describe a
semicircle. At M erect MP = AB per-
pendicular to MN, and draw PQ parallel to MN, intersecting
he circumference in Q. From Q let fall QR perpendicular
to MN; then MR and RN are the base and altitude of the

PROPOSITION XV.-PEOBLEM.

28. To find two straight lines in the ratio of the areas of two
given polygons, •

Let squares be found equal in area to o

the given polygons respectively (23 and /T^v.

24). Upon the sides of the right angle '^ [ ^_j,
ACB, take CA and CB equal to the sides
of these squares, join AB, and let fall CD perpendicular to

B N

144 ELEMENTS OF GEOMETRY.

AB. Then, by (III., 26), we have lU' = AD X AB, VF^
= JDB X AS- Hence

IV ^ AD .
VB' DB'

therefore AD and DB are in the ratio of the areas of the
given polygons.

EXERCISE.

Problem. — To find a square which shall he to a given square
in the ratio of two given straight lines, (y. 28.)

PROPOSITION XVli— PROBLEM.X)^

Q/n

29. To construct a polygon similar to a given polygon P and
equivalent to a given polygon Q.

Find M and iV, the sides of

squares respectively equal in area

to P and 0 (23 and 24).

I ^ 1 I ^ 1
Let a be any side of P, and find

a fourth proportional a' to Mj i^T, \ p' |

and a; upon a', as a homologous ^ — -f — '

side to a, construct the polygon P'

similar to P; this will be the required polygon. For, by

construction,

M_a

jsr a"

therefore, taking the letters P, Q, and P', to den0te the areas
of the polygons,

P^3P^a^

Q W' a'' '

BOOK IV.

145

But, the polygons P and P' being similar, we have, by
(Proposition IX,),

P' a'^ '

and comparing these equations, we have P' = Q.

Therefore the polygon P' is similar to the polygon P and
equivalent to the polygon Q, as required.

EXERCISE.

Problem. — To construct a polygon similar to a given polygon^
and whose area shall he in a given rhtio to that of the given
polygon, (v. 28, Exercise, an^III., 43.) ^

■J

G k ^^^

i.^

EXERCISES ON BOOK IV.

THEOREMS.

^r ^ 1. Two' triangles are equivalent if they have two sides of the

(^r'^^iiAM) one respectively equal to two sides of the other, and the included

At angle of the one equal to the supplement of the included angle

of the other.

-— ^^Vz. The two opposite triangles formed by joining any point in

the interior of a parallelogram 4o its four vertices are together

equivalent to one-half the parallelogram.

3. The triangle formed by joining the middle point of one of
the non-parallel sides of a trapezoid to the extremities of the
opposite side is equivalent to one-half the trapezoid, {v, I., Ex-
ercise 24.)

4. The figure formed by joining consecutively the four middle
points of the sides of any quadrilateral is equivalent to one-half
the quadrilateral, (v. I., Exercise 32.)

„ V. 6. If in a rectangle ABCD we draw
the diagonal AC^ inscribe a circle in the
triangle ABC^ and from its centre draw
OE and Oi^ perpendicular to AD and
DC respectively, the rectangle OD will
be equivalent to one-half the rectangle
ABCD.

C 6. The area of a triangle is equal to
one-half the product of its perimeter by
the radius of the inscribed circle.

7. The area of a rhonabus is one-half the product of the diag-
onals.

.\,w^j'«'

BOOK IV. 147

- '" 8. The straight line joining the middle points of the parallel
sides of a trapezoid divides it into two equivalent figures.

(4:'j ^. Any line drawn through the point of intersection of the diag-
onals of a parallelogram divides it into two equal quadrilaterals.

10. In an isosceles right triangle either leg is a mean propor-
tional between the hypotenuse and the perpendicular upon it
from the vertex of the right angle.

11. If two triangles have an angle in common, and have equal
areas, the sides about the equal angles are reciprocally propor-

V tional.
. .Ivt \'^' The perimeter of a triangle is to a side as the perpendicular
fronythe opposite vertex is to the radius of the inscribed circle.
! : (v.yExercise 6.)

f yi3. Two quadrilaterals are equivalent when the diagonals of ,
V one are respectively equal and parallel to the diagonals of the I
^o^her.
^ 14. The sum of the perpendiculars from any point within an

■ ♦ equilateral convex polygon upon the sides is constant.

y^ Suggestion. Join the point with the vertices of the polygon.
^ 15. The lines joining two opposite vertices of a parallelogram

^ with the middle points of the sides form a parallelogram whose
^^^area is one-third the a^ea of the given parallelogram.
A ^.^ 16. The sum of the squares on the segments of two j)erpendic-
ular chords in a circle is equivalent to the square on the diameter.
17. Let ABC be any triangle, and
"^ upon the sides AB, AC, construct
parallelograms AD, AF, of any mag-
nitude or form. Let their exterior
*^ sides DE, FG, meet in M; join MA,
^ and upon BC construct a parallelo-
gram BK, whose side BH is equal
and parallel to MA, Then the par-
allelogram BK is equivalent to the
sum of the parallelograms AD and
AF, {v. Proposition I.)
From this deduce the Pythagorean Theorem.
«y^^ V 18. Prove, geometrically, that th^ square described upon the
sum of two straight lines is equivalent to the sum of the squares
described on the two lines plits twice their rectangle.

Note. By the "rectangle of two lines" is here meant the rec-
tangle of which the two lines are the adjacent sides.
QV' 19. Prove, geometrically, that the square described upon the
difference of two straight lines is equivalent to the sum of the
squares described on the two lines minus twice their rectangle.

ELEMENTS OP GEOMETRY.

20. Prove, geometrically, that the rectangle of the sum and the

difference of two straight lines is equivalent to the difference of

the squares of those lines.

PROBLEMS.

21. To construct a triangle, given its angles and its area (eq
to that of a given square).

Suggestion. Construct any triangle having the given angles.
The problem then reduces to (29).

22. Given any triangle, to construct an isosceles triangle of the
same area, whose vertical angle is an angle of the given triangle.
{v. 19, Exercise.)

23. Given any triangle, to construct an equilateral triangle of
he same area. {v. Exercise 21.)

24. Bisect a given triangle by a parallel to one of its sides, (v.
Proposition VIII. and 28.)

25. Bisect a triangle by a straight line drawn through a given
point in one of its sides, (v. 19, Exercise.)

26. Inscribe a rectangle of a given area in a given circle.
Suggestion. Draw a diagonal of the rectangle. The problem

can then be reduced to inscribing in the given circle a right

;riangle of given area.
27. Given three points, A, B^ and (7, to find a fourth point P,
such that the areas of the triangles APB, APC, BPG, shall be
equal. (Four solutions.) (v. III., Exercise 19.)

;^*(..«'

' BOOK T.

REGULAR POLYGONS. MEASUREMENT ®P THE

CIRCLE.

1. Definition. A regular polygon is a polygon which is at
once equilateral and equiangular.

The equilateral triangle and the square are simple exam-
ples of regular polygons. The following theorem establishes
the possibility of regular polygons of any number of sides.

PROPOSITION I.— THEOREM.

2. If the circumference of a circle be divided into any number
of equal parts, the chords joining the successive points of division
form a regular polygon inscribed in the circle ; and the tangents
drawn at the points of division form a regular polygon circum^
scribed about the circle.

Let the circumference be divided into
the equal arcs AB, BC, GD^ etc. ; then,
1st, drawing the chords AB, BC, CD,
etc., ABGD, etc., is a regular inscribed
polygon. For its sides are equal, being
chords of equal arcs ; and its angles are
equal, being inscribed in equal segments.

2d. Drawing tangents at A, B, C, etc., the polygon GHK^
etc., is a regular circumscribed polygon. For, in the triangles
AGB, BHC, CKD, etc., we have AB = BC = CD, etc., and
the angles GAB, GBA, HBC, HOB, etc., are equal, since each

13* 149

150

ELEMENTS OF GEOMETRY.

is formed by a tangent and chord and is measured by half
of one of the equal parts of the circum-
ference (II., Proposition XY.) ; therefore
these triangles are all isosceles and equal
to each other. Hence we have the an-
gles a = H= K, etc., and AG = GB
=.BH = HG = CK, etc., from which,
by the addition of equals, it follows that
GH = HK, etc.

3. Corollary I. If the vertices of a regular inscribed poly-
gon are joined with the middle points of the arcs subtended by the
sides of the polygon, the joining lines will form a regular inscribed
polygon of double the number of sides.

4. Corollary II. If at the middle points of the arcs joining
adjacent points of contact of the sides of a regular circumscribed
polygon tangents are drawn, a regular circumscribed polygon of
double the number of sides will be formed.

5. Scholium. It is evident that the area of an inscribed
polygon is less than that of the inscribed polygon of double
the number of sides ; and the area of a circumscribed polygon
is greater than that of the circumscribed polygon of double
the number of sides.

EXERCISE.

Theorem. — If a regular polygon is inscribed in a circle, the
tangents drawn at the middle points of
the arcs subtended by the sides of the
inscribed polygon form a circumscribed
regular polygon, whose sides are par-
allel to the sides of the inscribed poly-
gon, and whose vertices lie on the radii
drawn to the vertices of the inscribed
polygon.

A'{

""~--^A

e/7 \

V"^'-^

^^\/

BOOK V. 151

PROPOSITION II.— THEOREM.

6. A circle may be circumscribed about any regular polygon ;
and a circle may also be inscribed in it.

Let ABCD ... be a regular polygon.
Through A' and B\ the middle points
of AB and BG^ draw perpendiculars,
and connect 0, their point of intersec-
tion, with all the vertices of the poly-
gon and with the middle points of all
the sides.

The triangles OA'B and OB'B are equal, by I., Proposition
X. OB'B and OB'G are equal, by I., Proposition YI. The
angle OBB' is one-half of ABG ; . • . 005' is one-half of the
equal angle BGD. Hence the triangles OB'G and OGG' are
equal, by I., Proposition YI. By continuing this process we
may prove all the small triangles equal. 0, then, is equidis-
tant from all the vertices, and therefore with 0 as a centre a
circle may be circumscribed about the polygon. 0 is also
equidistant from all the sides, and therefore with 0 as a
centre a circle may be inscribed in the polygon.

7. Definitions. The centre of a regular polygon is the common
centre, 0, of the circumscribed and in-
scribed circles.

The radius of a regular polygon is
the radius, OJ., of the circumscribed
circle.

The apothem is the radius, O-ff, of
the inscribed circle.

The angle at the centre is the angle, AOB, formed by radii
drawn to the extremities of any side.

152

ELEMENTS OF GEOMETRY.

8. The angle at the centre is equal to four right angles
divided by the number of sides of the

polygon.

9. Since the angle ABC is equal to
twice ABO, or to ABO + BAO, it fol-
lows that the angle ABC of the poly-
gon is the supplement of the angle at
the centre.

PROPOSITION in.— THEOREM.

10. Begular 'polygons of the same number of sides are similar,

Ijet ABCI)JE,A'B'C'iyE\
be regular polygons of the
same number of sides ; then
they are similar.

For, 1st, they ape mutu-
ally equiangular, since the
magnitude of an angle of

either polygon depends only on the number of the sides (8
and 9), which is the same in both.

2d. The homologous sides are proportional, since the ratio
AB : A'B' is the same as the ratio BG : B'C, or CD : CD',
etc.

Therefore the polygons fulfil the two conditions of simi-
larity.

11. Corollary. The perimeters of regular polygons of the
same number of sides are to each other as the radii of the cir-
cumscribed circles, or as the radii of the inscribed circles ; and
their areas are to each other as the squares of these radii, (y,
III., Proposition YIII., and lY., Proposition IX.)

BOOK V. 153

PROPOSITION IV.— THEOREM.

12. The area of a regular polygon is equal to half the product
of its perimeter and apothem.

For straight lines drawn from the centre to the vertices
of the polygon divide it into equal triangles whose bases are
the sides of the polygon and whose common altitude is the
apothem. The area of one of these triangles is equal to
half the product of its base and altitude ; therefore the sum
of their areas, or the area of the polygon, is half the product
of the sum of the bases by the common altitude; that is,
half the product of the perimeter and apothem.

EXERCISE.

Theorem. — The area of any polygon circumscribed about a
circle is half the product of its perimeter by the radius of the
circle.

PROPOSITION v.— THEOREM.

13. An arc of a circle is less than any line which envelops it
and has the same extremities.

Let AKB be an arc of a circle, AB its
chord ; and let ALB, AMB, etc., be any
lines enveloping it and terminating at A
and B.

Of all the lines AKB, ALB, AMB, etc.,
which can be drawn (each including between itself and the
chord AB the segment, or area, AKB), there must be at least
one minimum or shortest line, since all the lines are obviously
not equal. Now, no one of the lines ALB, AMB, etc., envel-
oping AKB, can be such a minimum ; for, drawing a tangent
CKL to the arc AKB, the ImeAGJCLB is less than ACLDB ;

154 ELEMENTS OF GEOMETRY.

therefore ALB is not the minimum ; and in the same way it
is shown that no other enveloping line can
be the minimum. Therefore the arc AKB
is the minimum.

14. Corollary. The circumference of a
circle is less than the perimeter of any poly-
gon circumscribed about it.

15. Scholium. The demonstration is applicable when AKB
is any convex curve whatever.

PKOPOSITION VI.— THEOREM.

16. If the number of sides of a regular polygon inscribed in a
circle be increased indefinitely, the apothem of the polygon will
approach the radius of the circle as its limit.

Let AB be a side of a regular polygon in-
scribed in the circle whose radius is OA;
and let OD be its apothem.

Whatever the number of sides of the poly-
gon OB < OA, by I., Proposition XYII.
OA < AD + OD (I., Axiom I.) • .. OA — OD <, AD, and
consequently OA — OD <^ AB.

The perimeter of the polygon is manifestly less than the
circumference of the circle. If n is the number of sides of
the polygon, AB is less than one-nth of the circumference.
Therefore, by taking a sufficiently great value of n, we can
make AB, and consequently OA — OD, as small as we please.

Since OA — OD can be made as small as we please by
increasing the number of sides of the polygon, but cannot be
made absolutely zero, OA is the limit of OD, as the number
of sides of the polygon is indefinitely increased (II., 39).

BOOK V.

155

PROPOSITION VII.— THEOREM.

17. The circumference of a circle is the limit which the perim-
eters of regular inscribed and circumscribed polygons approach
when the number of their sides is increased indefinitely ; and the
area of the circle is the limit of the areas of these polygons.

Let AB and CD be sides of a regular in-
scribed and a similar circumscribed polygon
(v. Proposition I., Exercise) ; let r denote the
apothem OE, R the radius OF^ p the perim-
eter of the inscribed polygon, P the perim-
eter of the circumscribed polygon. Then we
have (Proposition III., Corollary)

whence, by division (III., 10),
P — p __R — r

or P

Now, we have seen in Proposition YI. that by increasing

the number of sides of the polygons the diiference R — r may

be decreased at pleasure ; consequently, since — - does not in-

crease, — - X (-^ — r), or P — p, may be decreased at pleasure.
R

But P being always greater, and p always less, than the cir-
cumference of the circle, the difference between this circum-
ference and either P or ^ is less than the difference P -~ p^
and consequently may be made as small as we please by
increasing the number of sides of the polygons, and sinco it
obviously cannot be made absolutely zero, the circumference
is the common limit of P and p, as the number of sides of
the polygons is indefinitely increased.

IjiisaL.

156

ELEMENTS OF GEOMETRY.

Again, let s and S denote the areas of two similar inscribed
and circumscribed polygons. The difference
between the triangles COD and A OB is the
trapezoid CABD, the measure of which is
^ (CD + AB) X ^^; therefore the difference
between the areas of the polygons is

C F

±\^

^\/«

consequently,

S-s<PXiB-r).

Now, by increasing the number of sides of the polygons the
quantity P X {^ — 0) ^^^ consequently also S — 5, may be
decreased at pleasure. But S being always greater, and 5
always less, than the area of the circle, the difference between
the area of the circle and either S ov s is less than the differ-
ence S — 5, and consequently may also be made as small as
we please by increasing the number of sides of the polygons,
and since it obviously cannot be made absolutely zero, the
area of the circle is the common limit of S and 5, as the
number of sides of the polygons is indefinitely increased.

PROPOSITION VIII.— THEOREM.

18. The circumferences of two circles are to each other as their
radii, and their areas are to each other as the squares of their
radii.

Let B and B' be the radii
of the circles, C and 0' their
circumferences, S and S' their
areas.

Inscribe in the two circles
similar regular polygons of any
arbitrarily chosen number, n, of sides ; let P and P' denote

BOOK V. 157

the perimeters, A and A' the areas, of these polygons ; then,
the polygons being similar, we have (Proposition III., Corol-
lary)

P ^ ^ A^ R^

P' J2" A' B'^'

no matter what the value of n. ^

' Up

As we change n, P and -- X P' change, but remain always

equal to each other.

As n is indefinitely increased, P approaches the limit 0, and

7? 7?

^^ X P' approaches the limit ^^ X C". Therefore, by II.,

Theorem of Limits, these limits are equal, and we have

c = |xo',

^ = ?L

C B!'

In the same way we may prove

K = ^
S' It''

19. Corollary I. The circumferences of circles are to each
other as their diameters, and their areas are to each other as the
squares of their diameters.

Suggestion. D = 2P, if D is the diameter and R the ra-
dius.

158 ELEMENTS OF GEOMETRY.

20. Corollary II. The ratio of the circumference of a circle

to its diameter is constant ; that is, it is the same for all circles.

O 2R
For, from — = — — , we have at once

2B 2B''

This constant ratio is usually denoted by tt, so that for any
circle whose diameter is 2R and circumference G we have

2R '

21. Scholium. The ratio r is incommensurable (as can be
proved by the higher mathematics), and can therefore bo
expressed in numbers only approximately. The letter tt,
however, is used to symbolize its exact value.

22. Definitions. Similar arcs are
those which subtend equal angles
at the centres of the circles to
which they belong.

Similar sectors are sectors whose /

bounding radii include equal angles. ^ /^'^^^ . Z^^-^^Ce^

EXERCISE.

Theorem. — Similar arcs are to each other as their radii, and
similar sectors are to each other as the squares of their radii.

Suggestion. The arcs are like parts of their respective cir-
cumferences, and the sectors like parts of their circles.

BOOK V. 159

PROPOSITION IX.— THEOREM.

23. The area of a circle is equal to half the product of its
circumference by its radius.

Let the area of any regular polygon cir-
cumscribed about the circle be denoted by
A, its perimeter by P, and its apothem,
which is equal to the radius of the circle, by
B ; and let S be the area and C the circum-
ference of the circle. Then A= ^F y^ R
(Proposition IY.)j ^^ matter what the number of sides of
the polygon. If we change the number of sides of the
polygon, A and i P X -K change, but remain always equal to
each other.

As the number of sides is indefinitely increased, A ap-
proaches the limit S^ and ^ P X ^ the limit ^C X R- There-
fore, by II., Theorem of Limits, these limits are equal, and
we have

>Sf=^(7xP. [1]

24. Corollary. The area of a circle is equal to the square
of its radius multiplied by the constant number tz.

Suggestion. If we substitute for C in [1] its value 27zR
(20), we have

EXERCISE.

Theorem. — The area of a sector is equal to half the product
of its arc by the radius.

Suggestion. Compare the sector with the whole circle.

160

ELEMENTS OF GEOMETRY.

PEOBLEMS OF COIS^STEUCTIOK AND COMPUTA-

TIOK

PKOPOSITION X.— PROBLEM.

25. To inscribe a square in a given circle.

Draw any two diameters A C, BJ), per-
pendicular to each other, and join their
extremities by the chords ABj BC, CD,
DA; then ABCD is an inscribed square.

26. Corollary. To circumscribe a square about a circle,
draw tangents at the extremities of two perpendicular diam-
eters AC, BD.

27. Scholium. In the right triangle ABO we have AB' =
'UA' + 'UB'=2'UA\ whence AB =OA.y^, by which the
side of the inscribed square can be computed, the radius
being given.

PROPOSITION XI.— PROBLEM.

28. To inscribe a regular hexagon in a given circle.

Suppose the problem solved, and let
ABCDEF be a regular inscribed hexa-
gon.

Draw the radii OA and OB. The
angle A OB is measured by \ of the cir-
cumference, and therefore contains 60°.

OAB and OB A are therefore together equal to 180° — 60°,
or 120° ; and, since they are equal, each is 60°, and the tri-
angle OAB is equilateral, and th^gfor^the side of the in-
scribed regular hexagon is equal to thfe radius of the circle.

BOOK V.

161

Consequently, to inscribe a regular hexagon, apply the
radius to the circle six times as a chord.

29. Corollary. To inscribe an equilateral triangle^ join the
alternate vertices of the regular hexagon.

30. Scholium. Since OB bisects the arc ABC, it bisects the

chord AG Bit right angles ; and since in the isosceles triangle

A OB, AH is perpendicular to the base, it bisects the base,

and

OII=WB = WA;

that is, the apothem of an inscribed regular triangle is equal
to one-half the radius.

In the right triangle AHO, AH^ = WC — OTT' = TJA^ —
{lOAy=^lUA\2^n^

whence AC =^ OA-^/S, by which the side of the inscribed tri-
angle can be computed from the radius.

The apothem of the regular inscribed hexagon is equal to

1/5-

PROPOSITION XII.— PROBLEM.

31. To inscribe a regular decagon in a given circle.

Suppose the problem solved, and let
ABC .,. . . Jv be a regular inscribed deca-
gon.

Join AFj BGr ; since each of these
lines bisects the circumference, they are
diameters and intersect in the centre 0.
Draw BK intersecting OA in M.

The angle AMB is measured by half
the sum of the arcs KF and AB (II., Proposition XYI.), —

I 14*

t'X^

ELEMENTS OF GEOMETRY.

that is, )by two divi^ns of the circumference ; the inscribed
angle^ikHJ^ is measured by half the arc
BF^ — that is, also, by two divisions;
therefore AMB is an isosceles triangle,
and MB = AB.

Again, the inscribed angle MBO is
measured by half the arc KG, — ^that is,
by one division ; and the angle MOB at
the centre has the safne measure ; there-
fore 0MB is an isosceles triangle, and OM = MB = AB.

The inscribed angle MBA, being measured by half the arc
AK, — that is, by one division, — is equal to the angle AOB.
Therefore the isosceles triangles AMB and A OB are mutually
equiangular and similar, and give the proportion

whence

OA'.AB^AB: AM;
OA XAM = A^=(m\-

that is, the radius OA is divided in extreme and mean ratio
at Jf (III., 41) ; and the greater segment OJf is equal to the
side AB of the inscribed regular decagon.

Consequently, to inscribe a regular decagon, divide the
radius in extreme and mean ratio (III., 42), and apply the
greater segment ten times as a chord.

32. Corollary. To inscribe a regular
pentagon J join the alternate vertices of
the regular inscribed decagon.

BOOK V. 163

PROPOSITION XIII.— PROBLEM.

33. To inscribe a regular pentedecagon in a given circle.
Suppose AB is the side of a ■ f

regular inscribed pentedecagon, ^n*-- - -y^

or that the arc AB is ^ of the ^'^^===*==>.-^i^i;:i^^^'''^

circumference.

JSTow, the fraction J^. == i. _ -i^ ; therefore the arc AB is the
difference between \ and -^ of the circumference. Hence,
if we inscribe the chord AG equal to the side of the regular
inscribed hexagon, and then CB equal to that of the regular
inscribed decagon, the chord AB will be the side of the
regular inscribed pentedecagon required.

34. Scholium. Any regular inscribed polygon being given,
a regular inscribed polygon of double the number of sides
can be formed by bisecting the arcs subtended by its sides
and drawing the chords of the semi-arcs (Proposition I., Cor-
ollary I.). Also, any regular inscribed polygon being given^
a regular circumscribed polygon of the same number of sides
can be formed (Proposition I.). Therefore, by means of the
inscribed square, we can inscribe and circumscribe, succes-
sively, regular polygons of 8, 16, 32, etc., sides ; by means of
the hexagon, those of 12, 24, 48, etc., sides ; by means of the
decagon, those of 20, 40, 80, etc., sides ; and, finally, by means
of the pentedecagon, those of 30, 60, 120, etc., sides.

Until the beginning of the present century, it was sup-
posed that these were the only polygons that could be con-
structed by elementary geometry ; that is, by the use of the
straight line and circle only. Gaxjss, however, in his Disqui-
sitiones Arithmeticce^ Lipsise, 1801, proved that it is possible,
by the use of the straight line and circle only, to construct
regular polygons of 17 sides, of 257 sides, and in general of

>)'' nxXcj__^M3 -u^^JM"

0 F E

G D

\ ^^^^P

"-^4

\ \ 1

\\i ,

164 -

any number oF'sT^Fg which can be^ expressed by 2" + !> ^
being an integer, provided that 2" + 1 is a prime number.

I^ROPOSITION XIV.— PROBLEM.

35. Given the perimeters of a regular inscribed and a similar
• circumscribed polygon, to compute the perimeters of the regular
inscribed and circumscribed polygons of double the number of
sides.

Let AB he a side of the given
inscribed polygon, and CD a side
of the similar circumscribed poly-
gon, tangent to the arc AB Sit its
middle point B. Join AE, and at
A and B draw the tangents AF
and BG ; then AE is a side of the

regular inscribed polygon of double the number of sides, and
FG is a side of the circumscribed polygon of double the
number of sides.

Denote the perimeters of the given inscribed and circum-
scribed polygons by p and P, respectively ; and the perimeters
of the required inscribed and circumscribed polygons of
double the number of sides by y and P', respectively.

Since 0(7 is the radius of the circle circumscribed about
the polygon whose perimeter is P, we have (Proposition III.,
Corollary)

?=oc^^qc

p OA OE'
and since OF bisects the angle COE, we have (III., 15, Ex-
ercise)

OC^CF,
OE FE'
therefore

P^CF
p FE'

BOOK V. 165

whence, by composition,

P + p ^ CF-\-FE^ CE
• 2p 2FE Fa'

I^ow, FG is a side of the polygon whose perimeter is P', and
is contained as many times in F' as GE is contained in P;
hence (III., 9)

CE P
FG F"

and therefore

P+P P^

whence

2p F'

F + p

Again, the right triangles AES and EFW are similar, since
their acute angles EAS and FEJ}^ are equal, and give

AH^EN
AE EF

Since AH and AE are contained the same number of times
in p and p\ respectively, we have

AH ^p ,
AE ^'

and since EN and EF are contained the same number of
times in »' and P', respectively, we have

^^ EW^p^,

EF P"

therefore we have

p'=y^^lxrT\ ^2^

Therefore, from the given perimeters p and P we compute
P' by the equation [1], and then with p and P' we compute
p' by the equation r2"|.

whence

166 ELEMENTS OF GEOMETRY.

PROPOSITION XV.— PROBLEM.

36. To compute the ratio of the circumference of a circle to its
diameter^ approximately.

Method op Perimeters. — In this method, we take the
diameter of the circle as given and compute the perimeters
of some inscribed and a similar circumscribed regular poly-
gon. We then compute the perimeters of inscribed and cir-
cumscribed regular polygons of double the number of sides,
by Proposition XIY. Taking the last-found perimeters as
given, we compute the perimeters of polygons of double the
number of sides by the same method ; and so on. Each com-
putation gives us, of course, a pair of values between which
the value of the circumference must lie. As we continue the
process, these values will come nearer and nearer to the
actual value of the circumference (Proposition YII.), and we
may thus obtain as close an approximation to that value as
we please.

Taking, then, the diameter of the circle as given = 1, let
us begin by inscribing and circumscribing a square. The"
perimeter of the inscribed square = 4 X ? X i/^= 2y^ (27) ;
J that of the circumscribed square = 4 ; therefore, putting

p = 2^2 =t 2.8284271, S r I '^ '

} we find, by Proposition X., for the perimeters of the circum-

scribed and inscribed regular octagons.

P' = ^P^^ = 3.3187085,
p' = yp xP' = 3.0614675.

BOOK V. 167

Then, taking these as given quantities, we put

P = 3.3137085, p = 3.0614675,
and find by the same formulae for the polygons of 16 sides

P' == 3.1825979, / = 3.1214452.

Continuing this process, the results will be found as in the
following

TABLE.*

Number

Perimeter of

of sides.

circumscribed polygon.

inscribed polygon.

4.0000000

2.8284271

3.3137085

3.0614675

3.182.5979

3.1214452

3.1517249

3.1365485

3.1441184

3.1403312

128

3.1422236

3.1412773

256

3.1417504

3.1415138

512

3.1416321

3.1415729

1024

3.1416025

3.1415877

2048

3.1415951

3.1415914

4096

3.1415933

3.1415923

8192

3.1415928

3.1415926

From the last two numbers of this table we learn that the
circumference of the circle whose diameter is unity is less
than 3.1415928 and greater than 3.1415926 ; and since, when
the diameter = 1, we have (7 == tt (20), it follows that

TT = 3.1415927

within a unit of the seventh decimal place.

* The computations have been carried out with ten decimal places in
order to insure the accuracy of the seventh place, as given in the table.

168 ELEMENTS OF GEOMETRY.

37. Scholium. Archimedes (bom 287 b.c.) was the first to
assign an approximate value of ;:. By a method similar to
that above, he proved that its value is between 3^ and 3^J,
or, in decimals, between 3.1428 and 3.1408; he therefore as-
signed its value correctly within a unit of the third decimal
place. The number 3^, or %^^ usually cited as Archimedes'
value of Tz (although it is but one of the two limits assigned
by him), is often used as a sufficient approximation in rough
computations.

Metius (a.d. 1640) found the much more accurate value
1^, which correctly represents even the sixth decimal place.
It is easily remembered by observing that the denominator
and numerator written consecutively, thus, 113 1 355, present
the first three odd numbers each written twice.

More recently, the value has been found to a very great
number of decimals, by the aid of series demonstrated by
the Differential Calculus. Clausen and Dase, of Germany
(about A.D. 1846), computing independently of each other,
carried out the value to two hundred decimal places, and
their results agree to the last figure. The mutual verifica-
tion thus obtained stamps their results as thus far the best
established value to the two-hundredth place. (See Schu-
macher's Astronomische Nachrichten, No. 589.) Other com-
puters have carried the value to over five hundred places,
but it does not appear that their results have been verified.

The value to fifteen decimal places is

TT = 3.141592653589793.

For the greater number of practical applications, the value
t: = 3.1416 is sufficiently accurate. *~— -

EXERCISES ON BOOK V.

THEOREMS.

^1. An equilateral polygon inscribed in a circle is regular.

^ 2. An equilateral polygon circumscribed about a circle is reg-
ular if the number of its sides is odd.

^ 3. An equiangular polygon inscribed in a circle is regular if the
number of its sides is odd.

\ 4. An equiangular polygon circumscribed about a circle is reg-
ular.

\i 6. The area of the regular inscribed triangle is one-half the area
of the regular inscribed hexagon.

/ 6. The area of the regular inscribed hexagon is three-fourths of
that of the regular circumscribed hexagon.

I 7. The area of the regular inscribed hexagon is a mean propor-
tional between the areas of the inscribed and circumscribed equi-
lateral triangles.

(L \ 8. A plane surface may be entirely covered (as in the construc-
tion of a pavement) by equal regular polygons of either three,
four, or six sides.

9. A plane surface may be entirely covered by a combination
of squares and regular octagons having the same side, or by
dodecagons and equilateral triangles having the same side.

10. If squares be described on the sides of a regular hexagon,
and their adjacent external vertices be joined, a regular dodeca-
gon will be formed.

11. The diagonals of a regular pentagon form a regular pentagon.

12. The diagonals joining alternate vertices of a regular hexagon
enclose a regular hexagon one-third as large as the original hex-
agon. . \

15 169

170

ELEMENTS OF GEOMETRY.

13. The area of the regular inscribed octagon is equal to the
product of the side of the inscribed square by the diameter.

Suggestion. A quarter of the octagon is the sum of two triangles
having as a common base the side of the inscribed square, and
having the radius as the sum of their altitudes.

14. The area of a regular inscribed
dodecagon is equal to three times the
square of the radius.

15. Prove the correctness of the following
construction :

If AB and CD are two perpendicular
diameters in a circle, and E the middle
point of the radius 0(7, and if EFis taken
equal to EA^ then OF is equal to the side
of the regular inscribed decagon, and AF
is equal to the side of the regular inscribed
pentagon, {v. III., 42.)

16. From any point within a regular polygon of n sides, per-
pendiculars are drawn to the several sides ; prove that the sum
of thea^ perpendiculars is equal to n times the apothem.

Suggestion. Join the point with the vertices of the polygon, and
obtain an expression for the area in terms of the perpendiculars :
then see Proposition IV.

17. The side of the regular inscribed triangle is equal to the
hypotenuse of a right triangle of which the sides of the inscribed
square and of the regular inscribed hexagon are the sides, {v.
IV., Proposition X.)

18. If a is the side of a regular decagon inscribed in a circle
whose radius is i?,

a = | (1/5-1).

Suggestion. By (31), - =

a H

BOOK V.

171-

19. If a = the side of a regular polygon inscribed in a circle
whose radius is i?, and a' = the side of the
regular inscribed polygon of double the
number of sides, then

a^2 =r{2R — VU^
a

Suggestion. ABC and ADO are similar.
Hence ^ = ^ and ^UfVbut 37)^

= {2RY — a'\

7 a"

\ '''

20. If a = the side of a regular pentagon inscribed in a circle
whose radius is i?, then

21. If a = the side of a regular octagon inscribed in a circle
whose radius is B^ then

a = BV2 — y±

22. If a = the side of a regular dodecagon inscribed in a circle
whose radius is i2, then

=ri2l/2-l/5.

23. The side of the regular inscribed pentagon is equal to the
hypotenuse of a right triangle whose sides are the radius and the
side of the regular inscribed decagon.

24. The area of a ring bounded by two concentric circumfer-
ences is equal to the area of a circle having for its diameter a
chord of the outer circumference tangent to the inner circumfer-
ence.

25. If on the legs of a right triangle,
as diameters, semicircles are described
external to the triangle, and from the
whole figure a semicircle on the hypot-
enuse is subtracted, the remainder is
equivalent to the given triangle.

I^*r%.r

172

ELEMENTS OF GEOMETRY.

26. If on the two segments into which a diameter of a given
circle is divided by any point, as diameters,
semi-circumferences are described lying on op-
posite sides of the given diameter, the sum of
their lengths is equal to the length of a semi-
circumference of the given circle, and a line
which they form divides the circle into two
parts whose areas are to each other as the seg-
ments of the given diameter.

27. If a diameter of a given circle
is divided into n equal parts, and
through each point of division a
curved line of the sort described in
the last problem is drawn, these lines
will divide the circle into n equivar
lent parts.

28. If a circle rolls around the
circumference of a circle of twice
its radius, the two circles being
always tangent internally, the
locus of a fixed point on the cir-
cumference of the rolling circle
is a diameter of the fixed circle.

^^vsIfUA-^ygasguareis subdivided into n^
equal squares, nbSngany^^i^vm^number, and
in each of these smaller squares a circle is in-
scribed, the sum of their areas is equal to the
area of the circle inscribed in the original
square.

MISCELLANEOUS EXERCISES

PLANE GEOMETRY.

THEOREMS.

1. The sum of the three straight lines drawn from
any point within a triangle to the three vertices is
less than the sum and greater than the half sum
of the three sides of the triangle.

2. If one of the acute angles of a right triangle is double the
other, the hypotenuse is double the shortest side.

3. If from any point within an equilateral
triangle perpendiculars to the three sides are
drawn, the sum of these lines is constant, and
equal to the perpendicular from any vertex
upon the opposite side.

4. Lines drawn from one vertex of a
parallelogram to the middle points of the
opposite sides trisect a diagonal.

6. The bisectors of the angles con-
tained by the opposite sides (produced)
of an inscribed quadrilateral intersect
at right angles.

15*

174

ELEMENTS OF GEOMETRY.

6. If AOB is any given angle at the
centre of a circle, and if BC can be
drawn meeting AO produced in C, and
the circumference in Z>, so that CD shall
be equal to the radius of the circle, then
the angle C will be equal to one-third the
angle AOB.

Note, There is no method known of drawing J5(7, under these
conditions, and with the use of straight lines and circles only,
AOB being any given angle ; so that the trisection of an angle^ in
general, is a problem that cannot be solved by elementary geom-
♦*try.

7. If through P, one of the points
of intersection of two circumfer-
ences, any two secants, APB^ CPD^
are drawn, the straight lines, AC,
DB, joining the extremities of the
secants, make a constant angle PJ,
equal to the angle 3fPN formed by
the tangents at P.

8. If a figure is moved in a plane, it may be brought from one
position to any other by revolving it about a certain fixed point ;
that is, by causing each point of the figure to move in the circum-
ference of a circle whose centre is the fixed point.

9. If a square DEFO is inscribed in a right triangle ABC, so
that a side DE coincides with the hypotenuse BC (the vertices F
and O being in the sides AC and AB), then the side DE is a
mean proportional between the segments BD and EC of the
hypotenuse.

10. If the middle points of the sides of a triangle are joined by
straight lines, the medial lines of the triangle thus formed are
the medial line^ of the original triangle, and the perpendiculars
from, the vertices upon the opposite sides are the perpendiculars
at the middle poihts of the sides of the original triangle.

11. If O is the centre of the circle circumscribed about a triangle
ABC, and P is the intersection of the perpendiculars from the
angles upon the opposite sides, the perpendicular from O upon
the side JSC is equal to one-half the distance AP.

MISCELLANEOUS EXERCISES.

175

12. In any triangle, the centre of the circumscribed circle, the
intersection of the medial lines, and the intersection of the per-
pendiculars from the angles upon the opposite sides, are in the
same straight line ; and the distance of the first point from the
second is one-half the distance of the second from the third.

13. If two circles intersect in the points A and B, and through
A any secant CAD is drawn terminated by the circumferences at
C and i), the straight lines BC and BD are to each other as the
diameters of the circles.

14. If through the middle point of each diagonal of any quad-
rilateral a parallel is drawn to the other diagonal, and from the
intersection of these parallels straight lines are drawn to the
middle points of the four sides, these straight lines divide the
quadrilateral into four equivalent parts.

15. If three straight lines Aa, Bb, Ce^
drawn from the vertices of a triangle ABC
to the opposite sides, pass through a com-
mon point O within the triangle, then

Oa , Ob ,0g ^.
Aa"^ Bb~^ Cc

16. If from any point O within a tri-
angle ABC any three straight lines, Oa,
Ob, Oc, are drawn to the three sides, and
through the vertices of the triangle three
straight lines, Aa^, Bb\ Co', are drawn
parallel respectively to Oa, Ob, Oc, then

Oa ^^\\ Q^ —I
Aa^ T Bb' "^ Cc'

17. The area of a circle is a mean proportional between the
areas of any two similar polygons, one of which is circumscribed
about the circle and the other isoperimetrical with the circle.
{Galileo^ 8 Theorem.)

18. Two diagonals of a regular pentagon, not drawn from a
common vertex, divide each other in extreme and mean ratio.

176

ELEMENTS OF GEOMETRY.

LOCL

19. The angle ACB is any inscribed an-
gle in a given segment of a circle ; ^C is
produced to P, making CP equal to GB ;
find the locus of P.

20. The hypotenuse of a right triangle is given in magnitude
and position ; find the locus of the centre of the inscribed circle.

21. The base BC of a triangle ABC is given in position and
magnitude, and the vertical angle J. is of a given magnitude ;
find the locus of the centre of the inscribed circle.

22. From a given point O, any straight line OA
is drawn to a given straight line MN^ and OP is
drawn making a given angle with OA^ and such
that OP is to OA in a given ratio ; find the locus
of P.

With the same construction, if OP is so taken
that the product OP. OA is equal to a given con-
stant ; find the locus of P.

23. From a given point O, any straight
line OA is drawn to a given circumfer-
ence, and OP is drawn making a given
angle with OA, and such that OP is to
OA in a given ratio ; find the locus of P.

With the same construction, if OP is
so taken that the product OP. OA is
equal to a given constant ; find the locus of P.

24. One vertex of a triangle whose angles are given is fixed,
while the second vertex moves on the circumference of a given
circle ; what is the locus of the third vertex ?

25. Through A, one of the points of intersection of two given
circles, any secant is drawn cutting the two circumferences in the
points B and C; find the locus of the middle point of BG.

MISCELLANEOUS EXERCISES. 177

PROBLEMS.

26. Describe a circle through two given points which lie outside
a given line, the centre of the circle to be in that line. Show
when no solution is possible.

27. In a given circle, inscribe a chord of a given length which
produced shall be tangent to another given circle.

28. Through P, one of the points of intersection of two circum-
ferences, draw a straight line, terminated by the circumferences,
which shall be bisected in P.

29. Through one of the points of intersection of two circum-
ferences draw a straight line, terminated by the circumferences,
which shall have a given length.

30. In a given triangle ABC, to inscribe
a parallelogram DEFO, such that the adja-
cent sides DE and DO shall be in a given
ratio and contain a given angle.

31. Construct a triangle, given its base, the ratio of the other
two sides, and one angle.

32. To determine a point in a given arc of a circle, such that
the chords drawn from it to the extremities of the arc shall have
a given ratio.

33. To find a point within a given triangle, such that the three
straight lines drawn from it to the vertices of the triangle shall
make three equal angles with each other.

34. Inscribe a trapezoid in a given circle, knowing its area and
the common length of its inclined sides.

35. To construct a triangle, given one angle, the side opposite
to that angle, and the area (equal to that of a given square).

36. Divide a given circle into a given number of equivalent
parts, by concentric circumferences.

Also, divide it into a given number of parts proportional to
given lines, by concentric circumferences.

178 ELEMENTS OF GEOMETRY.

37. A circle being given, to find a given number of circles whose
radii shall be proportional to given lines, and the sum of whose
areas shall be equal to the area of the given circle.

38. In a given equilateral triangle, inscribe three equal circles
tangent to each other and to the sides of the triangle.

Determine the radius of these circles in terms of the side of the
triangle.

39. In a given circle, inscribe three equal circles tangent to each
other and to the given circle.

Determine the radius of these circles in terms of the radius of
the given circle.

NUMERICAL EXAMPLES.

Note.— The following approximate values are close enough for
ordinary purposes : tt = 2^2^ ^2 = ||, |/8 = |f, /S = ff. Radius
of earth = 3960 miles.
\^' 40. The vertical angle of an isosceles triangle is 36°, and the
length of the base is 2 feet ; find the base angles, the length of
the bisector of a base angle, and the length of a side of the given
triangle. Ans. TZ°, 2 feet, (1 + V5) feet.

I 41. One angle of a triangle is 60°, the including sides are 8 feet
and 8 feet ; find the area and the third side.

Ans. 61/3 square feet, 7 feet.
.^ ' V 42. The three sides of a triangle are 9 inches, 10 inches, and 17
Lo inches, its area is 36 square inches ; find the area of the inscribed
^^ circle. Ans. 47r.

"^ (/ 43. The adjacent sides of a parallelogram are 12 feet and 14 feet,
the area is 120 square feet ; find the long diagonal.

Ans. 24 feet.

44. The area of a right triangle is 6 square feet, the length of
the hypotenuse is 5 feet ; find the other sides.

Ans. 3 feet, 4 feet.

45. ©btain a formula connecting the length of a chord ^, its
distance from the centre c?, and the radius r.

Ans. - = r^ — cP,
4

46. Obtain a formula for the length ^ of a common tangent to
two circles, given tlie radii r, r'', and the distance between the
centres d. Ans. {r — r^y 4-^ = ^2 for external tangent.

[r -f r^f + ^ = c^ for internal tangent.

47. Through what angle does the hour-hand of a clock move in
1 hour? in 1 minute? Through what angle does the minute-
hand move in 1 minute ?

What angle do the hands of a clock make with each other at
ten minutes past three ? at quarter of six ? Ans. 35°, 97° 30^.

/ 179

180 ELEMENTS OF GEOMETRY.

48. Two secants cut each other without a circle, the intercepted
arcs are 12° and 48° ; what is the angle between the secants ?

Two chords intersect within the circle, a pair of opposite inter-
cepted arcs are 12° and 48° ; what is the angle between the chords ?

49. Tw^o tangents make with each other an angle of 60° ; re-
quired the lengths of the arcs into which their points of contact
divide the circle, given radius equals 7 inches.

Ans. 14| inches, 29^ inches.

60. A swimmer whose eye is at the surface of the water can
just see the top of a stake a mile distant ; the stake proves to be
8 inches out of water ; required the radius of the earth.

Am. 3960 miles.

61. A passenger standing on the deck of a steamer about to
start observes that his eye is on a level with the top of the wharf,
which he knows to be 12 feet high ; when they have steamed 8^
miles the wharf disappears below the horizon ; required the radius
of the earth. Ans. 3974 miles.

62. How many miles is the light of a light-house 150 feet high
visible at sea ? A7is. 15.

63. On approaching Portland from the sea, Mount Washington
is first visible 12 miles from shore ; Portland is 85 miles from
Mount Washington ; required the height of the mountain.

Ans. 6270 feet.

64. The latitude of Leipsic is 51° 21^ that of Venice 45° 26^ and
Venice is due south of Leipsic ; how many miles are they apart?
Use 4000 miles as the earth's radius. Ans. 413 miles.

66. The latitude of the Peak of Teneriffe is about 30° N. ; the
rising sun shines on its summit on the 21st of March 9 minutes
before it shines on its base ; required the height of the mountain.

Ans. About 12,000 feet.

66. A quarter-mile running-track 10 feet wide, with straight
parallel sides and semicircular ends, is to be laid out in a rectan-
gular field 220 feet wide. How long must the field be in order
that a runner, keeping in the middle of the track, may have one-
quarter of a mile to cover? how much can he gain by keeping
close to the inner edge of the track ? what is the area of the field ?
of the portion encircled by the track? of the track itself?

Ans. 550 feet ; 31f feet ; 121,000 square feet ; 97,428^ square feet ;
13,200 square feet.

67. The fly-wheel of an engine is connected by a belt with a
smaller wheel driving the machinery of a mill. The radius of the

NUMERICAL EXAMPLES.

181

fly-wheel is 7 feet ; of the small wheel, 21 inches. How many
revolutions does the small wheel make to one of the fly-wheel ?
The distance between the centres of the two wheels is 10^ feet.
What is the length of the connecting band ?

Ans. 51 feet 2 inches.

58. If from each vertex of a regular polygon as a centre, with
a radius equal to one-half the side, an arc is described outward
from side to side of the polygon, an ornamental figure much
used in architecture is formed. Such a figure formed on a polygon
of numerous sides is often used as a rose-window.

The figure bounded by three arcs is called a trefoil ; by four
arcs, a quatre-foil ; by five arcs, a cinque-foil.

Find the area of a tre-
foil, given the distance be-
tween the centres of adja-
cent arcs equal to 21 inches.

Ans. 7.338 square feet.

69. A rose-window
of six lobes is to be
placed in a circular
space 42 feet in diam-
eter. How many
square feet of glass
will it contain ?

Ans. 1123.8 square
feet.

SYLLABUS OF PLANE GEOMETRY.

POSTULATES, AXIOMS, AISTD THEOKEMS.

BOOK I.
Postulate I.

Through any two points one straight line, and only one, can

be drawn.

Postulate II.

Through a given point one straight line, and only one, can be
drawn having any given direction.

Axiom I.

A straight line is the shortest line that can be drawn between
two points.

Axiom II.

Parallel lines have the same direction.

Proposition I.

At a given point in a straight line one perpendicular to the line
can be drawn, and but one.

Corollary. Tiirough tlie vertex of any given angle one straight
line can be drawn bisecting the angle, and but one.

Proposition II.
All right angles are equal. ' -^

Proposition III.

The two adjacent angles which one straight line makes with
another are together equal to two right angles.

Corollary I. The sum of all the angles having a common ver-
tex, and formed on one side of a straight line, is two right angles.

Corollary II. The sum of all the angles that can be formed
about a point in a plane is four right angles.

182

SYLLABUS OF PLANE GEOMETRY. 183

Proposition IV.

If the sum of two adjacent angles is two right angles, their
exterior sides are in the same straight line.

Proposition V.

If two straight lines intersect each other, the opposite (or ver-
tical) angles are eq^ual.

Proposition VI.

Two triangles are equal when two sides and the included angle
of the one are respectively equal to- two sides and the included
angle of the other.

Proposition VII.

Two triangles are equal when a side and the two adjacent
angles of the one are respectively equal to a side and the two
adjacent angles of the other.

Proposition VIII.

In an isosceles triangle the angles opposite the equal sides are
equal.

Corollary. The straight line bisecting the vertical angle of an
isosceles triangle bisects the base, and is perpendicular to the base.

Proposition IX.

Two triangles are equal when the three sides of the one are
respectively equal to the three sides of the other.

Proposition X.

Two right triangles are equal when they have the hypotenuse
and a side of the one respectively equal to the hypotenuse and a
side of the other.

Proposition XI.

If two angles of a triangle are equal, the sides opposite to them
are equal, and the triangle is isosceles.

Proposition XII.

If two angles of a triangle are unequal, the side opposite the
greater angle is greater than the side opposite the less angle.

184 ELEMENTS OF GEOMETRY.

Proposition XIII.

If two sides of a triangle are unequal, the angle opposite the
greater side is greater than the angle opposite the less side.

Proposition XIV.

If two triangles have two sides of the one respectively equal to
two sides of the other, and the included angles unequal, the tri-
angle which has the greater included angle has the greater third
side.

Proposition XV.

If two triangles have two sides of the one respectively equal to
two sides of the other, and the third sides unequal, the triangle
which has the greater third side has the greater included angle.

Proposition XVI.

From a given point without a straight line one perpendicular
can be drawn to the line, and but one.

Proposition XVII.

The perpendicular is the shortest hne that can be drawn from
a point to a straight line.

Proposition XVIII.

If a perpendicular is erected at the middle of a straight line,
then every point on the perpendicular is equally distant from the
extremities of the line, and every point not on the perpendicular
is unequally distant from the extremities of the line.

Proposition XIX.

Every point in the bisector of an angle is equally distant from
the sides of the angle ; and every point not in the bisector is un-
equally distant from the sides of the angle ; that is, the bisector
of an angle is the locus of the points within the angle and equally
distant from its sides.

Proposition XX,

A convex broken line is less than any ether line which envelops
it and has the same extremities.

SYLLABUS OF PLANE GEOMETRY. 185

Proposition XXI.

If two oblique lines drawn from a point to a line meet the line
at unequal distances from the foot of the perpendicular, the more
remote is the greater.

Proposition XXII.

Two straight lines perpendicular to the same straight line are
parallel.

Proposition XXIII.

Through a given point one line, and only one, can be .drawn
parallel to a given line.

Proposition XXIV.

When two straight lines are cut by a third, if the alternate-
interior angles are equal, the two straight lines are parallel.

Corollary I. When two straight lines are cut by a third, if a
pair of corresponding angles are equal, the lines are parallel.

Corollary II. When two straight lines are cut by a third, if the
sum of two interior angles on the same side of the secant line is
equal to two right angles, the two lines are parallel.

Proposition XXV.

If two parallel lines are cut by a third straight line, the alter-
nate-interior angles are equal.

Corollary I. If two parallel lines are cut by a third straight
line, any two corresponding angles are equal.

Corollary II. If two parallel lines are cut by a third straight
line, the sum of the two interior angles on the same side of the
secant line is equal to two right angles.

Proposition XXVI.

The sum of the three angles of any triangle is equal to two
right angles.

Corollary. If one side of a triangle is extended, the exterior
angle is equal to the sum of the two interior opposite angles.

Proposition XXVII.

The sum of all the angles of any convex polygon is equal to
twice as many right angles, less four, as the figure has sides.

16*

186 ELEMENTS OF GEOMETRY.

Proposition XXVIII.

Two parallelograms are equal when two adjacent sides and the
included angle of the one are equal to two adjacent sides and the
included angle of the other.

Corollary. Two rectangles are equal when they have equal
bases and equal altitudes.

Proposition XXIX.

The opposite sides of a parallelogram are equal, and the oppo-
site angles are equal.

Proposition XXX. '

If two opposite sides of a quadrilateral are equal and parallel,
the figure is a parallelogram.

Proposition XXXI.

If the opposite sides of a quadrilateral are equal, the figure is a
parallelogram.

Proposition XXXII.

The diagonals of a parallelogram bisect each other.

BOOK II.
PROPOSITIONS.

Postulate.

A circumference may be described with any point as centre and
any distance as radius.

Proposition I.

Two circles are equal when the radius of the one is equal to the
radius of the other.

Proposition II.

Every diameter bisects the circle and its circumference.

Proposition III.

In equal circles, or in the same circle, equal angles at the centre
intercept equal arcs on the circumference.

Corollary. Conversely, in the same circle, or in equal circles,
equal arcs subtend equal angles at the centre.

SYLLABUS OF PLANE GEOMETRY. 187

Proposition IV.

In equal circles, or in the same circle, equal arcs are subtended
by equal chords.

Corollary. Conversely, in equal circles, or in the same circle,
equal chords subtend equal arcs.

Proposition V.

In equal circles, or in the same circle, the greater of two un-
equal arcs is subtended by the greater chord, the arcs being each
less than a semi-circumference.

Corollary. Conversely, in equal circles, or in the same circle
the greater of two unequal chords subtends the greater arc.

Proposition VI.

The diameter perpendicular to a chord bisects the chord and
the arcs subtended by it.

Corollary I. The perpendicular erected at the middle point of
a chord passes through the centre of the circle.

Corollary II. When two circumferences intersect, the straight
line joining their centres bisects their common chord at right
angles.

Proposition VII.

In the same circle, or in equal circles, equal chords are equally
distant from the centre ; and of two unequal chords the less is at
the greater distance from the centre.

Corollary. Conversely, in the same circle, or in equal circles,
chords equally distant from the centre are equal ; and of two
chords unequally distant from the centre, that is the greater
whose distance from the centre is the less.

Proposition VIII.
A straight line cannot intersect a circle in more than two points.

Proposition IX.

A straight line tangent to a circle is perpendicular to the radius
drawn to the point of contact.

Corollary I. A perpendicular to a tangent line drawn through
the point of c(mtact must pass through the centre of the circle.

Corollary II. If two circumferences are tangent to each other,
their centres and their point of contact lie in the same straight
line.

188 ELEMENTS OF GEOMETRY.

Proposition X.

When two tangents to the same circle intersect, the distances
from their point of intersection to their points of contact are
equal.

Proposition XI.

Two parallels intercept equal arcs on a circumference.

Doctrine of Limits. — Theorem.

If two variables dependent upon the same variable are so re-
lated that they are always equal, no matter what value is given
to the variable on which they depend, and if, as the independent
variable is changed in some specified way, each of them ap-
proaches a limit, the two Umits must be absolutely equal.

Proposition XII.

In the same circle, or in equal circles, two angles at the centre
are in the same ratio as their intercepted arcs.

Proposition XIII.

The numerical measure of an angle at the centre of a circle is
the same as the numerical measure of its intercepted arc, if the
unit of angle is the angle at the centre which intercepts the
adopted unit of arc.

Proposition XIV.

An inscribed angle is measured by one-half its intercepted arc.
Corollary. An angle inscribed in a semicircle is a right angle.

Proposition XV.

An angle formed by a tangent and a chord is measured by one-
half the intercepted arc.

Proposition XVI.

An angle formed by two chords intersecting within the circum-
ference is measured by one-half the sum of the arcs intercepted
between its sides and between the sides of its vertical angle.

Proposition XVII.

An angle formed by two secants intersecting without the cir-
cumference is measured by one-half the difference of the inter-
cepted arcs.

SYLLABUS OF PLANE GEOMETRY. 189

Proposition XVIII.

An angle formed by a tangent and a secant is measured by one-
half the difference of the intercepted arcs.

Corollary. An angle formed by two tangents is measured by
one-half the difference of the intercepted arcs.

BOOK III.

THEOREMS.

Proposition I.

A parallel to the base of a triangle divides the other two sides

proportionally.

Proposition II.

If a straight line divides two sides of a triangle proportionally,
it is parallel to the third side.

Proposition III.
Two triangles are similar when they are mutually equiangular.

Proposition IV.

Two triangles are similar when an angle in the one is equal to
an angle in the other, and the sides including these angles are
proportional.

Proposition V.

Two triangles are similar when their homologous sides are
proportional.

Proposition VI.

If two polygons are composed of the same number of triangles,
similar each to each and similarly placed, the polygons are
similar.

Proposition VII.

Two similar polygons may be decomposed into the same num-
ber of triangles, similar each to each and. similarly placed.

Proposition VIII.

The perimeters of two similar polygons are in the same ratio as
any two homologous sides.

190 ELEMENTS OF GEOMETRY.

Proposition IX.

If a perpendicular is drawn from the vertex of the right angle
to the hypotenuse of a right triangle :

1st. The two triangles thus formed are similar to each othei
and to the whole triangle ;

2d. The perpendicular is a mean proportional between the seg^
ments of tlie hypotenuse ;

3d. Each side about the right angle is a mean proportional be-
tween the whole hypotenuse and the adjacent segment.

Corollary. If from any point in the circumference of a circle a
perpendicular is let fall upon a diameter, the perpendicular is a
mean proportional between the segments of the diameter.

Proposition X.

The square of the length of the hypotenuse of a right triangle
is the sum of the squares of the lengths of the other two sides,
the three lengths being expressed in terms of the same unit.

Proposition XI.

If two chords intersect within a circle, their segments are re-
ciprocally proportional.

Proposition XII.

If two secants intersect without a circle, the whole secants and
their external segments are reciprocally proportional.

Corollary. If a tangent and a secant intersect, the tangent is a
mean proportional between the whole secant and its external
segment.

BOOK lY.

THEOREMS.

Proposition I.

Parallelograms having equal bases and equal altitudes are
equivalent.

Corollary. Any parallelogram is equivalent to a rectangle
having the same base and the same altitude.

SYLLABUS OF PLANE GEOMETRY. 191

Proposition II.

Two rectangles having equal altitudes are to each other as their
bases.

Corollary. Two rectangles having equal bases are to each other
as their altitudes.

Proposition III.

Any two rectangles are to each other as the products of their
bases by their altitudes.

Proposition IV.

The area of a rectangle is equal to the product of its base and
altitude.

Proposition V.

The area of a parallelogram is equal to the product of its base
and altitude.

Proposition VI.

The area of a triangle is equal to half the product of its base
and altitude.

Corollary I. A triangle is equivalent to one-half of any paral-
lelogram having the same base and the same altitude.

Corollary II. Triangles having equal bases and equal altitudes
are equivalent.

Corollary III. Triangles having equal altitudes are to each
other as their bases, and triangles having equal bases are to each
other as their altitudes.

Proposition VII.

The area of a trapezoid is equal to the product of its altitude by
half the sum of its parallel bases.

Proposition VIII.

Similar triangles are to each other as the squares of their homol-
ogous sides.

Proposition IX.

Similar polygons are to each other as the squares of their homol-
ogous sides.

Proposition X.

The square described upon the hypotenuse of a right triangle
is equivalent to the sum of the squares described on the other
two sides.

192 ELEMENTS OF GEOMETRY.

BOOK V.

THEOREMS.

Proposition I.

If the circumference of a circle be divided into any number of
equal parts, the chords joining the successive points of division
form a regular polygon inscribed in the circle ; and the tangents
drawn at the points of division form a regular polygon circum-
scribed about the circle.

Corollary I. If the vertices of a regular inscribed polygon are
joined with the middle points of the arcs subtended by the sides
of the polygon, the joining Unes will form a regular inscribed
polygon of double the number of sides.

Corollary II. If at the middle points of the arcs joining ad-
jacent points of contact of the sides of a regular circumscribed
polygon tangents are drawn, a regular circumscribed polygon
of double the number of sides will be formed.

Proposition II.

A circle may be circumscribed about any regular polygon, and
a circle may also be inscribed in it.

Proposition III.

Regular polygons of the same number of sides are similar.

Corollary. The perimeters of regular polygons of the same
number of sides are to each other as the radii of the circumscribed
circles, or as the radii of the inscribed circles ; and their areas are
to each other as the squares of these radii.

Proposition IV.

The area of a regular polygon is equal to half the product of
Its perimeter and apothem.

Proposition V.

An arc of a circle is less than any line which envelops it and
has the same extremities.

Corollary, The circumference of a circle is less than the perim-
eter of any polygon circumscribed about it.

SYLLABUS OF PLANE GEOMETRY. 193

Proposition VI.

If the number of sides of a regular polygon inscribed in a circle
be increased indefinitely, the apothem of the polygon will ap-
proach the radius of the circle as its limit.

Proposition VII.

The circumference of a circle is the Umit which the perimeters
of regular inscribed and circumscribed polygons approach when
the number of their sides is increased indefinitely ; and the area
of the circle is the limit of the areas of these polygons.

Proposition VIII.

The circumferences of two circles are to each other as their
radii, and their areas are to each other as the squares of their
radii.

Corollary I. The circumferences of circles are to each other as
their diameters, and their areas are to each other as the squares
of their diameters.

Corollary II. The ratio of the circumference of a circle to its
diameter is constant.

Proposition IX.

The area of a circle is equal to half the product of its circum-
ference by its radius.

Corollary. The area of a circle is equal to the square of its
radius multiplied by the constant number tt.

GEOMETRY OE SPACE.

In Plane Geometry we have considered merely figures com-
posed of lines and points, all of which are supposed to lie in
the same plane (v. Introduction, 5 and 6), and in the proposi-
tions and definitions of the preceding five books it has been
tacitly assumed that the figures in question are plane figures.
In many of the propositions and definitions this limitation is
essential to the truth of the proposition ; for example. Propo-
sitions I. and XXII., Book I., and Definition 20, Book II. In
others the demonstration given is inconclusive without the
limitation in question, although the proposition is true even
when the limitation is removed; for example. Exercise 1,
Proposition XXIII., Book I. While in propositions concern-
ing equal polygons, which depend for their proof directly or
indirectly upon a superposition of one polygon upon the
other, the limitation is obviously of no importance; for ex-
ample. Propositions YI., YII., and IX., Book I. It is, then,
important, when we use the theorems of Plane Geometry in
proving theorems of the Geometry of Space, to satisfy our-
selves that they are still true in the figures with which we are
concerned.

194

BOOK YI

THE PLANE. POLYBDRAL ANGLES.

1. Definition. A plane has already been defined as a surface
such that the straight line joining any two points in it lies
wholly in the surface.

JIT

Thus, the surface ikfiV" is a plane, if, A /

and B being any two points in it, the /—4- ?-

straight line AB lies wholly in the sur-
face.

The plane is understood to be indefinite in extent, so that,
however far the straight line is produced, all its points lie
in the plane. But to represent a plane in a diagram, we
are obliged to take a limited portion of it, and we usually
represent it by a parallelogram supposed to lie in the plane.

2. Definition. A plane is said to be determined by given lines
or points when one plane, and only one, can be drawn con-
taining the given lines or points.

PKOPOSITION I.— THEOREM.

K^^Through any given straight line a plane may be passed ;
hut the line will not determine the plane.

Let AB be a given straight line.
A straight line may be drawn in
any plane, and the position of that
plane may be changed until the

line drawn in it is brought into coincidence with AB. We
shall then have a plane passed through AB ; and this plane

195

r ~ J

!.-''

aJ. /I

L- 1

r|o ^ ELEMENTS OF GEOMETRY.

may be turned upon AB as an axis, and made to occupy-
as many different positions as we
choose, and as in each of these
positions it is a plane through AB,

we may have as many planes as lL

we choose through AB ; conse-
quently AB does not determine (2) a plane.

PBOPOSITION II.— THEOKEM.

/ 4. ^ plane is determined^ 1st, by a straight line and a point
/ without that line ; 2d, by two intersecting straight lines ; 3d, by
7 three points not in the same straight line ; 4th, by two parallel
! straight lines.

Ist. Through a given line AB a
plane may be passed, and may then / <? ^

be turned upon AB as an axis, until
it contains a given point G. If it is
then turned by the smallest amount

in either direction, it ceases to contain C. Therefore on©
plane, and only one, can be drawn containing a given line
and a given point without that line.

2d. Through one of the lines ABj and a point C of the
other, one plane, and only one, can be
drawn. This plane will contain AG
(1), and no other plane through AB
will.

3d. If three points. A, Bj 0, are
given, any plane containing them

must contain one of them and the line joining the other two
(1) J and one, and only one, such plane can be drawn.

BOOK VI. 197

4th. Two parallels, AB, CD, lie in the same plane, by Defi-
nition (I., 5) ; there is, then, one plane
containing them. There is only one,
for through AB and a point B of

CD only one plane can be passed. ^ J ^

^fo) Corollary. The intersection of

two planes is a straight line. For if two points of the in-
tersection be joined by a straight line, that line must lie in
both planes^ by (1) ; and no point outside of this line can be
common to the two planes, by Proposition II. ; therefore the
straight line in question is the line of intersection of the two
planes. ^

PERPENDICULARS AND OBLIQUE LINES TO PLANES.

6. Definition. A straight line is perpendicular to a plane
when it is perpendicular to every straight line drawn in the
plane through its foot ; that is, through the point in which it
meets the plane.

In the same case, the plane is said to be perpendicular to

the line.

PROPOSITION III.— THEOREM.

Cy From a given point without a plane, one perpendicular to

the plane can he drawn, and hut one ; and the perpendicular is

the shortest line that can he drawn from the point to the plane.

Let A be the given point, and JOT the
plane. Consider the various lines that can
be drawn from A to MN. These lines are
obviously not all of the same length ; there
must, then, be among them either one
minimum line, or a set of equal shortest
lines. There cannot be a set of equal
shortest lines. For, suppose that AB and AB' are two such

17*

. y

198

ELEMENTS OF GEOMETRY.

- J

lines. Join BB'. Then, since AB and AB'' Sire equal lines
drawn from A to BB\ they cannot be per-
pendicular to BB' (I., 'Proposition XYI.),
and consequently they are longer than the
perpendicular AC from A to BB\ by 1.,
Proposition XYII., which is contrary to
the hypothesis that they were shorter
than any other lines that could be drawn ^

from A to J/iV. There is therefore one,
and but one, minimum line from A to the plane/ Let AP be
that minimum line ; then AP is perpendicular to any straight
line BF drawn in the plane through its foot P. For, in the
plane of the lines AP and JEF, ^P is the shortest line that
can be drawn from A to any point in FF, since it is the
shortest line that can be drawn from J. to any point in the
plane MN; therefore AP is perpendicular to FF (I., Propo-
sition XYII.). Thus AP is perpendicular to any^ that is, to
every, straight line drawn in the plane through its foot, and
is therefore perpendicular to the plane.

There can be no other perpendicular from A to the plane
MN; for, if there were, both lines would be perpe-ndicular
to the line joining the points where they met the plane, and
we should have two perpendiculars from a point to a line,
which is contrary to I., Proposition XYI.
vj) Corollary. At a given point in a plane j one perpendicular
can be erected to the plane, and but one.

Let ilOTbe the plane and P
the point.

Let Jif'iV' be any other
plane, A' any point without it,
and A'P' the perpendicular
from A' to this plane. Suppose the plane Al'N^ to be applied

BOOK VI.

199

to the plane MN with the point P' upon P, and let AP be
the position then occupied by the perpendicular A'P'. We
then have one perpendicular, J.P, to the plane MN^ erected
at P. There can be no other : for let PB be another perpen-
dicular at P. Then AP and PB are both perpendicular to
PC, the line of intersection of MN with the plane deter-
mined by the two lines AP and BP^ at the same point, and
lie in the same plane with^ P(7, ^and this is contrary to I., j j

Proposition I. (^ ^lr^li' ^x^::..--^rar^)

9. Scholium. By the distance of a point from a plane is
meant the shortest distance; hence it is the perpendicular
distance from the point to the plane.

EXERCISES.

1. Theorem. — Oblique lines drawn from a point to a plane,
and meeting the plane at equal distances from the foot of the
perpendicular, are equal i and of two oblique lines meeting the
plane at unequal distances from the foot of the perpendicular
the more remote is the greater.

2. Theorem. — Equal oblique lines
from a point to a plane meet the
plane at equal distances from the foot
of the perpendicxdar ; and of two un-
equal oblique lines the greater meets
the plane at the greater distance from
the foot of the perpendicular.

200 ELEMENTS OF GEOMETRY.

PROPOSITION IV.— THEOREM.

(^O) If a straight line is perpendicular to each of two straight
lines at their point of intersection^ it is perpendicular to the plane
of those lines.

Let AP be perpendicular to PB and
P(7, at their intersection P ; then AP is
perpendicular to the plane MN which
contains those lines.

For, let PD be any other straight
line drawn through P in the plane MN.
Draw any straight line BDG intersect-
ing PB, PC, PD, in J5, (7, D; produce W
AP to A', making PA' = PA, and join a^
A and A' to each of the points B, C, D,

Since BP is perpendicular to AA', at its middle point, we
have BA = BA' (I., Proposition XYIII.), and for a like
reason CA = GA' ; therefore the triangles ABC, A'BC, are
equal (I., Proposition IX.), and the angle ABD is equal to
the angle A'BD. The triangles ABD and A'BD are equal
(I., Proposition YI.), and AD = A'D. Hence the triangles
APD and A'PD are equal (I., Proposition IX.). Therefore
the adjacent angles APD and A'PD are equal, and PD is
perpendicular to AP. AP, then, is perpendicular to any^
that is, to every, line passing, through its foot in the plane
MNp and is consequently perpendicular to the plane.
\l|t Corollary I. At a given point of g, straight line one
plane can he drawn perpendicular to the line, and hut one.

Let AP be the line, and P the point. Through AP pass
two planes, and in each of these planes draw through P a
line perpendicular to AP. The plane determined by these

BOOK VI. 201

two lines is perpendicular to AF at P, by Proposition

No other perpendicular plane can
be drawn through P, for, if it could,
a plane containing AF would inter-
sect the two perpendicular planes
in lines which would lie in the same
plane with J.P, and be perpendic-
ular to AF at the same point, ^ " * ^ 4^**v -^"^
which is contrary to I., Proposi-( \ yj^^-^tj^ ok.^^v^"'^'
tion I. * ^-^'H *
X QS^ Corollary II. Through a given point without a straight
line one plane can he drawn perpendicular to the line, and but
one.

In the plane determined by the point and the line draw a
perpendicular from the point to the line, and through the
foot of this perpendicular draw, in any second plane passing
through the given line, a second perpendicular to the line.
The plane of these two perpendiculars is obviously a plane
passing through the given point and perpendicular to the
given line.

No second perpendicular plane can be drawn through thtj
given point, for the plane determined by the line and the
point would cut the two perpendicular planes in lines which
would be two perpendicular lines from the given point to the
given line, which is contrary to I.. Proposition XYL . , ,

EXERCISES.

1. Theorem. — All the perpendiculars that can be drawn to a-
straight line at the same point lie in a plane perpendicular to the
line at the point.

202 ELEMENTS OF GEOMETRY.

2. Theorem. — If from the foot of a
perpendicular to a plane a straight line
is drawn at right angles to any line of
the plane, and its intersection with that
line is joined to any point of the perpen-
dicular, this last line will be perpendic-
ular to the line of the plane.

PARALLEL STRAIGHT LINES AND PLANES.

13. Definitions. A straight line is parallel to a plane when it
cannot meet the plane, though both be indefinitely produced.

In the same case, the plane is said to be parallel to the line.

Two planes are parallel when they do not meet, both being
indefinite in extent.

Jy. PROPOSITION v.— THEOREM.

(14) Two lines in space having the same direction are par-
allel.

Let AB and CD be two lines having

. the same direction. Through AB and any
point E of CD pass a plane, and in this ^ e d

plane draw through E a line parallel to
y'^^.y^ '^^' This line will have the same direction as AB (I.,
'^^^^/^^jSp^iom II.), and consequently the same direction as CD, and
^>-^^ must therefore coincide with CD, by I., Postulate 11. Hence,- ^
AB and CD are parallel, (f "^^ T''- ff • ^ ^•■' f '^-^. tT^l^

15. Corollary. Two lines paraltet to the seme lirie are par-
allel to each other. For they have the same direction.

^..^^ PROPOSITION VI.— THEOREM.

<ISJ If two straight lines are parallel, every plane passed
through one of them and not coincident with the plane of the
parallels is parallel to the other.

BOOK VI.

203

Let AB and CD be parallel lines, and MW any plane passed
through CD ; then the line AB and
the plane MN are parallel.

For the parallels AB, CD, are in
the same plane, A CDB, which in-
tersects the plane MN in the line
CD; and since AB cannot leave

this plane, if it meets MN at all it must meet it in some
point common to the two planes; that^i^, in some point of
CD, which is contrary to the hypothesis that AB and CD
are parallel.

17. Corollary I. Through any given straight line a plane
can be passed parallel to any other given straight line.

Let UK and AB be the two given
lines. In the plane determined by AB
and any point S of UK let HL be
drawn parallel to AB ; then the plane
MNj determined by SK and HL, is par-
allel to AB, by Proposition YI.

18. Corollary II. Through any given point a plane can Ir*
passed parallel to any two given straight lines in space.

Let 0 be the given point, and AB
and CD the given straight lines. In
the plane determined by the given
point 0 and the line AB let a 06 be
drawn through 0 parallel to AB ;
and in the plane determined by the
point 0 and the line CD let cOd be
drawn through 0 parallel to CD;

then the plane determined by the lines ab and cd is parallel
to each of the lines AB and CD, by Proposition VI.

--A

204 ELEMENTS OF GEOMETRY.

EXERCISE.

Theorem. — If a straight line and a plane are parallel^ the
intersection of the plane and a plane passed through the given
line is parallel to the given line. (v. Figure of Proposition VI.)

PROPOSITION VII.— THEOREM.

(10} Planes perpendicular to the same straight line are parallel
to each other.

The planes MW, PQ, perpendicular to the jn
same straight line AB, cannot meet ; for, if /
they met, we should have through a point L
of their intersection two planes perpendic-
ular to the same straight line, which is im- ^
possible (Proposition IV., Corollary II.); /
therefore these planes are parallel. ;^ I q \(\

PROPOSITION VIII.— THEOREM.

^2J The intersections of two parallel planes with any third
plane are parallel.

Let MW and PQ be parallel planes, r y\ y

and AD any plane intersecting them \ /^^ '\ \

in the lines AB and CD; then AB A \~

and CD are parallel. p \ \i)

For the lines AB and CD cannot \ \ / \

meet, since the planes in which they ^ ^ ^

are situated cannot meet, and they

are lines in the same plane AD ; therefore they are parallel.

EXERCISE.

Theorem. — Parallel lines intercepted between parallel planes
are equal.

BOOK VI. 205

PROPOSITION IX.— THEOREM.

21. A straight line perpendicular to one of two parallel planes
is perpendicular to the other.

Let MW and FQ be parallel planes, and let
the straight line AB be perpendicular to PQ;
then it will also be perpendicular to MN.

For through A draw any straight line AG
in the plane JOT, pass a plane through AB
and AC, and let BD be the intersection of
this plane with PQ. Then AC and BD are
parallel (Proposition YIII.) ; but AB is per-
pendicular to BD (6), and consequently also to AC; there-
fore AB, being perpendicular to any line AC which it meets
in the plane JfiV", is perpendicular to the plane JOT.

22. Corollary. Through any given point one plane can be
passed parallel to a given plane, and but one.

Suggestion. Drop a perpendicular line from the point to the

plane, and then pass a plane through the point perpendicular

to this line.

PROPOSITION X.— THEOREM.

23. If two angles, not in the same plane, have their sides
respectively parallel and lying in the same direction, they are
BQual and their planes are parallel.

Let BAC, B'A'C, be two angles lying
in the planes MN, M'N' ; and let AB,
AC, be parallel respectively to A'B',
A!C', and in the same directions.

1st. The angles BAC and B'A!C' are
equal. Take the distances A!B' and AB
equal, and A!C' and AC equal, and join
BC and B' C . Draw now the lines AA',
BB', and Cp'. The quadrilaterals AB' and AC are parallel-
is

206

ELEMENTS OF GEOMETRY.

ograms, by I., Proposition XXX. Therefore BB' and CC

are equal and parallel to AA'. They are

then equal and parallel to each other

(Proposition Y., Corollary), and BC is a

parallelogram. Hence BC = B'C\ and

the triangles A5(7and A'B'C are equal,

by I., Proposition IX. Consequently

the angles BAG dmd. B'A'C are equal.

2d. The planes MN and M'N' are
parallel. For MN is parallel to A'B'
and A'C\ by Proposition YI., and therefore, since if it met
M'N' the line of intersection would have to cut one or the
other of the intersecting lines A!B'^ A!G\ it is parallel to
M'W,

?. 1

^ A'

PROPOSITION XI.— THEOREM.

Yf one of two parallel lines is perpendicular to a plane,
the other is also perpendicular to that plarie.

Let AB^ A'B\ be parallel lines, and
let AB be perpendicular to the plane
MN; then A'B^ is also perpendicular
to MN.

For, let A and A' be the intersections
of these lines with the plane ; through

A' draw any line A'C in the plane MN, and through A draw
AG parallel to A'G' and in the same direction. The angles
BAG, B'A'G', are equal (Proposition X.) ; but BAG m 2i right
angle, since BA is perpendicular to the plane; hence B'A'G'
is a right angle ; that is, B'A' is perpendicular to any line
A'G' drawn through its foot in the plane MN, and is conse-
quently perpendicular to the plane. q

vi^..

V("

»*\- V ,' •-'^•

\ ■-^V^-.■c.I3t>v^5r-v

BOOK VI. 207

25. Corollary. Two straight lines perpendicular to the same
plane are parallel to each other,

EXERCISE.

Theorem. — Two parallel planes are everywhere equally distant,

BIEDRAL ANGLES.— ANGLE OF A LINE AND PLANE,

ETC.

26. Definition. When two planes meet and are terminated
by their common intersection, they form a diedral angle.

Thus, the planes AE, AF^ meeting in AB^ and
terminated by AB^ form a diedral angle.

The planes AE^ AF, are called the faceSy and
the line AB the edge, of the diedral angle.

A diedral angle may be named by four letters,
one in each face and two on its edge, the two
on the edge being written between the other two ; thus, the
angle in the figure may be named DABC.

When there is but one diedral angle formed at the same?
edge, it may be nariied by two letters on its edge ,- thus, in
the preceding figure, the diedral angle DABC may be named
the diedral angle AB.

27. Definition. The angle CAD formed hj two straight lines
AC, AD, drawn, one in each face of the diedral angle, per-
pendicular to its edge AB at the same point, is called the
plane angle of the diedral angle.

EXERCISES.

1. Theorem. — All plane angles of the same diedral angle are
equal, (v. Proposition X.)

2. Theorem. — If a plane is drawn perpendicidar to the edge
uj a diedral angle, its intersections with the faces of the diedral
angle form the plane angle of the diedral angle.

208

ELEMENTS OF GEOMETRY.

28. A diedral angle DABG may be conceived to be gener-
ated by a plane, at first coincident with a fixed plane AE,
revolving upon the line A ^ as an axis until it comes into the
position AF. In this revolution a straight line CA, perpen-
dicular to AB, generates the plane angle CAD.

29. Definition. Two diedral angles are equal when they
can be placed so that their faces shall coincide.

Thus, the diedral angles CABD,
G'A'B'D', are equal, if, when the
edge A'B' is applied to the edge
AB and the face A'F' to the face
AF^ the face A'E' also coincides
with the face AE.

Since the faces continue to co-
incide when produced indefinitely, it is apparent that the
magnitude of the diedral angle does not depend upon the
extent of its faces, but only upon their relative position.

30. Definition. Two diedral angles GABD, DABEj which
have a common edge AB and a common plane

BD between them, are called adjacent.

Two diedral angles are added together by
placing them adjacent to each other. Thus, the
diedral angle GABE is the sum of the two diedral
angles GABD and DABE.

,,^

31. Definition. Two planes are perpen-
dicular when the plane angle of the di-
edral angle which they form is a right
angle. The diedral angle is then called
a right diedral angle.

BOOK VI.

209

PROPOSITION XII.— THEOREM.

(^ Two diedral angles are equal if their plane angles are
equal.

Let the plane angles CAD and
C'A'D' of the diedral angles CABD,
C'A'B'D', be equal; then are the
diedral angles equal.

For, superpose C'A'B'iy upon
CABD, making the plane angle

C'A'D' coincide with its equal CAD ; then the planes of
these angles will coincide (Proposition II.). A'B' and AB^
being now perpendicular to the same plane at the same point,
must coincide (Proposition III., Corollary) ; and, finally, the
planes B'C and 50 will coincide, and B'D' and BD (Propo-
sition 11.) . Therefore the diedral angles are equal.

PROPOSITION XIII.— THEOREM.

331 Two diedral angles are in the same ratio as. their
angles.

Let CABD and GEFH be two '
diedral angles; and let CAD and
GEH be their plane angles.

Suppose the plane angles have
a common measure, contained m
times in CAD and n times in
GEH; we have, then,

CAD ^m
GEH n

plane

V- ~

1 1 II
1 III
1 ill
1 1 1 1
1 1 II

- -1

^^^^

-ij

Apply this measure to CAD and GEH^ and through the
lines of division and the edges of the given diedral angles

o 18*

210

ELEMENTS OF GEOMETRY.

pass planes, thus dividing CABD into m and GEFII into n
smaller diedral angles. Each of
these small diedral angles has one
of the parts into which CAD is
divided, or one of the parts into
which GEH is divided, as its
plane angle, because AB is per-
pendicular to the plane of CAD,
and EF to the plane of GEH, by

Proposition lY. These small diedral angles are, then, all
equal, by Proposition XII., and we have

?Fr::rl

^^-\

■\i

Therefore

CABD ^m
GEFH n'

CABD ^ CAD
GEFH GEH'

The proof is extended to the case where the given planej
angles are incommensurable, by the method exemplified in
the proof of II., Proposition XII., of III., Proposition I., and
of IV., Proposition II.

34. Scholium. Since the diedral angle is proportional to its
plane angle (that is, varies proportionally with it), the plane
angle is taken as the measure of the diedral angle, just as an
arc is taken as the measure of a plane angle. Thus, a diedral
angle will be expressed by 45° if its plane angle is expressed
by 45°, etc.

PROPOSITION XIV.— THEOREM.

V^Sa If a straight line is perpendicular to a plane, every plane
passed through the line is also perpendicular to that plane.

BOOK

211

Let AB be perpendicular to the plane MJSf; then any plane
PQ, passed through AB, is also perpendic-
ular to JfiV.

For, at B draw BC^ in the plane MWj
perpendicular to the intersection BQ. Since
AB is perpendicular to the plane 3IW, it is
perpendicular to BQ and BG; therefore
the angle ABC is the plane angle of the
diedral angle formed by the planes PQ and MN; and since
the angle ABC is a right angle, the planes are perpendicular
to each other.

PROPOSITION XV.-THEOBEM.

(^g) If two planes are perpendicular to each other, a straight
line drawn in one of them, perpendicular to their intersection, is
perpendicular to the other.

Let the planes PQ and JfZV be perpendic-
ular to each other ; and at any point B of
their intersection BQ let BA be drawn, in
the plane PQ, perpendicular to BQ; then
BA is perpendicular to the plane 3fN.

For, drawing BC, in the plane MW, per-
pendicular to BQ, the angle ABC is a right angle, since it is
the plane angle? of the right diedral angle formed by the two
planes ; therefore AB, perpendicular to the two straight lines
BQ, BC, is perpendicular to their plane MW (Proposition
IV.).

37. Corollary I. If two planes are perpendicular to each
other, a straight line drawn through any point of their intersec-
tion perpendicular to one of the planes will lie in the other, (v.
Proposition III., Corollary.)

212

ELEMENTS OF GEOMETRY.

38. Corollary II. If two planes are perpendicular^ a straight
line let fall from any point of one plane perpendicular to the
other will lie in the first plane, (v. Proposition III.)

PROPOSITION XVI.— THEOREM.

39. If two intersecting planes are each perpendicular to a
third plane, their intersection is also perpendicular to that plane.

Let the planes PQ, BS, intersecting in
the line AB, be perpendicular to the plane
MN ; then AB is perpendicular to the
plane M]Sf.

For, if from any point A of A B a per-
pendicular be drawn to MNj this perpen-
dicular will lie in each of the planes PQ
and BS (Proposition XY., Corollary II.),
and must therefore be their intersection AB.

EE3

PROPOSITION XVII.— THEOREM.

fwi Through any given straight line a plane can be passed
perpendicular to any given plane.

Let AB be the given straight line, and
MW the given plane. From any point A
of AB let AC he drawn perpendicular to
MN, and through AB and A C pass a plane
AD. This plane is perpendicular to MN
(Proposition XIY.).

Moreover, since, by Proposition XY., Corollary IL, any
plane passed through AB perpendicular to MN must contain
the perpendicular A C, the plane AD is the only plane per-
pendicular to MN that can be passed through AB, unless AB
is itself perpendicular to MN, in which case every plai^
through AB is perpendicular to MN. ^

BOOK VI.

213

EXERCISE.

Theorem. — The locus of the
points equally distant from
two given -planes is ttie plane
bisecting the diedral angle be-
tween the given planes, (v.
I., Proposition XIX.)

41. Definitions. The projection of a
point A upon a plane MN is the foot a
of the perpendicular let fall from A
upon the plane.

The projection of a line ABODE . . .
upon a plane MN is the line abcde . . .
containing the projections of all the
points of the line ABODE . . . upon the
plane.

PROPOSITION XVIII.— THEOREM.

(42^ The projection of a straight line upon a plane is a straight
line.

Let AB be the given straight line,
and MI^ the given plane. The plane
Ab, passed through AB perpendicular
to' the plane MN, contains all the per-
pendiculars let fall from points of AB

upon MN (Proposition XY., Corollary /V ^ i^u\t i *^

II.); therefore these perpendiculars all t^t^^ . ^^^'^a-y.^AJ

meet the plane MN in the intersection ab of the perpeiidic- i , V^
ular plane with MN. The projection of AB upon the plane
MN is, consequently, the straight line ab.

214

ELEMENTS OF GEOMETRY.

43. Scholium. The plane Ah is called the projecting plane of
the straight line AB upon the plane MN.

PKOPOSITION XIX.— THEOREM.

44. The acute angle which a straight line makes with its own
projection upon a plane is the least angle which it makes with
any line of that plane.

Let Ba be the projection of the straight
line BA upon the plane MN, the point B
being the point of intersection of the line
BA with the plane ; let BC be any other
straight line drawn through B in the
plane; then the angle ABa is less than
the angle ABC.

For, take BG = Ba, and join AC. In the triangles ABa,
ABC, we have AB common, and Ba = BC; but Aa <i AC,
since the perpendicular is less than any oblique line ; therefore
the angle ABa is less than the angle ABC(1., Proposition XY.).

45. Definition. The acute angle which a straight line makes
with its own projection upon a plane is called the inclination
of the line to the plane, or the angle of the line and plane.

46. Definition. Two straight lines AB,
CD, not in the same plane, are regarded
as making an angle with each other which
is equal to the angle between two straight
lines Ob, Od, drawn through any point O
in space, parallel respectively to the two
lines and in the same directions.

47. From the preceding definition, it follows that when a
straight line is perpendicular to a plane, it is perpendicular to all
the lines of the plane, whether the lines pass through its foot
or not.

BOOK VI. 215

POLYEDRAL ANGLES.

48. Definition. When three or more planes meet in a com-
mon point, they form a polyedral angle.

Thus the figure S-ABCD, formed by the
planes ASB, BSC, CSD, DSA, meeting in
the common point S^ is a polyedral angle.

The point S is the vertex of the angle ;
the intersections of the planes SA^ SB,
etc., are its edges; the portions of the
planes included between the edges are its faces ; the angles
ASB^ BSG, etc., formed by the edges, are its face angles.

A triedral angle is a polyedral angle having but three faces,
which is the least number of faces that can form a polyedral
angle.

49. In a polyedral angle every pair of adjacent edges form
a face angle, and every pair of adjacent faces form a diedral
angle. These face angles and diedral angles are the parts of
the polyedral angle.

50. Definition. Two polyedral angles are equal when their
faces and edges can be made to coincide, if one angle is suit-
ably superposed upon the other.

Of course it follows that corresponding parts of two equal
polyedral angles are equal.

51. Definition. Two polyedral angles are symmetrical if the
parts of one are respectively equal to the parts of the other ;
but the corresponding parts succeed each other in the two
angles in inverse order. When two polyedral angles are
(symmetrical, it is impossible to superpose one upon the other
in such a way as to bring corresponding parts together. One
figure is, so to speak, right-handed and the other left-handed.

216 ELEMENTS OF GEOMETRY.

52. Definition. A polyedral angle S-ABCD is convex, when
any section, ABCD, made by a plane cutting all its faces, is
a convex polygon (I., 54).

PROPOSITION XX.— THEOREM.

)-The sum of any two face angles of a triedral angle is
greater than the third.

The theorem requires proof only when the third angle
considered is greater than each of the others.

Let S-ABC be a triedral angle in which
the face angle ASC is greater than either
ASB or BSC; then ASB + BSG > ASC.

For in the face ASC draw SD making the
angle ASD equal to ASB, and through any
point D of SB draw any straight line ADC
cutting SA and SG; take SB = SD, and join AByBG.

The triangles ASD and ASB are equal, by the construction
(I., Proposition YI.), whence AD = AB. Now, in the tri-
angle ABC, we have

AB + BC>AG,

and, subtracting the equals AB and AD,

BC > DC;

therefore, in the triangles BSC and DSC, we have the angle
BSC > DSC (I., Proposition XY.), and adding the equal
angles ASB and ASD, we have ASB + BSC >-ASC.

BOOK VT. 217

PROPOSITION XXI.— THEOREM.

[5^ The sum of the face angles of any convex polyedral angle
is less than four right angles.

Let the polyedral angle S be cut by a plane, making the
section ABCDE^ by hypothesis, a convex
polygon. From any point 0 within this
polygon draw OA, OB, OC, OB, OK

The sum of the angles of the triangles
ASB, BSC, etc., which have the common
vertex S, is equal to the sum of the an-
gles of the same number of triangles
A OB, BOO, etc., which have the common
vertex 0. But in the triedral angles formed at A, B, 0, etc.,
by the faces of the polyedral angle and the plane of the
polygon, we have (Proposition XX.)

SAB + SAB > EAB,
SBA + SBC> ABC, etc. j

hence, taking the sum of all these inequalities, it follows that
the sum of the angles at the bases of the triangles whose
vertex is S is greater than the sum of the angles at the
bases of the triangles whose vertex is 0; therefore the sum
of the angles at >S^ is less than the sum of the angles at 0;
that is, less than four right angles.

218

ELEMENTS OF GEOMETRY.

*>C?

PROPOSITION XXII.-THEOREM.

55. If two triedral angles have the three face angles of the one
respectively equal to the three face angles of the other^ the cor-
responding diedral angles are equal.

In the triedral angles S and s, let ASB = ash, ASC = asc,
and BSC = bsc ; then the diedral angle SA is equal to the
diedral angle sa.

.^ On the edges of these angles take the six equal distances

SA, SB, SC, sa, sb, sc, and draw AB, BC, AC, ah, he, ac. The
isosceles triangles SAB and sab are equal, having an equal
angle included by equal sides, hence AB = ah ; and for the
same reason, BG = hc, AC = ac ; therefore the triangles
ABC and abc are equal.

At any point D in SA, draw DE in the face ASB and DF
in the face ASC, perpendicular to SA; these lines meet AB
and AC, respectively, for, the triangles ASB and ASC being
isosceles, the angles SAB and SAC are acute; let E and F
be the points of meeting, and join EF. Now on sa take sd
= SB, and repeat the same construction in the triedral
angle s.

The triangles ADE and ade are equal, since AB = ad, and
the angles at A and B are equal to the angles at a and d;
hence AE = ae and BE = de. In the same manner we have

BOOK VI. 219

AF = of and DF = df. Therefore the triangles AEF and
aef are equal (I., Proposition YI.)j and we have FF = ef.
Finally, the triangles FDF and edf, being mutually equilat-
eral, are equal; therefore the angle FDF, which measures
the diedral angle SA, is equal to the angle edf, which meas-
ures the diedral angle sa, and the diedral angles SA and sa
are equal (Proposition XII.). In the same manner it may
be proved that the diedral angles SB and SO are equal to
the diedral angles sb and sc, respectively.

Scholium. It follows that the polyedral angles >S^ and s are
either equal or symmetrical. Both cases are represented in
the figure.

EXERCISES ON BOOK YI.

THEOREMS.

1. If a straight line AB is parallel to a plane JfJV, any plane
perpendicular to the line AB is perpendicular to the plane 3lJS\
(v. Proposition VI., Exercise.)

2. If a plane is passed through one of the diagonals of a paral-
lelogram, the perpendiculars to this plane from the extremities
of the other diagonal are equal.

3. If the intersections' of a number of planes are parallel, all the
perpendiculars to these planes, drawn from a common point in
space, lie in one plane.

Suggestion. Through the common point pass a plane perpen-
dicular to one of the intersections, {v. Proposition XV., Corol-
lary II.)

4. If the projections of a number of points on a plane are in a
straight line, these points are in one plane.

5. If each of the projections of a line AB upon two intersecting
planes is a straight line, the line AB is a straight line.

6. Two straight lines not in the same plane being given : 1st, a
common perpendicular to the two lines can be drawn ; 2d, the
common perpendicular is the short- .

est distance between the two lines. o e d

Suggestion. Let AB and CD be
the two given lines. Pass through
AB a plane MN parallel to CX>, and
through AB and CD pass planes
perpendicular to MN. Their inter-
section Co is the required common
perpendicular. CD and cd are par-
allel, by 18, Exercise. N

2d. Any other line BF joining
AB and CD is greater than JEIT, the perpendicular from B to cd
(Proposition XV.), and therefore greater than Cc.
220

,1 1

l'\

/i i /

V "

' J

BOOK VI.

7. If two straight lines are intersected by
three parallel planes, their corresponding
segments are proportional, {v. Proposition
VIII.)

8. A plane passed through the middle point of the common
perpendicular to two straight lines in space, and parallel to both
these lines, bisects every straight line joining a point of one of
these Hues to a point of the other, {v. Exercise 7.)

9. In any triedral angle, the three planes bisecting the three
diedral angles intersect in the same straight line. {v. 40, Exer-
cise.)

10. In any triedral angle, the three planes
passed through the edges and the bisectors
of the opposite face angles respectively in-
tersect in the same straight line.

Suggestion. Lay off equal distances SAj
SB, SC, on the three edges, and pass a plane
through Aj B, C. The intersections of the
three planes in question with ABCsive the

medial lines of ABC, and have a common intersection, and the line
joining this common intersection with S lies in the three planes.

11. In any triedral angle, the three planes passed through the
bisectors of the face angles, and perpendicular to these faces
respectively, intersect in the same straight line.

Suggestion. Use the same construction as in Exercise 10. Then
the intersections of the three planes with ABC are perpendicular
to the sides of ABC at their middle points, and have a common
intersection.

12. In any triedral angle, the three planes
passed through the edges, perpendicular to
the opposite faces respectively, intersect in
the same straight line.

Suggestion. At any point A of one of the
edges, draw a plane ABC perpendicular to
the edge SA. The intersections of the three
planes with ABC are the perpendiculars
from the vertices of ABC^ upon the oppo-
site sides, and have a common intersection, {v. Proposition XVI.)

19*

222 ELEMENTS OF GEOMETRY.

LOCI.

13. Find the locus of the points in space which are equally
distant from two given points.

14. Locus of the points which are equally distant from two
given straight lines in the same plane.

15. Locus of the points which are equally distant from three
given points.

16. Locus of the points which are equally distant from three
given planes, {v. 40, Exercise.)

17. Locus of the points which are equally distant from three
given straight lines in the same plane.

\ 18. Locus of the points which are equally distant from the three
edges of a tried ral angle (Exercise 11).

19. Locus of the points in a given plane which are equally dis-
tant from two given points out of the plane.

20. Locus of the points which are equally distant from two
given planes, and at the same time equally distant from two
given points.

PROBLEMS.

In the solution of problems in space, we assume, — 1st, that a
plane can be drawn passing through three given points (or two in-
tersecting straight lines) and its intersections with given straight
lines or planes determined; and, 2d, that a perpendicular to a
given plane can be drawn at a given point in the plane, or from
a given point without it. The actual graphic construction of the
solutions belongs to Descriptive Geometry.

21. Through a given straight line, to pass a plane perpendicular
to a given plane, (v. Proposition XVII.)

22. Through a given point, to pass a plane perpendicular to a
given straight line.

Suggestion. If the given point is in the given line, pass two
planes through the given line, and draw in each of them, through
the given point, a line perpendicular to the given line. The plane
determined by these lines is the perpendicular plane required.
{v. Proposition IV.)

If the given point is not in the given line, pass a plane through
it and the given line, and in this plane, through the given point,
draw a line parallel to the given line. A plane through the given
point, perpendicular to this second line, is the plane required, (v.
Proposition XI.)

BOOK VI. 223

23. Through a given point, to pass a plane parallel to a given
plane, (v. Proposition IX., Corollary.)

24. To determine that point in a given straight line which is
equidistant from two given points not in the same plane with
the given line. (v. Exercise 13.)

25. To find a point in a plane which shall be equidistant from
three given points in space.

26. Through a given point in space, to draw a straight line
which shall cut two given straight lines not in the same plane.

Suggestion. Pass a plane through the given point and through
one of the given lines ; the line through the given point and the
point where the plane cuts the second given line is the solution
required.

27. Through a given point, to draw a straight line which shall
meet a given straight line and the circumference of a given circle
not in the same plane. (Two solutions in general.)

28. In a given plane and through a given point of the plane, to
draw a straight line which shall be perpendicular to a given line
in space.

Suggestion. Draw a plane through the given point and perpen-
dicular to the given line. Its intersection with the given plane is
the solution required.

29. Through a given point ^ in a plane, to draw a straight line
AT in that plane, which shall be at a given distance PT from a
given point P without the plane.

Suggestion. Drop a perpendicular from P to the plane, and with
the foot of this perpendicular as a centre, and with a radius equal
to a side of a right triangle whose hypotenuse is PT, and whose
other side is the length of the perpendicular, describe a circum-
ference in the plane. A tangent from A to this circumference is
the solution required, {v. 12, Exercise 2.)

30. Through a given point A, to draw to a given plane M a
straight line which shall be parallel to a given plane N and of a
given length.

BOOK YIL

POLYBDRONS.

1. Definition. A polyedron is a geometrical solid bounded
by planes.

The bounding planes, by their mutual intersections, limit
each other, and determine the faces (which are polygons),
the edgeSj and the vertices of the polyedron. A diagonal of a
polyedron is a straight line joining^ any two of its vertices
not in the same face.

The least number of planes that can form a polyedral
angle is three ; but the space within the angle is indefinite in
extent, and it requires a fourth plane to enclose a finite por-
tion of space, or to form a solid ; hence the least number of
planes that can form a polyedron is four.

2. Definition. A polyedron of four faces is called a tetra-
edron ; one of six faces, a hexaedron ; one of eight faces, an
octaedron ; one of twelve faces, a dodecaedron ; one of twenty
faces, an icosaedron.

3. Definition. A polyedron is convex when the section
formed by any plane intersecting it is a convex polygon.

All the polyedrons treated of in this work will be under-
stood to be convex.

4. Definition. The volume of any polyedron is the numer-
ical measure of its magnitude, referred to some other poly-
edron as the unit. The polyedron adopted as the unit is
called the unit of volume.

To measure the volume of a polyedron is, then, to find its
ratio to the unit of volume.
224

BOOK VII. 225

The most convenient unit of volume is the cube whose
edge is the linear unit.

5. Definition. Equivalent solids are those which have equal
volumes.

PEISMS AND PARALLEL0PIPED8.

6. Definitions. A prism is a polyedron two of whose oppo-
site faces, called hases^ are in parallel planes,

and whose lateral edges (that is, the edges
intersecting the bases) are all parallel to the
same line.

From this definition werreadily deduce the
following consequences :

1st. Any two lateral edges of a prism are
parallel (VI., Proposition Y., Corollary).

2d. All the lateral faces of a prism are parallelograms (YI.,
Proposition YIII.). Hence all the lateral edges are equal.

3d. The bases of a prism are equal polygons (YI., Proposi-
tion X.).

The lateral faces of' a prism constitute its lateral or convex
surface.

The altitude of a prism is the perpendicular distance be-
tween the planes of its bases (YI., Proposition IX.)

A triangular prism is one whose base is a triangle ; a quad-
rangular prism, one whose base is a quadrilateral ; etc.

7. Definitions. A right prism is one whose lateral
edges are perpendicular to the planes of its faces
(YI., Proposition XI.)

In a right prism, any lateral edge is equal to
the altitude.

The lateral faces of a right prism are perpen-
dicular to the bases (YI., Proposition XIY.)

226

ELEMENTS OF GEOMETRY.

An oblique prism is one whose lateral edges are oblique to
the planes of its bases.

In an oblique prism, a lateral edge is greater than the altitude.

8. Definition. A regular prism is a right prism whose bases
are regular polygons.

8. Definition. Ifaprism,A5CZ)jE^-JF;
is intersected by a plane GK, not
pj^rallel to its base, the portion of
the prism included between the base
and this plane, namely, \A 5 (7D.&-
QHIKLy is called a truncated prism.

10. Definition. A right section of a prism is the section made
by a plane passed through the prism perpendicular to one of
its lateral edges.

A right section is perpendicular to all the lateral edges
(YI., Proposition XI.) and to all the lateral faces (YI., Propo-
sition XIY.) of the prism.

11. Definition. A parallelopiped is a prism
whose bases are parallelograms. It is there-
fore a polyedron all of whose faces are par-
allelograms.

From this definition and YI., Proposi-
tion X., it is evident that any two oppo-
site faces of a parallelopiped are equal parallelograms.

12. Definition. A right parallelopiped is a par-
allelopiped whose lateral edges are perpendic-
ular to the planes of its bases. Hence, by YI.,
6, its lateral faces are rectangles ; but its bases
may be either rhomboids or rectangles.

BOOK VII.

227

A rectangular parallelopiped is a right parallelepiped whose
bases are rectangles. Hence it is a parallelopiped all of
whose faces are rectangles.

13. Definition. A cube is a rectangular parallelo-
piped whose edges are all equal. Hence its faces
are all squares.

PROPOSITION I.— THEOREM.

14. The sections of a prism made by parallel planes are equal
polygons.

For the portion of the prism in-
cluded between the two sections is a
new prism (6). Therefore its bases,
which are the sections in question,
are equal.

15. Corollary. Any section of a
prism made by a plane parallel to the
base is equal to the base.

EXERCISE.

Theorem. — In a rectangular parallelopiped, the four diagonals
are equal to each other ; and the square of a diagonal is equal
to the sum of the squares of the three edges which meet at a
common vertex.

228

ELEMENTS OF GEOMETRY.

PROPOSITION II.— THEOREM.

16. The lateral area of a prism is equal to the product of the
perimeter of a right section of the prism by a lateral edge.

Let AD' be a prism, and GHIKL a
right section of it; then the area of
the convex surface of the prism is equal
to the perimeter GHIKL multiplied by
a lateral edge AA'.

For, the sides of the section GHIKL^
being perpendicular to the lateral edges
AA', BB' (YI., 6), etc., are the altitudes
of the parallelograms which form the
convex surface of the prism, if we take as the bases of these
parallelograms the lateral edges AA\ BB\ etc., which are all
equal (6). Hence the area of the sum of these parallelo-
grams is

GH X AA' -\-HIX BB' + etc.
= {GH-\-HI+ etc.) X AA'.

17. Corollary. The lateral area of a right prism is equal
to the product of the perimeter of its base by its altitude.

PROPOSITION III.— THEOREM.

18. Two prisma are equal, if three faces including a triedral
angle of the one are respectively equal to three faces similarly
'placed including a triedral angle of the other,—

Let the triedral angles A
and a of the prisms ABCDE-
A\ abcde-a', be contained by-
equal faces similarly placed,
namely, ABCDE equal to
abcde, AB' equal to ab', and
AE' equal to aef ; then the
prisms are equal.

BOOK VII. 229

For, superpose the second prism upon the first, making the
base ahcde coincide with the equal base ABCDE. Since the
diedral angles ah and AB are equal and ae and AE are equal
(YI., Proposition XXII.), the plane ah' will coincide with the
plane AB'^ and the plane ae' with the plane AE', Hence the
intersection aa' will fall along the intersection AA', As the
faces ah' and AB' are equal, and have now been suitably
superposed, they must coincide throughout^ and a'h' will
coincide with A'B'. For the same reason,^ a'e' will coincide
with A'E'. Consequently, the plane determined by a'h' and
aV, namely, the plane of the upper base of the second prism,
will coincide with the plane of the upper base of the first
prism. Any lateral edge, as eef^ will fall along the corre-
sponding lateral edge EE'^ for they are now parallel to the
same line AA'^ and have a point e of one coinciding with a
point E of the other. They have thus the same direction
^ and a point in common, and must coincide throughout^ by I.,
Postulate II.

Since all the lateral edges of the second prism coincide
with the corresponding lateral edges of the first,, the planes
of all the corresponding lateral faces must coincide. There-
fore, as all the corresponding faces of the two prisms coincide
(the bases included), the prisms are equal.

19. Corollary I. Two truncated prisms are equal, if three
faces including a triedral angle of the one are respectively equal
to three faces similarly placed including a triedral angle of the
other. For the preceding demonstration applies whether the
planes A'B'C'D'E' and a'h'c'dle are parallel or inclined to the
lower bases.

20. Corollary II. Two right prisms are equals if they have
equal bases and equal altitudes.

230

ELEMENTS OF GEOMETRY.

In the case of right prisms, it is not necessary to add the
condition that the faces shall be
similarly placed ; for if the two
right prisms ABC-A! ^ dbc-a\ can-
not be made to coincide by placing
the base ABC upon the equal base
ahc ; yetj by inverting one of the
prisms and applying the base
ABC to the base a'6V, they will
coincide.

PROPOSITION IV.— THEOREM.

V 21. Any oblique prism is equivalent to a right prism whose
base is a right section of the oblique prism, and whose altitude is
equal to a lateral edge of the oblique prism.

JjQtABCDE-A^hG the oblique prism.
At any point F in the edge AA\ pass
a plane perpendicular to AA' and
forming the right section FGHIK.
Produce AA' to F\ making FF' z=
AA\ and through F' pass a second
plane perpendicular to the edge AA',
intersecting all the faces of the prism
produced, and forming another right
section F'G'H'I'K' parallel and equal

to the first. The prism FGHIK-F' is a right prism whose
base is the right section and whose altitude FF' is equal to
the lateral edge of the oblique prism.

The solid ABCDE-F is a truncated prism which is easily
shown to be equal to the truncated prism A'B'C'D'E'-F'
(Proposition III., Corollary I.). Taking the first away from
the whole solid ABCDE-F', there remains the right prism ;

BOOK VII.

231

taking the second away from the same solid, there remains
the oblique prism ; therefore the right prism and the oblique
prism have the same volume ; that is, they are equivalent.

^V^..^^ PROPOSITION v.— THEOREM.

26. Any parallelopiped is equivalent to a rectangular parallel-
opiped of the same altitude and an equivalent base.

Let ABGD^A' be
any oblique parallel-
opiped whose base is
A BCD J and altitude
B'O.

Produce the edges
AB,A'B',I)C,D'C';
in AB produced
take FG = AB, and
through F and G

pass planes FF'FI, GG'H'H, perpendicular to the produced
edges ; then the given parallelopiped and the right parallelo-
piped FF'FI-H are equivalent, by Proposition lY.

Produce, now, the edges of this second parallelopiped ZF,
FF\HG, H'G'; in IF produced take NK^IF, and through
.N and K pass planes KLL'K' smd NMM'N' perpendicular
to the produced edges. Then the second parallelopiped and
the parallelopiped NMM'W-K are equivalent, by Proposi-
tion lY. Consequently, the given parallelopiped and the
parallelopiped NMM'N'-K are equivalent. The last-named
parallelopiped is a right parallelopiped, by construction, since
the face KLL'K' was drawn perpendicular to the lateral
edges. Moreover, as the planes KL' and KN' are perpendic-
ular, the first to KI, the second to AG, they are perpendicular
to the plane AHK, by YI., Proposition XIY., and their inter-

232

ELEMENTS OF GEOMETRY.

section KK' is perpendicular to AHK(Y1.^ Proposition XYI.)>
and therefore to KL

(YL, 6). Hence the ^ ^' '' ^'

base KLL'K' is a
rectangle, and the
par allelopiped
Xii'X'-iV^is a rec-
tangular parallelopi-
ped. If, now, we
take KIjMN as its
base, its altitude is

equal to that of the given parallelopiped, since the planes
AUK and A'H'K' are parallel; and its base is equivalent
to ABGD^ since each of them is equivalent to FGHI (lY.,
Proposition I.).

^ PROPOSITION VI.— THEOREM.

23. The plane passed through two diagonally opposite edges
of a parallelopiped divides it into two equivalent triangular
prisms.

Let ABCD-A' be any parallelopi-
ped ; the plane A CCA', passed through
its opposite edges AA' and CC, divides
it into two equivalent triangular prisms
ABC- A' and ADC-A\

Let FGHI be any right section of
the parallelopiped, made by a plane
perpendicular to the edge A A'. The

intersection, FH, of this plane with the plane AC is the
diagonal of the parallelogram FGHI, and divides that par-
allelogram into two equal triangles, FGH and FIH. The
oblique prism ABG-A' is equivalent to a right prism whose

BOOK VII.

233

base is the triangle FGH and whose altitude is AA' (Propo-
sition lY.); and the oblique prism ADC- A' is equivalent to
a right prism whose base is the triangle FIH and whose alti-
tude is AA\ The two right prisms are equal (Proposition
III., Corollary II.) ; therefore the oblique prisms, which are
respectively equivalent to them, are equivalent to each other.

i- PKOPOSITION VII.— THEOREM.

24. Tioo rectangular parallelopipeds having equal bases are to
each other as their altitudes.

Let P and Q be two rectangular
parallelopipeds having equal bases,
and let AB and CD be their alti-
tudes.

Ist. Suppose the altitudes have
a common measure, which is con-
tained m times in AB and n times
in CD. Then we have

AB ^m
CD~ n'

Apply this measure to AB and CJ), and through the poinds
of division draw planes perpendicular to AB and CD. P
will thus be divided into m, and Q into n, smaller parallelo-
pipeds, all of which will be equal, by Proposition I., Corol-
lary, and Proposition III., Corollary II. Hence

\ N

\ !\

\ K

\ I\

\ \

Therefore

P^AB
i Q CD'

The proof may be extended to the case where the altitudes
are incommensurable, by the method exemplified in the proof

20*

234

ELEMENTS OF GEOMETRY.

of II., Proposition XII., of III., Proposition I., and of lY.,
Proposition II.
^'25. Scholium, The three edges of a rectangular parallelo-
$ piped which meet at a common vertex are called its dimen-
sions j and the preceding theorem may be expressed as follows :
Two rectangular parallelopipeds which have two dimensions
in common are to each other as their third dimensions.

f PROPOSITION VIII.— THEOREM.

^6. T.0 rectangular parallelopi,e,s Kaun, e.ual aUitudes
are to each other as their bases.

Let a, b, and c be the three dimen-
sions of the rectangular parallelo-
piped P; m, w, and c those of the
rectangular parallelopiped Q; the
dimension c, or the altitude, being
common.

Construct B, sl third rectangular
parallelopiped, having the dimensions
m, b, and c.

If a and m are taken as the alti-
tudes of P and JR, their bases are
equal, and, by Proposition YII.,

P^ a
1^ m*

If b and n are taken as the altitudes of R and §, theii
bases are equal, and, by Proposition VII.,

R^b, .
Q n'

and, multiplying these ratios together,

P^ a X b '
Q m X n

1
J

BOOK vir.

awRft

235

But a y^ b is the area of the base of P, awFm X w is the
area of the base of Q; therefore P and Q are in the ratio
of their bases.

27. Scholium. This proposition may also be expressed as
follows :

Two rectangular parallelopipeds which have one dimension in
common, are to each other as the products of the other two
dimensions.

PROPOSITION IX.— THEOREM.

28. Any two rectangular parallelopipeds are to each other as
the products of their three dimensions.

Let a, by and c be the three dimen-
sions of the rectangular parallelo-
piped P; m, n, and p those of the
rectangular parallelopiped Q.

Construct P, a third rectangular
parallelopiped, having the dimen-
sions a, 6, and p. By Proposition
YII. we have

\ K

C 1

L._P_ !

P
P

and

by Proposition

YIII.,

_ axb ,

m Xn'

and,

multiplying these ratios together,

axb X c

mXnXp

\ K

236 ELEMENTS OF GEOMETRY.

PROPOSITION X.— THEOREM. /4^

The volume of a rectangular parallelopiped is equal to the
product of its three dimensions, the unit of volume being the cube
whose edge is the linear unit.

Let a, b, c, be the three dimen-
sions of the rectangular parallelo-
piped P; and let Q be the cube
whose edge is the linear unit. The
three dimensions of Q are each
equal to unity, and we have, by
the preceding proposition,

\ K

p
Q

ax b X c
1 X 1 X 1

= aXbXc.

P.,

Now, Q being takeA as the unit of volume, - is the numer-

ical measure, or volume of P, in terms of this unit (4);
therefore the volume of P is equal to the product ^ X ^ X <^-

30. Scholium I. Since the product a X b represents the
base, when c is called the altitude, of the parallelopiped, this
proposition may also be expressed as follows :

The volume of a rectangular parallelopiped is equal to the
product of its base by its altitude.

31. Scholium II. When the three dimen-
sions of the parallelopiped are each ex-
actly divisible by the linear unit, the truth
of the proposition is rendered evident by
dividing the solid into cubes, each of which
is equal to the unit of volume. Thus, if
the three edges which meet at a common

vertex A are, respectively, equal to 3, 4, and 5 times the linear
unit, these edges may be divided respectively into 3, 4, and 5

t\ ^ "^^^

V. ^ v. ^

BOOK VII. 237

equal parts, and then planes passed through the several points
of division at right angles to these edges will divide the solid
into cubes, each equal to the unit cube, the number of which
is evidently 3x4x5-

But the more general demonstration, above given, includes
also the cases in which one of the dimensions, or two of
them, or all three, are incommensurable with the linear unit.

32. Scholium III. If the three dimensions of a rectangular
parallelopiped are each equal to <2, the solid is a cube whose
edge is a, and its volume is a X <^ X <^ = <^' ; or, the volume
of a cube is the third power of its edge. Hence it is that in
arithmetic and algebra the expression " cube of a number"
has been adopted to signify the " third power of a number."

PROPOSITION XI.— THEOREM.

33. The volume of any parallelopiped is equal to the product
of the area of its base by its altitude.

For, by Proposition Y., the volume of any parallelopiped is
equal to that of a rectangular parallelopiped having an equiv-
alent base and the same altitude (30).

PROPOSITION XII.— THEOREM. '

34. The volume of a triangular prism is equal to the product
of its base by its altitude.

Let ABC-A' be a triangular prism.
In the plane of the base complete the
parallelogram ABGD, and then through
D draw a line DD' parallel to AA\ and
through DD' and CO', and DD' and
AA\ pass planes, thus constructing the

238

ELEMENTS OF GEOMETRY.

parallelopiped ABCD-D'. The given prism is half of the
parallelepiped, by Proposition YI., and
it has the same altitude.

The volume of the parallelopiped is
equal to its base BD multiplied by its
altitude (Proposition XI.) ; therefore
the volume of the triangular prism is
equal to its base ABC, the half of BD,
multiplied by its altitude.

. 35. Corollary. The volume of any prism is equal to the
product of its base by its altitude.

Let ABCDE-A' be any prism. It may
be divided into triangular prisms by planes
passed through a lateral edge AA' and the
several diagonals of its base. The volume
of the given prism is the sum of the vol-
umes of the triangular prisms, or the sum
of their bases multiplied by their common ''-

altitude, which is the base ABODE of the given prism multi-
plied by its altitude.

PYKAMIDS.

36. Definitions. A pyramid is a polyedron bounded by a
polygon and triangular faces formed by
the intersections of planes passed through b

the sides of the polygon and a common
point out of its plane ; as S-ABGDE.

The polygon, ABODE, is the base of
the pyramid; the point, S, in which the
triangular faces meet, is its vertex; the
triangular faces taken together constitute
its lateral, or convex, surface; the area of

BOOK VII. 239

this surface is the lateral area; the lines SA^ SB^ etc., in
which the lateral faces intersect, are its lateral edges. The
altitude of the pyramid is the perpendicular distance SO from
the vertex to the base.

A triangular pyramid is one whose base is a triangle; a
quadrangular pyramid, one whose base is a quadrilateral ; etc.

A triangular pyramid, having but four faces (all of which
are triangles), is a tetraedron ; and any one of its faces may
be taken as its base.

37. Definitions. A regular pyramid is one whose base is a
regular polygon, and whose vertex is in the
perpendicular to the base erected at the centre A
of the polygon. This perpendicular is called
the axis of the regular pyramid.

From this definition it can be readily shown
that the lateral edges of a regular pyramid are /
all equal, and hence that the lateral faces are /''
equal isosceles triangles, ^"^L^^

The slant height of a regular pyramid is the
perpendicular from the vertex to the base of any one of its
lateral faces. It is the common altitude of all the lateral
faces.

38. Definitions. A truncated pyramid is the portion of a
pyramid included between its base and a plane cutting all its
lateral edges.

When the cutting plane is parallel to the base, the trun-
cated pyramid is called a frustum of a pyramid. The altitude
of a frustum is the perpendicular distance between its bases.

In a frustum of a regular pyramid, the lateral faces are
equal trapezoids; and the perpendicular distance between
the parallel sides of any one of these trapezoids is the slant
height of the frustum.

240

ELEMENTS OF GEOMETRY.

PEOPOSITION XIII.— THEOREM.

39. If a pyramid is cut by a plane parallel to its base : 1st,
the edges and the altitude are divided proportionally ; 2d, the
section is a polygon similar to the base.

Let the pyramid S-ABCDE, whose alti-
tude is SOj be cut by the plane abode par-
allel to the base, intersecting the lateral
edges in the points a, 6, c, d^ e, and the alti-
tude in 0 ; then

1st. The edges and the altitude are dir
voided proportionally.

Pass a plane through the altitude SO
and any lateral edge SA, cutting the base
in AO and the section in ao. By YI.,
Proposition YIII., _,ab, be, cd, . . . ao are parallel respectively
to AB, BC, CD, ...AO. Therefore, by III., Proposition I.,

SD"

' SO'

2d. The section abode and the base are similar. For they
are mutually equiangular, by YI., Proposition X., and by
similar triangles we have

ab Sa be Sb cd Sc

AB ~'SA' BC~ 'SB' CD ~ SC

whence

ab
AB

bc_
BC

cd
CD

and the homologous sides of the polygons are proportional.
Therefore the section and the base are similar.

BOOK VII.

241

40. Corollary 1. If a pyramid is cut by a plane parallel to
its base, the area of the section is to the area of the base as the
square of its distance from the vertex is to the square of the
altitude of the pyramid.

For

abcde ^ a^ ^^ jy^ Proposition IX. ;

ABODE

Z3^' -^ ' "^

but

ab , Sa So

AB~~ SA~ SO

Therefore

abcde So

ABCDE SO'

41. Corollary II. If two pyramids have equal altitudes and
equivalent bases, sections made by planes parallel to their bases
and at equal distances from their vertices are equivalent.

PROPOSITION XIV.— THEOREM.

42. The lateral area of a regular pyramid is equal to the
product of the perimeter of its base by one-half its slant
height.

For, let S- ABCDE be a regular pyr-
amid ; the lateral faces SAB, SBC, etc.,
being equal isosceles triangles, whose
bases are the sides of the regular poly-
gon ABCDE and whose common alti-
tude is the slant height SH, the sum
of their areas, or the lateral area of
the pyramid, is equal to the sum of
AB, BC, etc., multiplied by \SH.

L g 21

242

ELEMENTS OF GEOMETRY.

43. Corollary. The lateral area of the frustum of a regular
pyramid is equal to the half sum of the perimeters of its bases
multiplied by the slant height of the frustum.

PROPOSITION XV.— THEOREM.

44. If the altitude of any given triangular pyramid is divided
into equal parts, and through the points of division planes are
passed parallel to the base of the pyramid, and on the sections
made by these planes as upper bases prisms are described having
their edges parallel to an edge of the pyramid and their altitudes
equal to one of the equal parts into which the altitude of the
pyramid is divided, the total volume of these prisms will approach
the volume of the pyramid as its limit as the number of parts
into which the altitude of the pyramid is divided is indefinitely
increased.

Let S-ABC be the given trian-
gular pyramid, whose altitude is
A T. Divide the altitude A T into
any number of equal parts Ax,
xy, etc., and denote one of these
parts by h. Through the points
of division x, y, etc., pass planes
parallel to the base, cutting from
the pyramid the sections DBF,
GUI, etc. Upon the triangles

DEF, GHI, etc., as upper bases, construct prisms whose
lateral edges are parallel to SA, and whose altitudes are each
equal to h. This is effected by passing planes through EF,
HI, etc., parallel to SA. There will thus be formed a series
of prisms DEF- A, GSI-Dj^tc, inscribed in the pyramid.

/(■^ rror^ tW

IvI^A '

BOOK VII. 243

Again, upon the triangles ABC^ DEF, GHI^ etc., as lower
bases, construct prisms whose lateral edges are parallel to
SAj and whose altitudes are each equal to h. This also is
effected by passing planes through BG, JSF, HI^ etc., parallel
to SA. There will thus be formed a series of prisms ABC-D,
DEF-G, etc., which may be said to be circumscribed about
the pyramid.

The total volume of the inscribed prisms is obviously less
and the total volume of the circumscribed prisms is obviously
greater than the volume of the pyramid.

Each inscribed prism is equivalent to the circumscribed
prism immediately above it, since they have the same base
and equal altitudes. Consequently, the difference between
the total volume of the inscribed prisms and the total volume
of the circumscribed prisms is the volume of the lowest cir-
cumscribed prism ABC-D, and therefore the difference be-
tween the total volume of the inscribed prisms and the
volume of thp pyramid is less than the volume of ABC-D.
"By increasing at pleasure the number of parts into which
the altitude AT is divided, we can make the volume of
ABC-D as small as we please, since we diminish its altitude
at pleasure without changing its base. Therefore we can
make the difference between the total volume of the in-
scribed prisms and the volume of the pyramid as small as
we please ; but, as we have seen above, we cannot make it
absolutely zero. Hence the volume of the pyramid is the
limit of the total volume of the inscribed prisms, as the
number of parts into which the altitude AT ia divided is
indefinitely increased.

244 ELEMENTS OF GEOMETRY.

PROPOSITION XVI.— THEOREM.

45. Two triangular pyramids having equivalent bases and
equal altitudes are equivalent.

Let S-ABC and 8'-A'B^C' be two triangular pyramids

G A!

having equivalent bases ABC^ A'B'C'^ in the same plane, and
a common altitude A T.

Divide the altitude A T into any arbitrarily chosen number
n of equal parts, Ax^ xy^ yz^ etc., and through the points of
division pass planes parallel to the plane of the bases, inter-
secting the two pyramids. In the pyramid S-ABG inscribe
a series of prisms whose upper bases are the sections BEF,
GHIj etc., and in the pyramid S'-A'B'C inscribe a series of
prisms whose upper bases are the sections D'B'F'., G'HT,
etc. Since the corresponding sections are equivalent (Propo-
sition XIII., Corollary II.), the corresponding prisms, having
equivalent bases and equal altitudes, are equivalent; there-
fore the sum of the prisms inscribed in the pyramid S-ABG
is equivalent to the sum of the prisms inscribed in the
pyramid S'-A'B'C; that is, if we denote the total volumes
of the two series of prisms by V and F', we have

BOOK VII.

245

no matter what the value of n. If we vary w, V and F'
obviously vary.

If n is indefinitely increased, V has the volume of the pyr-
amid S-ABG, and F' the volume of the pyramid S'-A'B'C,
as its limit (Proposition XY.). Therefore, by III., Theorem
of Limits, these volumes are equal.

PROPOSITION XVII.— THEOREM.
i|fp A triangular pyramid is one-third of a triangular prism
of the same base and altitude.

Let S-ABC be a triangular pyramid. Through A and C
draw the lines AE and CD parallel to BS,
Through AE and CD, which are parallel, by
VI., Proposition V., Corollary, pass a plane, and
through S pass a second plane parallel to ABC.
The prism ABC-E has the same base and alti-
tude as the given pyramid, and we are to prove
that the pyramid is one-third of the prism.

Taking away the pyramid S-ABC from the
prism, there remains a quadrangular pyramid whose base is
the parallelogram ACDE and vertex S. The plane SEC^
passed through SE and SC^ divides this pjrramid into two
triangular pyramids, S-AEC and S-ECD, which are equiva-
lent to each other, since their triangular bases AEC and ECD
are the halves of the parallelogram A CDEj and their common
altitude is the perpendicular from S upon the plane ACDE
(Proposition XVI.). The pyramid S-ECD may be regarded
as having ESD as its base and its vertex at C; therefore it
is equivalent to the pyramid S-ABC, which has an equivalent
base and the same altitude. Therefore the three pyramids
into which the prism is divided are equivalent to each other,
and the given pyramid is one-third of the prism.

21* ;

246 ELEMENTS OF GEOMETRY.

47. Corollary. The volume of a triangular pyramid is equal
to one-third of the product of its base by its altitude.

^^^ PEOPOSITION XVIII.— THEOREM.

48. The volume of any pyramid is equal to one-third of the
product of its base by its altitude.

For any pyramid, S-ABCDB, may be di-
vided into triangular pyramids by passing
planes through an edge SA and the diagonals
ABj ACj etc., of its base. The bases of these
pyramids are the triangles which compose
the base of the given pyramid, and their
common altitude is the altitude SO of the
given pyramid. The volume of the given ^
pyramid is equal to the sum of the volumes of the triangular
pyramids, which is one-third of the sum of their bases multi-
plied by their common altitude, or one-third of the product
of the base ABODE by tte altitude SO.

49. Scholium. The volume of any polyedron may be found
by dividing it into pyramids, and computing the volumes of
these pyramids separately. The division may be effected by
taking a point within the polyedron and joining it with all
the vertices. The polyedron will then be decomposed into
pyramids whose bases will be the faces of the polyedron, and
whose common vertex will be the point taken within it.

PROPOSITION XIX.— THEOREM.

60. A frustum of a triangular pyramid is equivalent to the
sum of three pyramids whose common altitude is the altitude of
the frustum, and whose bases are the lower base, the upper base,
and a mean proportional between the bases, of the frustum.

1 '^ '

BOOK vir. ;,'; 247

Let ABC-D be a frustum of a triangular pyramid, the
plane DEF being parallel to the base ABC.

Through the vertices A, B, and ^ J'

C pass a plane ^-E^C, and through // y^\/ \

the vertices E, D, and 0 pass a /y^ ' \y^\ \

plane ^DC, dividing the frustum /^ I / /' a\

into three pyramids. ^C^^/i'J!jf---j53^^'^

The first of these, ABG-E, has ^^^l^""'^'^

for its base the lower base of the (^^^

frustum, and for its altitude the altitude of the frustum ; the
second, DEF-C, has for its base the upper base of the frus-
tum, and for its altitude the altitude of the frustum. It
remains to show that the third, ACD-E, is equivalent to a
pyramid having for its altitude the altitude of the frustum,
and for its base a mean proportional between the bases of
the frustum.

Through E in the plane ABED draw a line EEf parallel to
AD, and through E', D, and C pass a plane. EE' is parallel
to the plane ACFD, by YI., Proposition YI. Therefore the
pyramids ACD-E and ACD-E' are equivalent, since they
have the same base and equal altitudes. If we take D as
the vertex and AE'C as the base of ACD-E' , it has for its
altitude the altitude of the frustum.

Through F in the plane ACFD draw FF' parallel to AD,
AE'F'-D is a prism, and consequently its bases DEF and
AE'F' are equal^ and E'F' is parallel to EFy and therefore

to 5a

AE'F' ^ AF'
AE'C AC'

since the triangles AE'F' and AE'C have the same altitude.

AE'C AE' n ..

= -j^-, for the same reason.
AJjC ab

248

ELEMENTS OF GEOMETRY.

AF' AE'

by III., Proposition I.
Therefore

DEF ^AE'C
AE'C ABC'

and the base of the pyramid AE'C-D is a mean proportional
between the bases of the frustum.

^1. Corollary. A frustum of any pyramid is equivalent to
the sum of three pyramids whose common altitude is the altitude
of the frustum, and whose bases are the lower base, the upper
base, and a mean proportional between the bases, of the frustum.

Suggestion, Let ABCDE-F be a frustum of any pyramid
S-ABCDE. Construct a triangular pyramid, S'-A'B'C,
having the same altitude as S-ABCDE, and a base A'B'C
equivalent to ABODE and in the same plane with it. Let
the plane of the upper base of the given frustum be pro-
duced to cut the triangular pyramid in F'G'T. The upper
bases of the frustums are equivalent, by Proposition XIII.)

BOOK VII. 249

Corollary II., and the frustums themselves are equivalent,
since the pyramids are equivalent and the pyramids above
the frustums are equivalent.

"^ PROPOSITION XX.— THEOREM.

52. A truncated triangular prism is equivalent to the sum of
three pyramids whose common base is the base of the prism^ and
whose vertices are the three vertices of the inclined section.

Let ABC-BEF be a truncated tri-
angular prism whose base is ABC and
inclined section BEF.

Pass the planes AEC and DEC,
dividing the truncated prism into the
three pyramids E-ABC, E-ACI), and
E-CBF.

The first of these pyramids, E-ABC^ has the base ABC
and the vertex E.

The second pyramid, E-A CD, is equivalent to the pyramid
B-ACB ; for they have the same base ACB, and the same
altitude, since their vertices E and B are in the line EB
parallel to this base. But the pyramid B-ACB is the same
as B-ABC ; that is, it has the base ABC and the vertex B.

The third pyramid, E-CBF, is equivalent to the pyramid
B-ACF ; for they have equivalent bases CBF and ACF in
the same plane, and also the same altitude, since their ver-
tices E and B are in the line EB parallel to that plane. But
the pyramid B-ACF is the same as F-ABC; that is, it has
the base ABC and the vertex F.

Therefore the truncated prism is equivalent to three pyra-
mids whose common base is ABC and whose vertices are E^
D, and F,

250

ELEMENTS OF GEOMETRY.

THE KEGULAR POLYEDRONS.

53. Definition. A regular polyedron is one whose faces are
all equal regular polygons and whose polyedral angles are all
equal to each other.

PROPOSITION XXI.— THEOREM.

64.. Only five regular (convex^ polyedrons are possible.

(^The faces of a regular polyedron must be regular polygons, \
and at least three faces are necessary to form a polyedral \
angle ; moreover, the sum of the face angles of a polyedral /
angle must be less than four right angles )(YI., Proposition
XXI.).

1st. The simplest regular polygon is the equilateral tri-
angle, and, since each angle of an equilat-
eral triangle is an angle of 60°, three equi-
lateral triangles «an be combined to form
a polyedral angle. It is probable, then,
that a regular polyedron can be formed
bounded by equilateral triangles and hav-
ing three at each vertex.

There is such a regular polyedron. It
has four faces, and is called the regular
tetraedron.

Since four angles of 60° are less than four right angles,
four equilateral triangles can be combined to form a polyedral

BOOK VII.

251

angle. It is probable, then, that a regu-
lar polyedron can be formed bounded by-
equilateral triangles and having four at
each vertex.

There is such a regular polyedron.
It has eight faces, and is called the regu-
lar octaedron.

Since five angles of 60° are less than four right angles,
five equilateral triangles can be combined
to form a polyedral angle. It is probable, then, that a reg-
ular polyedron can be formed bounded by
equilateral triangles and having five at each
vertex.

There is such a regular polyedron. It
has twenty faces, and is called the regular
icosaedron.

No regular polyedrons bounded by equi-
lateral triangles and having more than five at a vertex are
possible. For six or more angles of 60° cannot form a poly-
edral angle.

2d. The next regular polygon to the equilateral triangle,
in order of simplicity, is the square, each of whose angles is
a right angle.

Three right angles can be combined to form
a polyedral angle. It is probable, then, that a
regular polyedron can be formed bounded by
squares and having three at each vertex.

There is such a regular polyedron. It has
six faces, and is called the cuhe^ or the regular hexaedron.

No regular polyedrons bounded by squares and having
more than three at a vertex are possible. For four or more
right angles cannot form a polyedral angle.

252 ELEMENT8 OP GEOMETRY.

3d. The next regular polygon is the regular pentagon, each
of whose angles contains 108° (I., Proposition XXYII.).

Three angles of 108° each can be combined to form a poly-
edral angle. It is probable, then, that a
regular polyedron can be formed bounded
by regular pentagons and having three
at each vertex.

There is such a regular polyedron.
It has twelve faces, and is called the
regular dodecaedron.

!No regular polyedrons bounded by
regular pentagons and having more than three at a vertex
are possible. For four or more angles of 108° cannot be
combined to form a polyedral angle.

4th. Each angle of the regular hexagon contains 120°. No
regular polyedron can be formed bounded by hexagons. For
three or more angles of 120° cannot be combined to form a
polyedral angle.

No regular polyedron can be formed bounded by reg-
ular polygons of more than six sides. For it follows, from
I., 55, Exercise, that the greater the number of sides in a
regular polygon the greater the magnitude of its angles, and
since, as we have seen, the angles of a hexagon are too great
to allow the existence of a polyedral angle whose plane faces
are regular hexagons, those of any regular polygon of more
than six sides will be too great.

Therefore the only possible regular polyedrons are the five
we have figured.

55. Scholium I. It must be observed that we have not
attempted to prove that the five regular polyedrons are pos-
sible. This can be done by showing how to construct them ;
but the investigation is difficult and tedious.

BOOK VII.

253

56. Scholium II. The student may derive some aid in com-
prehending the preceding discussion of the regular polyedrons
by constructing models of them, which he can do in a very
simple manner, and at the same time with great accuracy, as
follows.

Draw on card-board the following diagrams ; cut them out
entire, and at the lines separating adjacent polygons cut the
card-board half through ; the figures will then readily bend
into the form of the respective surfaces, and can be retained
in that form by gluing the edges.

Tetraedron.

Hexaedron,

Octaedron.

Dodccaedron.

Tcosaedron.

EXERCISES ON BOOK VIL

THEOREMS.

1. The volume of a triangular prism is equal to the product of
the area of a lateral face by one-half the perpendicular distance
of that face from the opposite edge.

2. The lateral surface of a pyramid is greater than the base.
Suggestion. Join the projection of the vertex on the base with

the corners of the base.

3. At any point in the base of a regular pyramid a perpendic-
ular to the base is erected which intersects the several lateral faces
of the pyramid, or these faces produced. Prove that the sum of
the distances of the points of intersection from the base is con-
stant.

Suggestion. The distances in question are proportional to the
distances of the foot of the perpendicular from the sides of the
base, and these distances have a constant sum. {v. V., Exer-
cise 16.)

4. Two tetraedrons which have
a triedral angle of the one equal
to a triedral angle of the other,
are to each other as the products
of the three edges of the equal
triedral angles, {v. IV., 19, Ex-
ercise.)

6. In a tetraedron, the planes passed through the three lateral
edges and the middle points of the edges of the base intersect in
a straight line.
254

BOOK VII. 255

Suggestion. The intersections of the planes with the base are
medial lines of the base. Therefore they intersect in the line
joining the vertex with the point of intersection of the medial
lines of the base.

6. The lines joining each vertex of a tetraedron with the point
of intersection of the medial lines of the opposite face all meet in
a point, which divides each line in the ratio 1 : 4.

Note. This point is the centre of gravity
of the tetraedron.

Suggestion. If AF and DG are two of /

the lines in question,, they must intersect, /

since they both lie in the plane passed / g
through AD and the middle point E of ^V/
the opposite edge. Moreover, since EF M

= lED and EG = lEA (I., Exercise 38),
GFis, parallel to AD and is equal to ^AD. C

Whence HF = iHA and GH = iHD.

The lines through C and B will also each cut off \ of AF, Hence
the four lines have a common intersection.

7. The straight lines joining the middle points of the opposite
edges of a tetraedron all pass through the centre of gravity of
the tetraedron, and are bisected by the centre of gravity, {v. III.,

Exercise 7.) ^

8. The plane which bisects a diedral angle of a ie^raedron
divides the opposite edge into segments which are proportional
to the areas of the adjacent faces.

Suggestion. Consider the volumes of the two parts into which
the tetraedron is divided.

9. If a, 6, e, d, are the perpendiculars from the vertices of a
tetraedron upon the opposite faces, and a\ 6'', g\ d^, the perpen-
diculars from any point within the tetraedron upon the same
faces respectively, then

abed

Suggestion. Join the point in question with the vertices of the
tetraedron, and compare the volumes of the four tetraedrons thus
obtained with the volume of the given tetraedron.

256

ELEMENTS OF GEOMETRY.

10. The altitude of a regular tetraedron is equal to the sum of
the four perpendiculars let fall from any point within it upon the
four faces.

11. Any lateral face of a prism is less than the sum of the other
lateral faces, {v. Proposition II.)

PROBLEMS.

12. Given three indefinite straight lines in space which do not
intersect, to construct a parallelopiped which shall have three of
its edges on these lines, {v. VI., Exercise 8.)

13. Within a given tetraedron, to find a point such that planes
passed through this point and the edges of the tetraedron shall
divido^lie tetraedron into four equivalent tetraedrons. {v. Exer-
cise 6.) '

BOOK YIIL

THE THREE ROUND BODIES.

1. Op the various solids bounded by curved surfaces, but
three are treated of in Elementary Geometry, — namely, the
cylinder^ the cone^ and the sphere^ which are called the three

ROUND BODIES.

THE CYLINDER.

2. Definitions. A cylindrical surface is a curved surface gen-
erated by a moving straight line which continually touches a
given curve, and in all of its positions is parallel to a given
fixed straight line not in the plane of the curve.

Thus, if the straight line Aa moves so as continually to
touch the given curve ABCD, and so
that in any of its positions, as ^6,
Cc, Dd, etc., it is parallel to a given
fixed straight line 3im, the surface
ABCDdcba is a cylindrical surface.
If the moving line is of indefinite
length, a surface of indefinite extent
is generated.

The moving line is -called the generatrix ; the curve which
it touches is called the directrix. Any straight line in the
surface, as Bb^ which represents one of the positions of the
generatrix, is called an element of the surface.

To draw an element through any given point of a cylin-
drical surface, it is sufficient to draw a line through the point
parallel to the given fixed straight line, or parallel to ap
element (I., Postulate II.).

r . 22* 257

258

ELEMENTS OF GEOMETRY.

In this general definition of a cylindrical surface, the direc-
trix may be any curve whatever. Hereafter we shall assume
it to be a closed curve, and usually a
circle, as this is the only curve whose
properties are treated of in element-
ary geometry.

3. Definition. The solid Ad bounded
by a cylindrical surface and two par- .
allel planes, ABD and ahd^ is called
a cylinder ; its plane surfaces ABD,

abd, are called its bases; the curved surface is sometimes
called its lateral surface; and the perpendicular distance
between its bases is its altitude.

The elements of a cylinder are all equal.

A cylinder whose base is a circle is called a circular cylinder.

4. Definition. A right cylinder is one whose
elements are perpendicular to its base.

5. Definition. A right cylinder with a circular
base, as ABCa, is called a cylinder of revolution,
because it may be generated by the revolution
of a rectangle A Ooa about one of its sides, Oo,
as an axis ; the side Aa generating the curved ^
surface, and the sides OA and oa generating

the bases. The fixed side Oo is the axis of the cylinder.
The radius of the base is called the radius of the cylinder.

m\-

PROPOSITION I.— THEOREM.

6. Every section of a cylinder made by a plane passing through
an element is a parallelogram.

Let Bb be an element of the cylinder Ac ; then the section
BbdD, made by a plane passed through Bb, is a parallelogram.

BOOK VIII.

259

The line Dd in which the cutting plane intersects the
curved surface a second time is an ele-
ment. For, if through any point D of
this intersection a straight line is drawn
parallel to Bh^ this line, by the definition
of a cylindrical surface, is an element of
the surface, and it must also lie in the
plane Bd ; therefore this element, being
common to both surfaces, is their intersection.

The lines BD and hd are parallel (VI., Proposition YIII.),
and the elements Bh and Dd are parallel ; therefore Bd is a
parallelogram.

7. Corollary. Every section of a right cylinder made by a
plane perpendicular to its base is a rectangle.

PROPOSITION II.— THEOREM.

8. The bases of a cylinder are equal.

Let BD be the straight line joining two points of the
perimeter of the lower base, and let a
plane passing through BD and the ele-
ment Bb cut the upper base in the line
bd ; then BD = bd (Proposition I.).

Let A be any third point in the perim-
eter of the lower base, and Aa the corre-
sponding element. Through the parallels
Aa and Bb pass a plane, and through Aa *^

and Dd pass a plane. Then AB = ab and AD = ad (Propo-
sition I.) ; and the triangles ABD^ abdj are equal. Therefore,
if the upper base be applied to the lower base with the line
bd in coincidence with its equal BD^ the triangles will co-
incide and the point a will fall upon A ; that is, any point a
of the upper base will fall on the perimeter of the lower base,

260 ELEMENTS OF GEOMETRY.

and consequently the perimeters will coincide throughout.
Therefore the bases are equal.

9. Corollary I. Any two parallel sec- »w/.^^ /

tions of a cylindrical surface are equal. /-^^^^^i^V

For these sections are the bases of a A>-^^- /

cylinder. / .../^/'^

10. Corollary II. All the sections of a circular cylinder
parallel to its bases are equal circles; and the straight line
joining the centres of the bases passes through the centres of all
the parallel sections. This line is called the axis of the cylinder.

Suggestion. In the base draw two diameters, and through
these diameters and elements of the cylinder pass planes.
They will cut all the sections in diameters, and their line of
intersections will pass through all the centres.

11. Definition. A tangent plane to a cylinder is a plane which
passes through an element of the curved surface without cut-
ting the surface. The element through which it passes is
called the element of contact

THE CONE.

12. Definition. A conical surface is a curved surface gener
ated by a moving straight line which continually touches a
given curve, and passes through a given fixed point not in
the plane of the curve.

Thus, if the straight line SA moves so as continually to
touch the given curve ABGD, and in all its positions, SB, SC,
SBj etc., passes through the given fixed point aS, the surface
S-ABCD is a conical surface.

BOOK VIII.

261

The moving line is called the generatrix ; the curve which
it touches is- called the directrix.
Any straight line in the surface, as
SBj which represents one of the posi-
tions of the generatrix, is called an
element of the surface. The point S
is called the vertex.

The straight line joining any point
of a conical surface with the vertex
is obviously an element.

If the generatrix is of indefinite
length, as ASa^ the whole surface generated consists of two
symmetrical portions, each of indefinite extent, lying on
opposite sides of the vertex, as S-ABGD and S-abcd, which
are called nappes ; one the upper, the other the lower, nappe.

13. Definition. The solid S-ABCD, bounded by a conical
surface and a plane ABD cutting the surface, is called a cone ;
its plane surface ABD is its base, the point ;S^ is its vertex, and
the perpendicular distance SO from the vertex to the base is
its altitude.

A cone whose base is a circle is called a circular cone. The
straight line drawn from the vertex of a circular cone to the
centre of its base is the axis of the cone.

14. Definition. A right circular cone is a cir-
cular cone whose axis is perpendicular to its
base, as S-ABCD.

The right circular cone is also called a cone
of revolution, because it may be generated by
the revolution of a right triangle, SA 0, about
one of its perpendicular sides, SO, as an axis ;
the hypotenuse SA generating the curved surface, and the
remaining perpendicular side OA generating the base.

262

ELEMENTS OF GEOMETRY.

PROPOSITION III.— THEOREM.

15. Every section of a cone made by a plane passing through
its vertex is a triangle.

Let the cone S-ABCD be cut by a
plane SBC^ which passes through the
vertex S and cuts the base in the straight
line BC ; then the section SBC is a tri-
angle; that is, the intersections SB and
SG with the curved surface are straight
lines.

For the straight lines joining S with B and C are elements
of the surface, by the definition of a cone, and they also lie
in the cutting plane ; therefore they coincide with the inter-
sections of that plane with the curved surface ; and BG^ being
the intersection of two planes, is a straight line.

PROPOSITION IV.— THEOREM.

16. If the base of a cone is a circle^ every section made ty a
plane parallel to the base is a circle.

Let the section abc, of the circular
cone S-ABGj be parallel to the base.

Let 0 be the centre of the base, and
let 0 be the point in which the axis SO
cuts the plane of the parallel section.
Through SO and any number of ele-
ments SAj SBj etc., pass planes cutting
the base in the radii OA^ OB, etc., and
the parallel section in the straight lines oa, ob, etc.
is parallel to OA, and ob to OB, we have

Since oa

ofi So „„j ob So „i^ ^ oa ob

BOOK VIII. 263

But OA = OB, therefore oa = oh ; hence all the straight
lines drawn from o to the perimeter of the section are equal,
and the section is a circle.

17. Corollary. The axis of a circular cone passes through
the centres of all the sections parallel to the base,

18. Definition. A tangent plane to a cone is a plane which
passes through an element of the curved surface without
cutting this surface. The element through which it passes

is called the element of contact. Cv//^^^'^^^

THE SPHERE.

19. Definition. A sphere is a solid bounded by a surface all
the points of which are equally distant from a point within,
called the centre.

A sphere may be generated by the revolu-
tion of a semicircle ABC about its diameter
J. (7 as an axis ; for the surface generated
by the curve ABC will have all its points
equally distant from the centre 0.

A radius of the sphere is any straight
line drawn from the centre to the surface.
A diameter is any straight line drawn through the centre and
terminated both ways by the surface.

Since all the radii are equal and every diameter is double
the radius, all the diameters are equal.

20. Definition. It will be shown that every section of a
sphere made by a plane is a circle ; and, as the greatest pos--
sible section is one made by a plane passing through the
centre, such a section is called a great circle. Any section
made by a plane which does not pass through the centre is
called a small circle.

264 ELEMENTS OF GEOMETRY.

21. Definition. The poles of a circle of the sphere are the
extremities of the diameter of the sphere which is perpen-
dicular to the plane of the circle j and this diameter is called
the axis of the circle.

PROPOSITION v.— THEOREM.

22. Every section of a sphere made by a plane is a circle.

1st. If the -plane passes through
the centre of the sphere, the lines .^-^^---^

joining points on the perimeter of aJifs//.V}liyss^\c

the section with the centre 0 of the / ^^^vHI]^?'^ \
sphere are radii of the sphere, and ^F;---— --^------^

are therefore all equal. Consequently \ j J

it is a circle with its centre at 0. X,^^ I ^^/

2d. If the plane does not pass ^

through the centre of the sphere, as

abcj draw a diameter EOD pei*pendicular to the section and
meeting it at o. If points a, b, c, of the perimeter are joined
with 0, and also with 0, the triangles aeO, boO, coO, are all
equal (I., Proposition X.). Therefore ao, bo, co, etc., are all
equal, and the section is a circle with its centre at o.

23. Corollary I. The axis of a circle on a sphere passes
through the centre of a .circle.

24. Corollary II. All great circles of the same sphere are
equal.

25. Corollary III. Every great circle divides the sphere into
two equal parts.

Suggestion. Superpose one part upon the other, (v. II.,
Proposition II.)

BOOK VIII. 265

26. Corollary IY. Any two great circles on the same sphere
bisect each other ; for the common inter-
section AB of their planes passes through
the centre of the sphere and is a diame-
ter of each circle.

27. Corollary Y. An arc of a great
circle may he drawn through any two given
points of the surface of the sphere^ and,
unless the points are the opposite eoctremities

of a diameter, only one such arc can he drawn; for the two
points, together with the centre 0, determine the plane of a
great circle whose circumference passes through the points.

If, however, the two given points are the extremities A
and 5 of a diameter of the sphere, the position of the circle
is not determined, for the points J., 0, and B, being in the
same straight line, will not determine a plane (YI., Proposi-
tion I.).

28. Corollary YI. An arc of a circle may he drawn through
any three given points on the surface of the sphere; for the
three points determine a plane which cuts the sphere in a
circle.

EXEBCISE.

Theorem. — The greater the distance of the plane of a small

hsircle from the centre of the sphere, the less the circle.

li 23

266 ELEMENTS OF GEOMETRY.

PROPOSITION VI.— THEOREM.

29. All the points in the circumference of a circle of the sphere
J,- are equally distant from either of its poles.

Let abed be any circle'of the sphere, ..-s^^^--^

and PP' the diameter of the sphere „/^'Wt'^' — ^\j

perpendicular to its plane; then, by I %.
the definition (21), P and P' are the ^^^
poles of the circle ahcd^ and, by Prop- \ \^
osition Y., Corollary I., FT' passes \

through 0, the centre of ahcd. Join ^1^

P with any points, a, 6, c, on the cir-
cumference of the circle. Then Pa, P6, Pc, are equal, since
the triangles Poa, Tob^ Poc^ are equal, by I., Proposition VI.
Hence all the points of the circumference abed are equally
distant from the pole P. For the same reason, they are
equally distant from the pole P'.

30. Corollary I. All the arcs of great circles drawn from a
pole of a circle to points in its circumference are equal, since
their chords are equal chords in equal circles.

By the distance of two points on the surface of a sphere
is usually understood the arc of a great circle joining the two
points. The arc of a great circle drawn from any point of a
given circle abc, to one of its poles, as the arc Pa, is called
the polar distance of the given circle, and the distance from
the nearest pole, is usually understood.

31. Corollary II. The polar distance of a great circle is a
quadrant ; thus, PA, PB, etc., P'A, P'B, etc., polar distances
of the great circle ABCD, are quadrants ; for they are the
measures of the right angles A OP, BOP, AOP', BOP', etc.,
whose vertices are at the centre of the great circles PAP'j
PBP', etc.

BOOK VIII. 267

In connection with the sphere, by a quadrant is usually to
be understood a quadrant of a great circle.

32. Corollary III. If a point on the surface of a sphere is
at a quadrant's distance from each of two given points of the
surface which are not opposite extremities of a diameter, it is the
pole of the great circle passing through them.

Suggestion. Let P be at a quadrant's distance from B and
C; then FOB and POC are right angles, and FO is perpen-
dicular to the plane ABCD.

33. Scholium. By means of poles, arcs of circles may be
drawn upon the surface of a sphere with the same ease as
upon a plane surface. Thus, by revolving the arc Fa about
the pole P, its extremity a will describe the small circle abd ;
and by revolving the quadrant FA about the pole P, the
extremity A will describe the great circle ABD.

If two points, B and (7, are given on the surface, and it is
required to draw the arc ^C, of a great circle, between them,
it will be necessary first to find the pole P of this circle ; for
which purpose, take B and G as poles, and at a quadrant's
distance describe two arcs on the surface intersecting in P.
The arc BC can then be described with a pair of compasses,
placing one foot of the compasses on P and tracing the arc
with the other foot. The opening of the compasses (distance
between their feet) must in this case be equal to the chord
of a quadrant ; and to obtain this it is necessary to know the
radius of the sphere.

34. Definition. A plane is tangent to a sphere when it has
but one point in common with the surface of the sphere,

35. Definition. Two spheres are tangent to each other when
their surfaces have but one point in common.

268

ELEMENTS OF GEOMETRY.

PROPOSITION VII.— THEOREM.

36. A plane tangent to a sphere is perpendicular to the radius
drawn to the point of contact.

For any other line drawn from
the centre of the sphere to the
plane must reach beyond the sur-
face of the sphere, and therefore
must be greater than the radius.
The radius is, then, the shortest
line that can be drawn from the
centre of the sphere to the plane,

and is consequently perpendicular to the plane (YI., Proposi-
tion III.).

37. Corollary. Conversely, a plane perpendicular to a
radius of a sphere at its extremity is tangent to the sphere.

38. Scholium. Any straight line A T, drawn in the tangent
plane through the point of contact, is tangent to the sphere.

Any two straight lines, AT, A T\ tangent to the sphere at
the same point A, determine the tangent plane at that point.

PROPOSITION VIII.— THEOREM.

39. The intersection of two spheres is a circle whose plane is
perpendicular to the straight line joining the centres of the
spheres, and whose centre is in that line.

Through the centres 0 and (7 of
the two spheres let any plane be
passed, cutting the spheres in great
circles which intersect each other in
the points A and B ; the chord AB
is bisected at G by the line 0(7 at
right angles (II., Proposition YI., Corollary II.). If we now

Tv^

BOOK VIII. 269

revolve the plane of these two circles about the line 0(7, the
circles will generate the two spheres, and the point A will
describe the line of intersection of their surfaces. Moreover,
since the line AC will, during this revolution, remain perpen-
dicular to 0(7, it will generate a circle whose plane is per-
pendicular to 0(y, and whose centre is C.

SPHERICAL ANGLES.

40. Definition. The angle of two curves passing through the
same point is the angle formed by the two tangents to the
curves at that point.

This definition is applicable to any two intersecting curves
in space, whether drawn in the same plane or upon a surface
of any kind.

, PROPOSITION IX.— THEOREM.

M. The angle of two arcs of great circles is equal to the angle
of their planes^ and is measured by the arc of a great circle
described from its vertex as a pole and included between its sides
(produced if necessary).

Let AB and AB' be two arcs of >^ j,

great circles, A T and A T' the tan- / i \^V~-^

gents to these arcs at A^ and 0 the / j W \b

centre of the sphere. TA and T'A ^L^ 4*r~"f3)^
lie in the planes of their arcs, and \^^~~"^ \ I /
are perpendicular to the radius OA X^^ ! J/

drawn to their point of contact. ^

They form, then, the plane angle

measuring the diedral angle formed by the planes of the arcs ;
but, by (40), the angle which they form is equal to the angle
of the two arcs.

270

ELEMENTS OF GEOMETRY.

Now let CO' be the arc of a great circle described from A
as a pole aiid intersecting the arcs
AB^ AB' (produced if necessary),
in G and C. The radii OG and
OG' are perpendicular to J.0, since
the arcs AG, AG', are quadrants
(Proposition YI., Corollary II.);
therefore the angle GOG' is a plane
angle of the diedral angle A 0, and
is equal to TAT', or to BAB', and it is obviously measured
by the arc GG'.

42. Corollary. All arcs of great circles drawn through the
pole of a given great circle are perpendicular to its circumfer-
ence; for their planes are perpendicular to its plane (YI.,
Proposition XIY.).

SPHERICAL POLYGONS.

43. Definition. A spherical polygon is a portion of the sur-
face of a sphere bounded by three or more arcs of great
circles, as ABGD.

Since the planes of all great circles pass
through the centre of the sphere, the planes
of the sides of a spherical polygon form, at
the centre 0, a polyedral angle of which
the edges are the radii drawn to the ver-
tices of the polygon, the face angles are
angles at the centre measured by the sides of the polygon,
and the diedral angles are equal to the angles of the polygon
(Proposition IX.).

Since in a polyedral angle each face angle is assumed to be
less than two right angles, each side of a spherical polygon
will be assumed to be less than a semi-circumference.

BOOK viir. 271

A spherical polygon is convex when its corresponding poly-
edral angle at the centre is convex (VI., 52).

A diagonal of a spherical polygon is an arc of a great circle
joining any two vertices not consecutive.

44. Definition. A spherical triangle is a spherical polygon
of three sides. It is called right angled^ isosceles^ or equilat-
eral^ in the same cases as a plane triangle.

45. In consequence of the relation established between
polyedral angles and spherical polygons (43), it follows that
from any property of polyedral angles we may infer an anal-
ogous property of spherical polygons.

Eeciprocally, from any property of spherical polygons we
may infer an analogous property of polyedral angles.

The latter is in almost all cases the more simple mode of
procedure, inasmuch as the comparison of figures drawn on
the surface of a sphere is nearly if not quite as simple as the
comparison of plane figures.

46. Arcs of great circles on the same sphere can be super-
posed and made to coincide just as straight lines are super-
posed and made to coincide. We have merely to place one
point of the first arc on some given point of the second, and,
keeping this point fixed, to turn the first arc about it as a
pivot, until some second point in the arc not diametrically
opposite the fixed point falls on the second arc. The two
arcs must then coincide throughout, by Proposition Y., Corol-
lary Y.

Equal angles formed by arcs of great circles on the surface
of the same sphere can be superposed and made to coincide
just as equal plane angles are superposed and made to co-
incide ; that is, if the vertex of the first angle is placed upon
the vertex of the second, and one side of the first placed
upon the corresponding side of the second, the other side of

272 ELEMENTS OF GEOMETRY.

the first will coincide with the other side of the second. For,
if the two given angles are equal, their diedral angles are
equal (Proposition IX.). If the vertices of the angles co-
incide, the edges of the diedral angles coincide ; if a side of
the first angle is placed on a side of the second, one face of
the first diedral angle coincides with one face of the second.
The remaining faces of the diedral angles must then coincide,
and consequently the remaining sides of the given angles
coincide.

47. Definition. Two spherical triangles are symmetrical if
all the parts of one are respectively equal to the parts of the
other, but the corresponding parts are arranged in opposite
orders in the two triangles.

Two symmetrical triangles, as ABC^ ABC'j
in the figure cannot be made to coincide;
for, to bring the vertex C upon the corre-
sponding vertex C, the second triangle would
have to be turned over, and the two convex
surfaces would thus be brought together.

48;^here is, however, one exception to the
last remark, — namely, the case of symmetrical isosceles trian-
gles, Por, if ABC is an isosceles
spherical triangle and AB = AC,
then, in its symmetrical triangle,
we have A'B' = A!C\ and con-
sequently AB = A'C, AG =
A'B\ and, since the angles A and

A' are equal, if AB be placed on A'G\ AG will fall on its
equal A'B\ and the two triangles will coincide through-
out.

49. Definition. If from the vertices of a spherical triangle
as poles, arcs of great circles are described, these arcs form

/}l\ /

BOOK VIII.

273

by their intersection a second triangle, which is called the
polar triangle of the first.

Thus, if Aj B, and C are the poles of the
arcs of great circles, B'C'j A'C\ and A'B\
respectively, A'B'C is the polar triangle of
ABC.

Since all great circles, when completed,
intersect each other in two points, the arcs
jB'C", A'C, A'B\ if produced, will form three
other triangles ; but the triangle which is taken as the polar
triangle is that whose vertex A\ homologous to A, lies on
the same side of the arc BC as the vertex A; and so of the
other vertices.

PROPOSITION X.— THEOREM.

50y^f the first of two spherical triangles is the polar triangle
of the second, then, reciprocally, the second is the polar triangle
of the first.

Let A'B'C be the polar triangle of ABC;
then is ABC the polar triangle of A'B'C,

For, since A is the pole of the arc B'C,
the point 5' is at a quadrant's distance from
A ; and, since C is the pole of the arc A'B',
the point ^' is at a quadrant's distance from
C ; therefore B' is the pole of the arc AC
(Proposition YI., Corollary III.). In the same manner it is
shown that A' is the pole of the arc BC, and C" the pole of
the arc AB. Moreover, A and A' are on the same side of
B'C\ B and B' on the same side of A'C, C and C on the
same side of A'B' ; therefore ABC is the polar triangle of
A'B'C.

274

ELEMENTS OF GEOMETRY.

PROPOSITION XI.— THEOREM.

5lX/« t.o polar triangles, ea^U anyle of one is measure, l>y
the supplement of the side lying opposite to it in the other.

Let ABC and A'B'C be two polar tri-
angles.

Let the sides AB and AC, produced if
necessary, meet the side B'C in the points h
and c. The vertex A being the pole of the
arc he, the angle A is measured by the arc he
(Proposition IX.).

Now, B ' being the pole of the arc Ac, and
C the pole of the arc Ah^ the arcs B'c and G'h are quadrants ;
hence we have

B'C -\- he =^B'c -\- C'h = Si semi-circumference.

Therefore he, which measures the angle A, is the supplement
of the side B'G\

In the same manner it can be shown that each angle of
either triangle is measured by the supplement of the side
lying opposite to it in the other triangle.

.52. Scholium. Let the angles of the trian-
gle ABC be denoted by A, B, and C, and let
the sides opposite to them, namely, BC, AC,
and AB, be denoted by a, h, and c, respec-
tively. Let the corresponding angles and
sides of the polar triangle be denoted by A',
B', C, a', h', and c\ Also let both angles and
sides be expressed in degrees. Then the preceding theorem
gives the following relations :

A +a'=B -\-h':=^C +c'^ 180°.
A' +a :=:^B' -\rh =C' + c = 180°.

BOOK VIII. 275

PROPOSITION XII.— THEOREM.

53. Two triangles on the same sphere are either equal or sym-
metrical, when two sides and the included angle of one are re-
spectively equal to txoo sides and the included angle of the other.

In the triangles ^50 and DEF, let the
angle A be equal to the angle D, the side
AB equal to the side DE, and the side
A G equal to the side DF.

1st. When the parts of the two trian-
gles are in the same order, ABC can be
applied to DEF^ as in the corresponding
case of plane triangles (I., Proposition
VI.), and the two triangles will coincide ; therefore they are
equal.

2d. When the parts of the two
triangles are in inverse order, let
DE'F be the symmetrical triangle
of BEF, and therefore having its
angles and sides equal, respectively,
to those of DEF. Then, in the
triangles ABC and DE'F, we shall
have the angle BAC equal to the

angle E'DF, the side AB to the side DE', and the side
AG to the side DF, and these parts arranged in the same
order in the two triangles ; therefore the triangle ABG is
equal to the triangle DE'F, and consequently symmetrical
with DEF,

54. Scholium. In this proposition, and in the propositions
which follow, the two triangles may be supposed on the same
sphere, or on two equal spheres.

276

ELEMENTS OF GEOMETRY.

PKOPOSITION XIII.— THEOREM.

55. Two triangles on the same sphere are either equal or sym-
metrical^ when a side and two adjacent angles of one are equal
respectively to a side and two adjacent angles of the other.

Let the triangles ABC
and BEF have the side a
equal to the side d^ and the c' / "^ X6'
angles B^ C, equal respec-
tively to the angles E, F ;
then are the triangles equal.

Construct the polar tri-
angles of ABC and DEF. We have h' = e', c' = /', and A' ==
D\ by Proposition XI. Then A'B'C and D'E'F' are equal
or symmetrical, by Proposition XII. Therefore their polar
triangles ABC, DEF, are equal or symmetrical.

58. Scholium. The proposition might be proved by direct
superposition, as in I., Proposition YII.

PROPOSITION XIV.— THEOREM.

57. Two triangles on the same sphere are either equal or sym-
metrical, when the three sides of one are respectively equal to the
three sides of the other.

For if their vertices are joined with the centre of the
sphere, the triedral angles thus formed have the three face
angles of the one respectively equal to the three face angles
of the other, and consequently, by VI., Proposition XXII., their
corresponding diedral angles are equal. The given triangles
are, then, mutually equilateral and mutually equiangular, and
are equal or symmetrical.

BOOK vni. 277

58. Scholium. The proposition can be proved as in I., Prop-
osition IX.

PROPOSITION XV.— THEOREM.

59. If two triangles on the same sphere are mutually equi-
angular, they are also mutually equilateral, and are either equal
or symmetrical. .

Let the spherical trian- '^'"'^ ^^^^^

gles M and JV be mutually
equiangular.

Let M' be the polar tri-
angle of M, and iV' the polar triangle of iV. Since M and
N are mutually equiangular, their polar triangles M' and iV'
are mutually equilateral (Proposition XL); therefore, by
Proposition XIY., the triangles M^ and N' are mutually equi-
angular. But M' and iV' being mutually equiangular, their
polar triangles M and iV are mutually equilateral. Conse-
quently, M and N are either equal or symmetrical.

60. Scholium. It may seem to the student that the pre-
ceding property destroys the analogy which subsists be^^een
plane and spherical triangles, since two mutually equiangular
plane triangles are not necessarily mutually equilateral. But
in the case of spherical triangles the equality of the sides
follows from that of the angles only upon the condition that
the triangles are constructed upon the same sphere or on
equal spheres ; if they are constructed on spheres of diiferent
radii, the homologous sides of two mutually equiangular tri-
angles will no longer be equal, but will be proportional to the
radii of the sphere ; the two triangles will then be similar, as
in thje case of plane triangles.

278 ELEMENTS OF GEOMETRY.

EXERCISES.

1. Theorem. — In an isosceles spherical triangle the angles
opposite the equal sides are equal.

2. Theorem. — The arc drawn from the vertex of an isosceles
spherical triayigle to the middle point of the base is perpendicular
to the base, and bisects the vertical angle.

3. Theorem. — If two angles of a spherical triangle are equal,
the triangle is isosceles.

PROPOSITION XVI.-^THEOREM.

61. Any side of a spherical triangle is less than the sum of

the other two.

Let ABC be a spherical triangle;
then any. side, as AC, is less than the
sum of the other two, AB and BC. /

For, in the corresponding triedral /

angle formed at the centre 0 of the /^'l-"''''

sphere, we have the angle AOC less o**'
than the sum of the angles AOB and

BOC (VI., Proposition XX.) ; and since the sides of the tri-
angle measure these angles, respectively, we have AC <^ AB
+ BC.

EXERCISES.

1. Theorem. — If two angles of a spherical triangle are un-
equal, the side opposite the greater angle is greater than the side
opposite the less angle, (v. I., Proposition XII.)

2. Theorem. — If two sides of a spherical triangle are unequal,
the angle opposite the greater side is greater than the angle oppo-
site the less side. (v. L, Proposition XIII.)

BOOK VIIT.

279

PEOPOSITION XVII.— THEOEEM.

62. The sum of the sides of a convex spherical polygon is less
than the circumference of a great circle.

For the sum of the face angles of the corresponding poly-
edral angle at the centre of the sphere is less than four right
angles (YI., Proposition XXI.).

PEOPOSITION XVIII.— THEOEEM.

^ 63. The sum of the angles of a spherical triangle is greater
than two, and less than six, right angles.

For, denoting the angles of a spherical tri-
angle by A, B, C, and the sides respectively
opposite to them in its polar triangle by a', ^<
h', d, we have (Proposition XI.)

A = 180° — o!, J?=: 180° — 6', 0= 180° —c',

the sum of which is

A^ B -^C ^ 540° — {a! + 6' + c').

But a' ^y -\- c' <i 360° (Proposition XYII.) ; therefore A +
B -\- G ^ 180° ; that is, the sum of the three angles is
greater than two right angles. Also, since each angle is
less than two right angles, their sum is less than six right
angles.

64. Scholium. A spherical triangle may have two or even
three right angles j also two or even three obtuse angles.

280

ELEMENTS OF GEOMETRY.

PROPOSITION XIX.-THEOEEM.

65. Two symmetrical spherical triangles are equivalent.

Let ABG and A'B'C be symmetrical spherical triangles.
Let P be the pole of the
small circle passing through
A, B, and C (Proposition
Y., Corollary YI.). Then
the arcs PA, PB, PC, are
equal (Proposition YL, Cor-
ollary I.), and divide ABG
into three isosceles trian-
gles.

Through A' and B' in the triangle A'B'C draw arcs mak-
ing with A'B' angles equal respectively to PAB and PBA,
and join P\ their point of intersection, with G'. The isos-
celes triangle PAB is equal to the triangle P'A'B\ by Propo-
sition XIII. and (48). The isosceles triangle PBG is equal
to the triangle P'B'G\ by Proposition XII. and (48). The
isosceles triangle PGA is equal to the triangle P'G'A\ by
Proposition XIY. and (48). Hence ABG and A'B'C' are
equivalent.

If the pole P should fall
without the triangle ABG, the
triangle would be equivalent
to the sum of two of the isos-
celes triangles diminished by
the third ; but, as the same
thing would occur for the sym-
metrical triangle, the conclu-
sion would be the same.

■\p

pv— -/--

BOOK VIII.

281

66. Definition. If a spherical triangle ABC
has two right angles, B and (7, it is called a
hi-rectangular triangle ; and, since the sides AB
and J. (7 must each pass through the pole of BO
(Proposition IX., Corollary), the vertex A is
that pole, and therefore AB and A C are quadrants.

67. Definition. A lune is a portion of the sur-
face of a sphere included between two semi-
circumferences of great circles ; as AMBNA.

The two angles of a lune are equal, since
each is equal to the diedral angle formed by
the planes of the arcs of the lune ; and the
lune is equal to the sum of two equal bi-rectan-
gular triangles, each of which has the angle of the lune for its
third angle.

EXERCISE.

Theorem. — Two lunes on the same sphere or on equal spheres
re equal if their angles are equal.

PROPOSITION XX.— THEOREM. ^

68. If two arcs of great circles intersect on the surface of a
hemisphere^ the sum of the opposite spherical triangles which they
form is equivalent to a lune whose angle is the angle between the
arcs in question.

Let the arcs ACA', BCB\ intersect
on the surface of the hemisphere
ABA'B'C. Then will the triangles
ABG^ AIB'C^ be together equivalent
to a lune whose angle is A GB.

For, continue the arcs ACA\ BCB\
until they intersect in C A' C = AC,
B'C ^^ BCy and A'B' = AB, since they subtend equal angles.

282

ELEMENTS OF GEOMETRY.

The triangles A'B'C and ABC are
then equal or symmetrical, by Propo-
sition XIY., and are in either case
equivalent (Proposition XIX.). There-
fore ABC and A'B'C are together
equivalent to A'B'C + A'B'C; that
is, to the lune CA'C'B\

MEASUEEMENT OF THE SUEFACES OF SPHEEICAL

FIGUEES.

69. Definition. A degree of spherical surface, or, more briefly,
u spherical degree, is -^ of the surface of a hemisphere. It
is a convenient unit in measuring the surfaces of spherical
figures. Like the degree of arcs, it is not a unit of absolute
magnitude, but depends upon the size of the sphere on which
the figures are drawn.

It may be conveniently conceived as a bi-rectangular spher-
ical triangle whose third angle is an angle of one degree.

PEOPOSITION XXI.— THEOEEM.

70. A lune is to the surface of the sphere as the angle of the
lune is to four right angles.

Let AWBMA be a lune, and let
MNP be the great circle whose poles
are the extremities of the diameter
AB.

Since the angle of the lune is meas-
ured by the arc MW, the angle of the
lune is to four right angles as the arc
MN is to the whole circumference MNPM.

Ist. Suppose that MN and the circumference have a com-

BOOK VIII. 283

mon measure which is contained m times in MN and n times
in MNPM. Then

MN _ m
MNPM~~ n

Apply the measure to the circumference, and through the
points of division and the axis AB pass planes; they will
divide the whole surface of the sphere into n equal lunes (67,
Exercise), of which the given lune ANBMA will contain m.

Therefore

ANBMA _m
surface of sphere w'
and we have

ANBMA MN-

surface of sphere MNFM'

2d. We can extend the proof to the case where MN and
MNPM are incommensurable by our usual method, (v. YII.,
Proposition YII.)

71. Corollary. The area of a lune is expressed by twice its
angle, the angular unit being the degree, and the unit of surface
the spherical degree.

For, by (69), the area of the surface of the sphere is 720
spherical degrees. We have, then, if S is the area and A

the angle of the lune,

S ^ A ,
720 360 '
whence

S=2A.

72. Scholium. If the angle A contains a whole number of
degrees, and each of the parts of the arc MN in the figure
above is one degree, each of the small lunes is made up of
two spherical degrees, and the lune AMBN obviously con-
tains twice as many spherical degrees as the arc MN contains
degrees of arc.

284 ELEMENTS OF GEOMETRY.

A- PROPOSITION XXII.— THEOREM.

73. The area of a spherical triangle is equal to the excess of
the sum of its angles over two right angles.

For, let ABC be a spherical triangle.
Complete the great circle ABA^B', and
produce the arcs AC and BC to meet
this circle in A' and B\

We have, by the figure,

ABC + A'BC = lune A,
ABC + AB'C=luneB,

and, by Proposition XX.,

ABC + A'B'C = lune C.

The sum of the first members of these equations is equal to
twice the triangle ABC, plus the four triangles ABC, A'BC,
AB'C, A'B'C, which compose the surface of the hemisphere,
whose area is 360 spherical degrees.

Therefore, denoting the area of the triangle ABC by T,
we have (Proposition XXI., Corollary)

2T + 360° = 2A-\-2B + 2(7,
T + 180° = A + B-^ C,
T = A-i- B + C— 180°.

74. Scholium. The excess of the sum of the angles of a
spherical triangle over two right angles is sometimes called
its spherical excess.

EXERCISE.

Theorem. — The area of a spherical polygon
is measured by the sum of its angles minus
the product of two right angles multiplied by
the number of sides of the polygon less two.

BOOK VIII. 285

75. Scholium. It must not be forgotten that Propositions
XXI. and XXII. merely enable us to express our areas in
spherical degrees ; that is, in terms of yj^ of the surface of
the whole sphere. If the area is required in terms of the
ordinary unit of surface (lY., 1), the area of the surface of
the sphere must first be given in terms of the unit in question.

SHORTEST LINE ON THE SURFACE OF A SPHERE
BETWEEN TWO POINTS.

PROPOSITION XXIII.— THEOREM.

76. The shortest line that can be drawn on the surface of a
sphere between two points is the arc of a great circle^ not greater
than a semi-circumference, joining the two points.

Let AB be an arc of a great circle, less
than a semi-circumference, joining any
two points A and B of the surface of a
sphere ; and let C be any arbitrary point
taken in that arc. Then we say that
the shortest line from A to B, on the sur-
face of the sphere, must pass through C.

From A and B as poles, with the polar distances A C and
BCj describe circumferences on the surface ', these circumfer-
ences touch at C and lie wholly without each other. For,
let Mhe any point other than C in the circumference whose
pole is A, and draw the arcs of great circles AM, BM, form-
ing the spherical triangle A MB. We have, by Proposition
XYI., AM + BM > AB, and subtracting from the two
members of this mequality the equal arcs AM and AC, wq
have BM > BC; therefore ikf lies without the circumference
whose pole is B.

286 ELEMENTS OF GEOMETRY.

Now let AFGB be any line from A to 5, on the surface of
the sphere, which does not pass through the point C, and
which therefore cuts the two circumfer-
ences in different points, one in F^ the
other in G. Then a shorter line can be
drawn from A to B^ passing through G.
For, whatever may be the nature of the
line AF^ an equal line can be drawn from
Ato C; since, if AC and AF be conceived
to be drawn on two equal spheres having
a common diameter passing through A, and therefor" having
their surfaces in coincidence, and if one of these spfh^res be
turned upon the common diameter as an axis, the point A
will be fixed and the point F will come into coincidence with
C; the surfaces of the two spheres continuing to coincide,
the line AF will then lie on the common surface between A
and C. For the same reason, a line can be drawn from B to
0, equal to BG. Therefore a line can be drawn from A to B,
through G, equal to the sum of AF and BG, and consequently
less than AFGB. The shortest line from A to B therefore
passes through C; that is, through any, or every, point in
AB. Consequently it must be the arc AB itself.

EXERCISES ON BOOK VIII.

THEOREMS.

1. A SPHERE can be circumscribed about any tetraedron.
Suggestion. The locus of the points

equally distant from A, B^ and C is the
perpendicular EM erected at the centre
of the circle circumscribed about ABC
(VI., Exercise 15.) The locus of the
points equally distant from B^ C, and D
is tne perpendicular FN^ and both EM
and FN lie in the plane perpendicular to
BC Sit its middle point, since that plane
contains all the points equally distant
from B and C. EM and FN therefore
intersect, and O, their point of intersec-
tion, is equally distant from the four ver-
tices of the tetraedron. There is only one such point. Therefore
only one sphere can be circumscribed about a tetraedron.

2. The perpendiculars erected at the centres of the four faces
of a tetraedron meet in a point.

3. A sphere can be inscribed in any tetra-
edron.

Suggestion, The locus of the points equally
distant from two faces of the tetraedron is
the plane bisecting the diedral angle be-
tween them.

4. The planes bisecting the six diedral angles of a tetraedron
intersect in a point.

287

288 ELEMENTS OF GEOMETRY.

LOCI.

6. Locus of the points in space which are at a given distance
from a given straight line.

6. Locus of the points which are at the distance a from a point
Ay and at the distance b from a point B.

7. Locus of the centres of the spheres which are tangent to
three given planes.

8. Locus of the centres of the sections of a given sphere made
by planes passing through a given straight line.

Suggestion. Pass a plane through the centre of the sphere per-
pendicular to the given straight line. Then see II., Exercise 24.

9. Locus of the centres of the sections of a given sphere made
by planes passing through a given point.

PROBLEMS.

10. Through a given point on the surface of a sphere, to pass a
plane tangent to the sphere, (v. Proposition VII., Corollary.)

11. Through a given straight line without a sphere, to pass a
plane tangent to the sphere.

Suggestion. Through the centre of the sphere pass a plane per-
pendicular to the given line. In this plane, from its point of
intersection with the line, draw a line tangent to the circle in
which the plane cuts the sphere. A plane through the tangent
line and the given line is the tangent plane required. (Two solu-
tions.)

12. Through a given point without a sphere, to pass a plane
tangent to the sphere.

13. To cut a given sphere by a plane passing through a given
straight line so that the section shall have a given radius.

SuggeMion. Pass a plane through the centre of the sphere per-
pendicular to the given line. Then v. II., Exercise 37.

BOOK VIII. 289

14. To construct a spherical surface with a given radius— Ist,
passing through three given points ; 2d, passing through two
given points and tangent to a given plane, or to a given sphere ;
3d, passing through a given point and tangent to two given planes,
or to two given spheres, or to a given plane and a given sphere ;
4th, tangent to three given planes, or to three given spheres, or to
two given planes and a given sphere, or to a given plane and
two given spheres.

15. Through a given point on the surface of a sphere, to draw a
great circle tangent to a given small circle.

Suggestion, With the pole of the small circle as a pole, and with
a polar distance equal to the polar distance of the small circle
plus a quadrant, describe a second small circle. With the given
point as a pole describe a great circle. A point of intersection of
this great circle with the second small circle will be the pole of
the great circle required.

16. To draw a great circle tangent to two given small circles.

17. At a given point in a great circle, to driaw an arc of a great
circle which shall make a given angle with the first.

BOOK IX.

MEASUREMENT OP THE THREE ROUND BODIES.
THE CYLINDER.

1. Definition. The area of the convex, or lateral, surface
of a cylinder is called its lateral area.

2. Definition. A prism is inscribed
in a cylinder when its base is inscribed
in the base of the cylinder and its
lateral edges are elements of the cyl-
inder. It follows that the upper base
of the prism is inscribed in the upper
base of the cylinder.

To inscribe, then, a prism of any
given number of lateral faces in a cylinder, we have merely to
inscribe in the base a polygon of the given number of sides,
and through the vertices of the polygon to draw elements
of the cylinder. Planes passed through adjacent elements
will form the lateral faces of the prism which is obviously
wholly contained in the cylinder.

3. Definition. A prism is circum-
scribed about a cylinder when its
base is circumscribed about the
base of the cylinder and its lateral
edges are parallel to elements of
the cylinder.

It follows that its lateral faces
are tangent to the lateral faces of
the cylinder (YIII., 11) ; for any

face, as AB'y contains the element bb\ since it contains the
290

BOOK IX.

291

parallel line AA^ and the point b (YI., Proposition II.), and,
by YIII., Proposition I., it cannot cut the surface of the cyl-
inder again jinless AB cuts the base again ; and that its upper
base is circumscribed about the upper base of the cylinder.
The cylinder is obviously wholly contained in the prism.

4. Definition. A right section of a
cylinder is a section made by a plane
perpendicular to its elements ; asabcdef.

The intersection of the same plane
with an inscribed or circumscribed
prism is a right section of the prism.

5. Definition. Similar cylinders of revolution are those which
are generated by similar rectangles revolving about homolo-
gous sides.

PKOPOSITION I.— THEOREM.

6. If a prism whose base is a regular polygon be inscribed
in or circumscribed about a given cylinder^ its volume will ap-
proach the volume of the cylinder as its limit, and its lateral sur-
face will approach the lateral surface of the cylinder as its limit
as the number of sides of its base is indefinitely increased.

For, if we could make the base
of the prism exactly coincide with
the base of the cylinder, the prism
and the cylinder would coincide
throughout, and their volumes
would be equal and their lateral
surfaces equal.

But, by increasing the number
of sides of the base of the prism.

292

ELEMENTS OF GEOMETRY.

we can make it come as near as we please to coinciding with
the base of the cylinder (Y., Propo-
sition YII.) ; we can then make
the prism and cylinder fail of co-
incidence by as small an amount
as we choose. Consequently, by
increasing at pleasure the number
of sides of the base of the circum-
scribed or inscribed prism, we can
make the difference between the
volumes of prism and cylinder,
and between the lateral surfaces

of prism and cylinder, as small as we choose, but cannot make
it absolutely zero.

7. Scholium. The proposition just proved is true when the
base of the prism is not a regular polygon ; but it is only for
the case of the regular polygon that a rigorous proof has
been given in Book V.

PEOPOSITION II.— THEOREM.

8. The lateral area of a cylinder is equal to the product of
the perimeter of a right section of the cylinder by an element of
the surface.

Let ABCBEF be the base and AA'
any element of a cylinder, and let the
curve abcdef be any right section of
the surface. Denote the perimeter of
the right section by P, the element
AA' by E^ and the lateral area of the
cylinder by S.

Inscribe in the cylinder a prism
ABCDEF A' of any arbitrarily chosen

BOOK rx.

293

number n of faces. The right section, abcdef, of this prism
will be a polygon inscribed in the right section of the cylin-
der formed by the same plane (4). Denote the lateral area
of the prism by s, and the perimeter of its right section by
p ; then, the lateral edge of the prism being equal to E, we
have (VII., Proposition II.)

s = pX E,

no matter what the value of n. If n is indefinitely increased,
s approaches the limit JS (Proposition I.), and p X E, the
limit P X E. Therefore, by III., Theorem of Limits,

S = P X E.

9. Corollary I. The lateral area of a cylinder of revolution
is equal to the product of the circumference of its base by its
altitude.

This may be formulated,

if R is the radius of the base and H the altitude

10. Corollary II. The lateral ^ ^

areas of similar cylinders of revo-
lution are to each other as the
squares of their altitudes, or as the
squares of the radii of their bases.

S _ 2t.R.H

Suggestion.

R H

r ' h
by (5).

^
r*

jff»

= -- = -—, smce - =

27:r.h
R H

25*

h<^

294

ELEMENTS OP GEOMETRY.

PBOPOSITION III.— THEOREM. ,

11. The volume of a cylinder is equal to the product of its
base by its altitude.

Let the volume of the cylinder be
denoted by F, its base by B, and its
altitude by S. Let the volume of an
inscribed prism be denoted by F', and
its base by J5'; its altitude will also be
-H, and we shall have (VII., Proposi-
tion XIL, Corollary)

no matter what the number of faces of the prism.

If the number of faces of the prism is indefinitely in-
creased, F' has the limit F, and ^' X -S" the limit B X S,
Therefore

v = bxb:.

12. Corollary I. For a cylinder of revolution this proposition
may be formulated^ V = izB^.H. (Y., Proposition IX., Corol-
lary.)

13. Corollary II. The volumes of similar cylinders of revo-
lution are to each other as the cubes of their altitudes^ or as the
cubes of their radii.

THE CONE.

14. Definition. The area of the convex, or lateral, surface
of a cone is called its lateral area.

BOOK IX.

15. Definition. A pyramid is inscribed
in a cone when its base is inscribed in
the base of the cone and its vertex
coincides with the vertex of the cone.

It follows that the lateral edges of
the pyramid are elements of the cone.

An inscribed pyramid is wholly con-
tained within the cone.

16. Definition. A pyramid is circum-
scribed about a cone when its base is circumscribed about the
base of the cone and its vertex co-
incides with the vertex of the cone.

Any lateral face, as SAB^ of the
pyramid is tangent to the cone ; for,
since it passes through a and S^ it
contains the element Sa^ and it can-
not cut the convex surface again
without cutting the perimeter of
the base again (YIII., Proposition

111.).

The cone is then wholly contained within the pyramid.

17. Definition. A truncated cone is
the portion of a cone included be-
tween its base and a plane cutting
its convex surface.

When the cutting plane is par-
allel to the base, the truncated cone
is called a frustum of a cone; as
ABCD-abcd. The altitude of a frus-
tum is the perpendicular distance
Tt between its bases.

If a pyramid is inscribed in the cone, the cutting plane

296

ELEMENTS OF GEOMETRY.

determines a truncated pyramid inscribed in the truncated
cone ; and if a pyramid is circumscribed about the cone, the
cutting plane determines a truncated pyramid circumscribed
about the truncated cone.

18. Definition. In a cone of revolution
all the elements are equal, and any ele-
ment is called the slant height of the
cone.

In a cone of revolution the portion
of an element included between the par-
allel bases of a frustum, as Aa^ or Bh^ is
called the slant height of the frustum.

19. Definition. Similar cones of revolu-
tion are those which are generated by similar right triangles
revolving about homologous sides.

PROPOSITION IV.— THEOREM.

20. If a 'pyramid he inscribed in or circumscribed about a
given cone, its volume will approach the volume of the cone as its
limit, and its lateral surface will approach the convex surface
of the cone as a limit, as the number of faces of the pyramid is
indefinitely increased.

The demonstration is precisely the same as that of Propo-
sition I., substituting cone for cylinder and pyramids for
prisms.

21. Corollary. A frustum of a cone is the limit of the in-
scribed and circumscribed frustums of pyramids, the number of
whose faces is indefinitely increased.

BOOK IX.

297

PROPOSITION v.— THEOREM.

22. The lateral area of a cone of revolution is equal to the
product of the circumference of its base by half its slant height.

Suggestion. Circumscribe a regular
pyramid about the cone, and then sup-
pose the number of its faces to be
indefinitely increased, (y. YII., Prop-
osition XIY.)

23. CoROLLAEY I. Thc proposition
may be formulated, S = tcRI/, where
E is the radius of the base and L the
slant height.

24. Corollary II. The lateral areas

of similar cones of revolution are to each other as the squares
of their slant heights, or as the squares of their altitudes, or as
the squares of the radii of their bases.

^^u:;^ PROPOSITION VI.— THEOREM.

25. The lateral area of a frustum of a cone of revolution is
equal to the half sum of the circumferences of its bases multi*
plied by its slant height.

Suggestion. Circumscribe the frus-
tum of a regular pyramid about the
frustum of the cone (17), and sup-
pose the number of its faces indefi-
nitely increased, (y. YII., Proposi-
tion XIY., Corollary.)

26. Corollary I. The proposition
may be formulated, S = t:{B -\- r)L,
if R and r are the radii of the bases
and L is the slant height.

298

ELEMENTS OF GEOMETRY.

27. Corollary II. The lateral area of a frustum of a cone
of revolution is equal to the circumfer-
ence of a section equidistant from its ]
bases multiplied by its slant height. jji

Suggestion, IKz= ^(om + OM). (v.
Exercise 24, Book I.)

PROPOSITION VII.— THEOREM.

28. The volume of any cone is equal to one-third of the product
of its base by its altitude.

Suggestion. Inscribe a pyramid in
the cone, and suppose the number of
its faces to be indefinitely increased.
(v. VII., Proposition XVIII.)

29. Corollary I. For a cone of
revolution, the proposition may be for-
mulated, V = InW^.H.

30. Corollary II. Similar cones of

revolution are to each other as the cubes of their altitudes, or as
the cubes of the radii of their bases.

exercise.

Theorem. — A frustum of any cone is
equivalent to the sum of three cones whose
common altitude is the altitude of the frus-
tum, and whose bases are the lower base,
the upper base, and a mean proportional
between the bases of the frustum, (v. VII., Proposition XIX.,
Corollary.)

BOOK IX. 299

THE SPHERE.

31. Definition. A spherical segment is a portion of a sphere
included between two parallel planes.

The sections of the sphere made by the parallel planes are
the bases of the segment ; the distance between the planes is
the altitude of the segment.

Let the sphere be generated by the revo-
lution of the semicircle EBF about the axis
EF; and let Aa and Bb be two parallels,
perpendicular to the axis. The solid gener-
ated by the figure ABba is a spherical seg-
ment ; the circles generated by Aa and Bb
are its bases ; and ab is its altitude.

If two parallels Aa and TE are taken, one
of which is a tangent at E^ the solid generated by the figure
EAa is a spherical segment having but one base, which is the
section generated by Aa. The segment is still included be-
tween two parallel planes, one of which is the tangent plane
at E, generated by the line ET.

32. Definition. A zone is a portion of the surface of a
sphere included between two parallel planes.

The circumferences of the sections of the sphere made by
the parallel planes are the bases of the zone ; the distance
between the planes is its altitude.

A zone is the curved surface of a spherical segment.

In the revolution of the semicircle EBF about EFy an arc
AB generates a zone ; the points A and B generate the bases
of the zone ; and the altitude of the zone is ab.

An arc, EA, one extremity of which is in the axis, gener-
ates a zone of one base, which is the circumference described
by the extremity A.

300

ELEMENTS OF GEOMETEY.

33. Definition. When a semicircle revolves about its diam-
eter, the solid generated by any sector of the semicircle is
called a spherical sector.

Thus, when the semicircle EBF revolves
about EFj the circular sector COD generates
a spherical sector.

The spherical sector is bounded by three
curved surfaces; namely, the two conical
surfaces generated by the radii 00 and OD,
and the zone generated by the arc CD. This
zone is called the base of the spherical sector.

OD may, however, coincide with OF, in which case the
spherical sector is bounded by a conical surface and a zone
of one base.

-Again, 00 may be perpendicular to OF, in which case the
spherical sector is bounded by a plane, a conical surface, and
a zone.

PROPOSITION VIII.— LEMMA.

34. The area of the surface generated by a straight line re-
volving about an axis in its plane, is equal to the projection of
the line on the axis multiplied by the circumference of the circle
whose radius is the perpendicular erected at the middle of the
line and terminated by the axis.

Let AB be the straight line revolving
about the axis J^Y; ab its projection on
the axis ; 01 the perpendicular to it, at its
middle point 7, terminating in the axis;
then area AB = ab X circ. 01.

For, draw IK perpendicular and AH
parallel to the axis. The area generated

BOOK IX. 301

by AB is that of the frustum of a cone ; hence (Proposition "
VI., Corollary II.)

area AB = AB X circ. IK.

The triangles ABH and lOK are sinxjlar, being mutually
equiangular, and we have rCj-^] /

but

an4

AH^IK ab_^IK,
AB or AB OV

circ^ = ^(Y., Proposition YIIL),

ah circ. IK

A B^ circ.OI'

ah X circ. 01 = AB X circ. IK.

au ;x^ cue. ui = jljd ;<, circ. in.. q

Therefore _^--^ ~~^ Jp V'-'x

area AB = ah X circ. 01 AN \ ' '

If AB meets XY^ the surface generated is a conical sur-
face ; but the proposition still holds, as may be easily proved.
(v. Proposition Y.)

If AB is parallel to the axis, the result is the same. (y.
Proposition II., Corollary I.)

PEOPOSITION IX.— THEOREM. ^

35. The area of a zone is equal to the product of its altitude
by the circumference of a great circle.

Let the sphere be generated by the revo- ^^.^^

lution of the semicircle EBF about the axis /(i

EF ; and let the arc AD generate the zone b/^-— X^

whose area is required. I

Jj^t the arc AD be divided into any num- ^V'

ber of equal parts. AB^ BG, CD, and draw ^^—
the chords AB, BC, etc. These chords are

302

ELEMENTS OF GEOMETRY.

all equal, since they subtend equal ares ; and the perpendic-
ulars at their middle points all pass through
the centre 0 of the semicircle, and are equal
(II., Proposition YII.).

Let abj he, etc., be the projections of these
chords on the axis. Then, by Proposition
YIII.,

area AB = ab X cire. 01,
area BC == be X cire. 01,
area CD =z ed X cire. 01.

Hence the sum of these areas, which is the area generated
by the broken line ABCD, is equal to

(ab + 6c + cd) X cire. 01;

that is, to ad X cire. 01.

Calling the area generated by the broken inscribed line, 8y

we have

S = ad X cire. 01,

no matter what the number of the equal parts into which the
arc AD is divided. If, now, we increase the number of parts
indefinitely, 01 will approach the radius of the sphere, and
cire. 01 the circumference of a great circle as its limit, and
S will approach the surface of the zone as its limit. There*
fore

surfaee of zone = ad X cireumferenee of great eirele.

36. Corollary. The proposition may be formulated,
S = 2tzR.H,

where R is the radius of the sphere and II Ihe altitude of
the zone.

A'- cJ-r K'J+

BOOK IX. 303

PEOPOSITION X.— THEOREM.

^ Sfr. The area of the surface of a sphere is equal to the product
of its diameter by the circumference of a great circle%

This follows directly from Proposition IX., since the sur«
face of the whole sphere may be regarded as a zone whose
altitude is the diameter of the sphere.

38. Corollary I. This may be formulated^

S = 27: B X2R = 4:nR\

._ Hence the surface of a sphere is equivalent to four great circles. '

39. Corollary II. The surfaces of two spheres are to each
other as the squares of their diameters^ or as the squares of their
radii.

40. Scholium. The area of a spherical degree on a sphere

whose radius is R is -^ (^IH., 69), and, by the aid of this
value, we may readily reduce the area of a spherical polygon
to ordinary square measure. .

PROPOSITION XI.— THEOREM.

41. The volume of a sphere is equal to the area of its surface
multiplied by one-third of its radius.

Circumscribe a polyedron about
the sphere. This may be done by
taking at pleasure points on the sur-
face of the sphere, and drawing
tangent planes at these points. The
circumscribed polyedron wholly con-
tains the sphere, and is greater than

the sphere. Join all the vertices of the polyedron with the
centre of the sphere, and pass planes through the edges of
the polyedron and these lines, thus dividing the polyedron

304 ELEMENTS OF GEOMETRY.

into pyramids, each of which has its vertex at the centre of
the sphere, and has a face of the poly-
edron as its base, and has, therefore,
the radius of the sphere for its alti-
tude (YIII., Proposition YII.). The

volume of any one of these pyramids ^ J^ \pJh

is then one-third of the product of a
face of the polyedron by the radius
of the sphere, and the sum of the

volumes of the pyramids, or the whole volume of the poly-
edron, is one-third of the product of the sum of the faces
of the polyedron by the radius of its sphere ; that is, one-third
of the product of the whole surface of the polyedron by the
radius of the sphere. Representing the surface of the poly-
edron by s, and its volume by u, we have

V = IBs,
and this equation holds no matter what the number of the
faces of the polyedron.

If, now, we increase the number of faces of the polyedron
by drawing additional tangent planes to the sphere, we de-
crease the volume v, for each new tangent plane cuts off a
corner of the polyedron. We may carry on indefinitely this
process of shaving down the polyedron, and may thus make
the difference between its volume and the volume of the
sphere as small as we please ; but we cannot make the two
volumes absolutely coincide. As the two volumes approach
coincidence, the two surfaces also approach coincidence. If,
now, S is the surface and V the volume of the sphere, S is the
limit of s and Y the limit of v, as the number of faces of the
circumscribed polyedron is indefinitely increased. Therefore,
by III., Theorem of Limits,

BOOK IX. 305

42. Corollary I. The result of this proposition may be for-

mulatedj

F = p'

43. Corollary II. The volumes of two spheres are to each
other as the cubes of their radii, or as the cubes of their diameters.

PROPOSITION XII.— THEOREM.

44. The volume of a spherical sector is equal to the area of
the zone which forms its base multiplied by one-third the radius
of the sphere.

The proof is analogous to the proof of Proposition XI.
The form of the circumscribed polyedron is, however, some-
what more complicated, as it will be bounded by a surface
made up of plane faces tangent to the zone of the spherical
sector, and by two pyramidal faces tangent to, or inscribed in,
the two conical surfaces of the spherical sector.

45. Definition. A spherical pyramid is a f. ^^

solid bounded by a spherical polygon and y^ \ /\
the planes of the sides of the polygon ; as ^ \ /\,'''
0-ABCD. The centre of the sphere is the \ // \ /

vertex of the pyramid j the spherical poly- • '^~ Y

gon is its base.

EXERCISE.

Theorem. — The volume of a spherical pyramid is equal to the
area of its base multiplied by one-third of the radius of the
sphere.

u 26*

306

ELEMENTS OF GEOMETRY.

PROPOSITION XIII.— PROBLEM.

46. To find the volume of a spherical segment.

Any spherical segment may be obtained from a spherical
sector by adding to it, or subtracting from it, cones having as
bases the bases of the segment.

For example, let us consider a segment of
two bases which does not contain the centre
of the sphere. The segment generated by
the revolution of ABGD about OC may be '
obtained by taking the cone generated by
OAD from the sum of the cone generated by
OBG and the spherical sector generated by OAB.

Call OC /, ODp, DC h, AD r, BC r', and OA E, and the
volume of the segment F. Then we have the simple relations

r' + p'

P —Pi
K*, r'^ + p""

The area of the zone of the segment is 27ri2.^ (Proposition
IX., Corollary). Hence , ^ ^^/{

V = ^T:hB^ + i7r/r"-— Upr" (Proposition XII., and Proposi-

^ - .T^~--^:;r2r:^^^ tion YII., Corollayg I.),

y = !;:(/ -"^ + %^^^i§^'^i - i^p(^ -%^

V = (/ - p)7:R' - UG" -^ p% [1]

a convenient formula when the distances of the bases of the
segment from the centre of the sphere are given.

Another convenient formula can be obtained by introducing
in [1] A, r, and / in place of j? and p'. We have

F=(/-;>)|[3iJ'

(y + /j' + i'0]-

BOOK IX.

Now
Hence

and

2/i> + ;>'-

,;^it+Jt^\

= f (iP - /' + iJ' - r") - 1 = 3JP - Kr" + /^ - 1',

and we have

[2]

This formula is convenient when the areas of the bases of
the segment are given, and it may be put into words as
follows :

The volume of a spherical segment is equal to the half sum
of its bases multiplied by its altitude plus the volume of a sphere
of which that altitude is the diameter, .

V^^

\ -.^^;

EXERCISES ON BOOK IX.

THEOREMS.

1. Give a strict proof of Proposition I. and Proposition IV. for
the volumes of cylinder and cone, by showing that the difference
between the volumes of the inscribed and circumscribed figures
can be decreased at pleasure.

2. Assuming that if a solid has a plane face the area of that face
is less than the rest of the surface of the solid, prove, first, that
if two convex solids have a plane face in common, and one solid
is wholly included by the other, its surface is less than that of the
other {v. V., 13), and then give a strict proof of Proposition I. and
Proposition IV. for the surfaces of cylinder and cone.

8. The volumes of a cone of revolution, a sphere, and a cylinder
of revolution are proportional to the numbers 1, 2, 3 if the bases
of the cone and cylinder are each equal to a great circle of the
sphere, and their altitudes are each equal to a diameter of the
sphere.

4. An equilateral cylinder (of revolution) is one a section of
which through the axis is a square. An equilateral cone (of
revolution) is one a section of which through the axis is an equi-
lateral triangle. These definitions premised, prove the following
theorems :

I. The total area of the equilateral cylinder inscribed in a
sphere is a mean proportional between the area of the sphere and
the total area of the inscribed equilateral cone. The same is true
of the volumes of these three bodies.

II. The total area of the equilateral cylinder circumscribed
about a sphere is a mean proportional between the area of the
sphere and the total area of the circumscribed equilateral cone.
The same is true of the volumes of these three bodies.

5. If h is the altitude of a segment of one base in a sphere whose
radius is r, the volume of the segment is equal to Trh'^H — ^h),

6. The volumes of polyedrons circumscribed about the same
sphere are proportional to their surfaces.

308 .

MISCELLANEOUS EXERCISES

ON THE

GEOMETKY OF SPACE.

1. A PERPENDICULAR let fall from the middle point of a line
upon any plane not cutting the line is equal to one-half the sum
of the perpendiculars let fall from the ends of the line upon the
same plane.

2. The perpendicular let fall from the point of intersection of
the medial lines of a given triangle upon any plane not cutting
the triangle is equal to one-third the sum of the perpendiculars
from the vertices of the triangle upon the same plane.

3. The perpendicular from the centre of gravity of a tetraedron
upon any plane not cutting the tetraedron is equal to one-fourth
the sum of the perpendiculars from the vertices of the tetraedron
upon the same plane.

4. The volume of a truncated triangular prism is equal to the
product of the area of its lower base by the perpendicular upon
the lower base let fall from the intersection of the medial lines of
the upper base.

6. The volume of a truncated parallelopiped is equal to the
product of the area of its lower base by the perpendicular from
the centre of the upper base upon the lower base.

6. If ABCD is any tetraedron, and O any point within it, and
if the straight lines AO^ BO^ CO^ DO, are produced to meet the
faces in the points a, 6, c, d, respectively, then

Aa'^ Bb~^ Cc "^ Dd

809

310 - ELEMENTS OF GEOMETRY.

/7. If the three face angles of the vertical triedral angle of a
tetraedron are right angles, and the lengths of the lateral edges
are represented by a, 6, and c, and of the altitude by p, then

8. If the three face angles of the vertical triedral angle of a
tetraedron are right angles, the square of the area of the base is
equal to the sum of the squares of the areas of the lateral faces.

9. The perpendicular from the middle point of the diagonal of
a rectangular parallelepiped upon a lateral edge bisects the edge,
and is equal to one-half of the projection of the diagonal upon
the base.

10. A straight line of a given length moves so that its extremi-
ties are constantly upon two given perpendicular but non-inter-
secting straight lines : what is the locus of the middle point of the
moving line ?

PROBLEMS.

11. To cut a given polyedral angle of four fac^s by a plane so
that the section shall be a parallelogram.

^\ 12. To cut a cube by a plane so that the section shall be a regular
hexagon.

13. To find the ratio of the volumes generated by a rectangle
revolving successively about its two adjacent sides.

SYLLABUS OP PROPOSITIONS

SOLID GEOMETRY.

BOOK VI.

THEOREMS.

Proposition I.

Through any given straight line a plane may be passed, but
the line will not determine the plane.

Proposition II.

A plane is determined, 1st, by a straight line and a point with-
out that line ; 2d, by two intersecting straight lines ; 3d, by three
points not in the same straight line ; 4th, by two parallel straight
lines.

Corollary, The intersection of two planes is a straight line.

Proposition III.

From a given point without a plane one perpendicular to the
plane can be drawn, and but one ; and the perpendicular is the
shortest line that can be drawn from the point to the plane.

Corollary. At a given point in a plane one perpendicular can be
erected to the plane, and but one.

Proposition IV.

If a straight line is perpendicular to each of two straight lines
at their point of intersection, it is perpendicular to the plane of
those lines.

Corollary I. At a given point of a straight line, one plane can
be drawn perpendicular to the line, and but one.

Corollary II. Through a given point without a straight line,
one plane can be drawn perpendicular to the line, and but one.

311

312 ELEMENTS OF GEOMETRY.

Proposition V.

Two lines in space having the same direction are parallel.
Corollary, Two lines parallel to the same line are parallel to
each other.

Proposition VI.

If two straight lines are parallel, every plane passed through
one of them and not coincident with the plane of the parallels is
parallel to the other.

Corollary I. Through any given straight line a plane can be
passed parallel to any other given straight line.

Corollary II. Through any given point a plane can be passed
parallel to any two given straight lines in space.

Proposition VII.

Planes perpendicular to the same straight line are parallel to
each other.

Proposition VIII.

The intersections of two parallel planes with any third plane
are parallel.

Proposition IX.

A straight line perpendicular to one of two parallel planes is
perpendicular to the other.

Corollary. Through any given point one plane can be passed
parallel to a given plane, and but one.

Proposition X.

If two angles, not in the same plane, have their sides respec-
tively parallel and lying in the same direction, they are equal
and their planes are parallel.

Proposition XI.

If one of two parallel lines is perpendicular to a plane, the
other is also perpendicular to that plane.

Corollary. Two straight lines perpendicular to the same plane
are parallel to each other.

Proposition XII.
Two diedral angles are equal if their plane angles are equal.

Proposition XIII.
Two diedral angles are in the same ratio as their plane angles.

SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 313

Proposition XIV.

If a straight line is perpendicular to a plane, every plane passed
through the line is perpendicular to the plane.

Proposition XV.

If two planes are perpendicular to each other, a straight line
drawn in one of them, perpendicular to their intersection, is
perpendicular to the other.

Corollary I. If two planes are perpendicular to each other, a
straight line drawn through any point of their intersection per-
pendicular to one of the planes will lie in the other.

Corollary II. If two planes are perpendicular, a straight line
let fall from any point of one plane perpendicular to the other
will lie in the first plane.

Proposition XVI.

If two intersecting planes are each perpendicular to a third
plane, their intersection is also perpendicular to that plane.

Proposition XVII.

Through any given straight line a plane can be passed perpen-
dicular to any given plane.

Proposition XVIII.
The projection of a straight line upon a plane is a straight line.

Proposition XIX.

The acute angle which a straight line makes with its own pro-
jection upon a plane is the least angle it makes with any line of
that plane.

Proposition XX.

The sum of any two face angles of a triedral angle is greater
than the third.

Proposition XXI.

The sum of the face angles of any convex polyedral angle is
less than four right angles.

Proposition XXII.

If two triedral angles have the three face angles of the one re-
spectively equal to the three face angles of the other, the corre-
sponding diedral angles are equal.
0 27

314 ELEMENTS OF GEOMETRY.

BOOK VII.

THEOREMS.
Proposition I.

The sections of a prism made by parallel planes are equal poly-
gons.

Corollary. Any section of a prism made by a plane parallel to
the base is equal to the base.

Proposition II.

The lateral area of a prism is equal to the product of the perim-
eter of a right section of the prism by a lateral edge.

Corollary. The lateral area of a right prism is equal to the
product of the perimeter of its base by its altitude.

Proposition III.

Two prisms are equal, if three faces including a triedral angle
of the one are respectively equal to three faces similarly placed
including a triedral angle of the other.

Corollary I. Two truncated prisms are equal, if three faces in-
cluding a triedral angle of the one are respectively equal to three
faces similarly placed including a triedral angle of the other.

Corollary II. Two ri^ht prisms are equal if they have equal
bases and equal altitudes.

Proposition IV.

Any oblique prism is equivalent to a right prism whose base is
a right section of the oblique prism, and whose altitude is equal
to a lateral edge of the oblique prism.

Proposition V.

Any parallelopiped is equivalent to a rectangular parallelopiped
of the same altitude and an equivalent base.

Proposition VI.

The plane passed through two diagonally opposite edges of a
parallelopii^ed divides it into two equivalent triangular prisms.

Proposition VII.

Two rectangular parallelepipeds having equal bases are to each
other as their altitudes.

SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 315

Proposition VIII.

Two rectangular parallelepipeds having equal altitudes are to
each other as their bases.

Proposition IX.

Any two rectangular parallelopipeds are to each other as the
products of their three dimensions.

Proposition X.

The volume of a rectangular parallelopiped is equal to tne
product of its three dimensions, the unit of volume being the
cube whose edge is the linear unit.

Proposition XI.

The volume of any parallelopiped is equal to the product of the
area of its base by its altitude.

Proposition XII.

The volume of a triangular prism is equal to the product of its
base by its altitude.

Corollary. The volume of any prism is equal to the product of
its base by its altitude.

Proposition XIII.

If a pyramid is cut by a plane parallel to its base, 1st, the edges
and the altitude are divided proportionally ; 2d, the section is a
polygon similar to the base.

Corollary I. If a pyramid is cut by a plane parallel to its base,
the area of the section is to the area of the base as the square of
its distance from the vertex is to the square of the altitude of the
pyramid.

Corollary II. If two pyramids have equal altitudes and equiva-
lent bases, sections made by planes parallel to their bases and at
equal distances from their vertices are equivalent.

Proposition XIV.

The lateral area of a regular pyramid is equal to the product of
the perimeter of its base by half its slant height.

Corollary. The lateral area of the frustum of a regular pyra-
mid is equal to the half sum of the perimeters of its bases multi-
plied by the slant height of the frustum.

316 ELEMENTS OF GEOMETRY.

Proposition XV.

If the altitude of any given triangular pyramid is divided into
equal parts, and through the points of division planes are passed
parallel to the base of the pyramid, and on the sections made by
these planes as upper bases prisms are described having their
edges parallel to an edge of the pyramid and their altitudes equal
to one of the equal parts into which the altitude of the pyramid
is divided, the total volume of these prisms will approach the
volume of the pyramid as its limit as the number of parts into
which the altitude of the pyramid is divided is indefinitely
increased.

Proposition XVI.

Two triangular pyramids having equivalent bases and equal
altitudes are equivalent.

Proposition XVII.

A triangular pyramid is one-third of a triangular prism of the
same base and altitude.

Corollary. The volume of a triangular pyramid is equal to one-
third of the product of its base by its altitude.

Proposition XVIII.

The volume of any pyramid is equal to one-third of the product
of its base by its altitude.

Proposition XIX.

A frustum of a triangular pyramid is equivalent to the sum of
three pyramids whose common altitude is the altitude of the frus-
tum, and whose bases are the lower base, the upper base, and a
mean proportional between the bases of the frustum.

Corollary. A frustum of any pyramid is equivalent to the sum
of three pyramids whose common altitude is the altitude of the
frustum, and whose bases are the lower base, the upper base, and
a mean proportional between the bases of the frustum.

Proposition XX.

A truncated triangular prism is equivalent to the sum of three
pyramids whose common base is the base of the prism and whose
vertices are the three vertices of the inclined section.

Proposition XXI.
Only five regular (convex) polyedrons are possible.

SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 317

BOOK Till.
THEOREMS.
Proposition I.

Every section of a cylinder made by a plane passing through
an element is a parallelogram.

Corollary. Every section of a right cylinder made by a plane
perpendicular to its base is a rectangle.

Proposition II.

The bases of a cylinder are equal.

Corollary I. Any two parallel sections of a cylindrical surface
are equal.

Corollary II. All the sections of a circular cylinder parallel to
its bases are equal circles, and the straight line joining the centres
of the bases passes through the centres of all the parallel sections.

Proposition III.

Every section of a cone made by a plane passing through its
vertex is a triangle.

Proposition IV.

If the base of a cone is a circle, every section made by a plane
parallel to the base is a circle.

Corollary. The axis of a circular cone passes through the centres
of all the sections parallel to the base.

Proposition V.

Every section of a sphere made by a plane is a circle.

Corollary I. The axis of a circle on a sphere passes through
the centre of the circle.

Corollary II. All great circles of the same sphere are equal.

Corollary III. Every great circle divides the sphere into two
equal parts.

Corollary IV. Any two great circles on the same sphere bisect
each other.

Corollary V. An arc of a great circle may be drawn through any
two given points on the surface of a sphere, and, unless the points
are the opposite extremities of a diameter, only one such arc can
be drawn.

Corollary VI. An arc of a circle may be drawn through any
three given points on the surface of a sphere.

27*

318 ELEMENTS OF GEOMETRY.

Proposition VI.

All the points in the circumference of a circle on a sphere are
equally distant from either of its poles.

Corollary I. All the arcs of great circles drawn from a pole of a
circle to points in its circumference are equal.

Corollary II. The polar distance of a great circle is a quadrant.

Corollary III. If a point on the surface of a sphere is at a quad-
rant's distance from each of two given points of the surface,
which are not opposite extremities of a diameter, it is the pole
of the great circle passing through them.

Proposition VII.

A plane tangent to a sphere is perpendicular to the radius
drawn to the point of contact.

Corollary. A plane perpendicular to a radius of a sphere at its
extremity is tangent to the sphere.

Proposition VIII.

The intersection of two spheres is a circle whose plane is per-^
pendicular to the straight line joining their centres, and whose
centre is in that line.

Proposition IX.

The angle of two arcs of great circles is equal to the angle of
their planes, and is measured by the arc of a great circle described
from its vertex as a pole and included between its sides (produced
if necessary).

Corollary. All arcs of great circles drawn through the pole of a
given great circle are perpendicular to its circumference.

Proposition X.

If the first of two spherical triangles is the polar triangle of the
second, then, reciprocally, the second is the polar triangle of the
first.

Proposition XI.

In two polar triangles, each angle of one is measured by the
supplement of the side lying opposite to it in the other.

Proposition XII.

Two triangles on the same sphere are either equal or symmet-
rical when two sides and the included angle of one are respectively
equal to two sides and the included angle of the other.

SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 319

Proposition XIII.

Two triangles on the same sphere are either equal or symmet-
rical when a side and the two adjacent angles of one are respec-
tively equal to a side and the two adjacent angles of the other.

Proposition XIV.

Two triangles on the same sphere are either equal or symmet-
rical when the three sides of one are respectively equal to the
three sides of the other.

Proposition XV.

If two triangles on the same sphere are mutually equiangular,
they are mutually equilateral, and are either equal or symmet-
rical.

Proposition XVI.

Any side of a spherical triangle is less than the sum of the
other two.

Proposition XVII.

The sum of the sides of a convex spherical polygon is less than
the circumference of a great circle.

Proposition XVIII.

The sum of the angles of a spherical triangle is greater than
two, and less than six, right angles.

Proposition XIX.
Two symmetrical spherical triangles are equivalent.

Proposition XX.

If two arcs of great circles intersect on the surface of a hemi-
sphere, the sum of the opposite spherical triangles which they
form is equivalent to a lune whose angle is the angle between the
arcs in question.

Proposition XXI.

A lune is to the surface of the sphere as the angle of the lune is
to four right angles.

Corollary. The area of a lune is expressed by twice its angle,
the angular unit being the degree, and the unit of surface the
spherical degree.

320 ELEMENTS OF GEOMETRY.

Proposition XXII.
The area of a spherical triangle is equal to the excess of the
sum of its angles over two right angles.

Proposition XXIII.
The shortest line that can be drawn on the surface of a sphere
between two points is the arc of a great circle, not greater than a
semi-circumference, joining the two points.

BOOK IX.

THEOREMS.
Proposition I.
If a prism whose base is a regular polygon be inscribed in or
circumscribed about a given cylinder, its volume will approach
the volume of the cylinder as its limit, and its lateral surface will
approach the lateral surface of the cylinder as its limit as the
number of sides of its base is indefinitely increased.

Proposition II.

The lateral area of a cylinder is equal to the product of the
perimeter of a right section of the cylinder by an element of the
surface.

Corollary I. The lateral area of a cylinder of revolution is equal
to the product of the circumference of its base by its altitude.
This may be formulated,

Corollary II. The lateral areas of similar cylinders of revolu-
tion are to each other as the squares of their altitudes, or as the
squares of the radii of their bases.

Proposition III.

The volume of a cylinder is equal to the product of its base by
its altitude.

Corollary I. For a cylinder of revolution this may be formu-
lated,

Corollary II. The volumes of similar cylinders of revolution
are to each other as the cubes of their altitudes, or as the cubes
of their radii.

SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 321

Proposition IV.

If a pyramid be inscribed in or circumscribed about a given
cone, its volume will approach the volume of the cone as its limit,
and its lateral surface will approach the convex surface of the
cone as its limit as the number of faces of the pyramid is indefi-
nitely increased.

Corollary, A frustum of a cone is the limit of the inscribed and
circumscribed frustums of pyramids, the number of whose faces
is indefinitely increased.

Proposition V.

The lateral area of a cone of revolution is equal to the product
of the circumference of its base by half its slant height.
Corollary I. This proposition may be formulated,

S = ^RL.

Corollary II. The lateral areas of similar cones of revolution
are to each other as the squares of their slant heights, or as the
squares of their altitudes, or as the squares of the radii of their
bases.

Proposition VI.

The lateral area of a frustum of a cone of revolution is equal to
the half sum of the circumferences of its bases multiplied by its
slant height.

Corollary I. This proposition may be formulated,

S=Tr{R-{.r)L.

Corollary II. The lateral area of a frustum of a cone of revolu-
tion is equal to the circumference of a section equidistant from its
bases multiplied by its slant height.

Proposition VII.

The volume of any cone is equal to one-third the product of its
base by its altitude.

Corollary I. For a cone of revolution this proposition may be
formulated,

V= ITER'S.

Corollary II. Similar cones of revolution are to each other aa
the cubes of their altitudes, or as the cubes of the radii of their

322 ELEMENTS OF GEOMETRY.

Proposition VIII.

The area of the surface generated by a straight line revolving
about an axis in its plane is equal to the projection of the line
on the axis multiplied by the circumference of the circle whose
radius is the perpendicular erected at the middle of the line and
terminated by the axis.

Proposition IX.

The area of a zone is equal to the product of its altitude by the
circumference of a great circle.

Corollary. This proposition may be formulated,

Proposition X.

The area of the surface of a sphere is equal to the product of^
its diameter by the circumference of a great circle.
Corollary I. This may be formulated,

S = 27ri2 X2B = ^TzB^,-

Hence the surface of a sphere is equivalent to four great circles.

Corollary II. The surfaces of two spheres are to each other aa
the squares of their diameters, or as the squares of their radii.

Proposition XI.

The volume of a sphere is equal to the area of its surface multi-
plied by one-third of its radius.
Corollary I. This proposition may be formulated,

Corollary II. The volumes of two spheres are to each other as
the cubes of their radii, or as the cubes of their diameters.

Proposition XII.

The volume of a spherical sector is equal to the area of the zone
which forms its base multiplied by one-third the radius of the
sphere. ^__

" Of THB

THE END. T ^ I T^ F

({

THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW

AN INITIAL FINE OF 25 CENTS

WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO SO CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.

JEP 27 1934

i.'^-. Uvti^fe

Nov-i^t94fl1yt

::-:Jan'49R^

LD21-100m-7,'33

YB 56844

'S'^Ut

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Full text of "Chauvenet's treatise on elementary geometry"

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