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ELEMENTARY TREATISE
GEOMETRY,
SIMPLIFIED FOR
BEGINNERS NOT VERSED IN ALGEBRA,
PART I
CONTAINING
PLANE GEOMETRY,
WITH ITS APPLICATION TO THE SOLUTION OP PROBLEMS.
BY FRANCIS J. GRUND.
L»
Effete 3E&ftfon, stereotype*
BOSTON:
CHARLES J. HENDEE,
AND
G. W. PALMER AND COMPANY,
1838.
DISTRICT OF MASSACHUSETTS, TO WIT:
District Clerk's Office.
Bjb it remembered, that on the fourth day of December, A. D. 1830, in the fifty-
fifth year of the Independence of the United States of America, Francis J
Grhnd, of the said district, lias deposited in this office the title of a book, too
right whereof he claims as author, in the words following, to wit :
Arc Elementary Treatise on Geometry, simplified for Beginners not versed m
Algebra. Part I, containing Plane Geometry, with its Application to the Solution of
Problems. By Francis J. Grund. Second Edition.
In conformity to the act of the Congress of the United States, entitled, " An
act for the encouragement of learning, by securing the copies of maps, charts,
and books, to the authors and proprietors of such copies, during the times therein
mentioned ;" and also to an act, entitled, " An act supplementary to an act,
entitled, ' An act for the encouragement of learning, by securing the copies of
maps, charts, and books, to the authors and proprietors of such copies during the
times therein mentioned ;' and extending the benefits thereof to the arts of de-
signing, engraving, and etching historical and other prints."
').;*- * • • • ,'JtooC W. DAVIS,
\ ,. •derkqf't^e'Di\1rict of Massachusetts.
CAJORI
J .
RECOMMENDATIONS.
From John Farrar, Professor of Mathematics and Natural Philosophy
at Harvard University.
Mr. Gruud's Elementary Treatise on Geometry contains much useful
matter, not generally to be found in English works of this description.
There is considerable novelty, also, in the style and arrangement. The
subject appears to be developed in a manner well suited to the younger
class of learners, and to such an extent, and with such illustrations, as
renders it a valuable introduction to the more extended works on Ge-
ometry.
JOHN FARRAR.
February 18th, 1830.
From G. B. Emerson, Principal of the English Classical School,
Boston.
Mr. Grund's Geometry unites, in an unusual degree, strictness of de-
monstration with clearness and simplicity. It is thus very well suited to
form habits of exact reasoning in young beginners, and to give them
favorable impressions of the science. I have adopted it as a text book
in my own school.
GEO. B. EMERSON.
February 18th, 1830.
»
From E. Bailey, Principal of the Young Ladies' High School, Boston.
Dear Sir — From the specimens of your work on Geometry which I
have seen, and especially from the sheets I have used in my school since
it went to the press, I have formed a high opinion of its merits. The
general plan of the work appears to be very judicious, and you have
executed it with great ability. Simplicity has been carefully studied, yet
not at the expense of rigid demonstration. In this respect, it seems ad-
mirably fitted for the use of common schools. Believing your work cal-
918317
4 RECOMMENDATIONS.
culated and destined to do much good, in a department of science which
has been too long neglected, I hope it may soon become generally known.
Very respectfully, yours, &c.
E. BAILEY.
February 17th, 1830.
From F. P. Leverett, Principal of the Latin School, Boston.
December 7th, 1830.
Dear Sir — I have looked with much satisfaction over the sheets of
the second edition of your ' First Lessons in Plane Geometry.' It is a
more simple and intelligible treatise on Geometry than any other with
which I am acquainted, and seems to me well adapted to the understand-
ings of young scholars.
I am, dear sir, respectfully yours,
F. P. LEVERETT.
From William B. Fowle, Principal of the Monitorial School, Boston.
Boston, February 17th, 1830.
Mr. Grund — Dear Sir— I have examined every page of your
'First Lessons in Plane Geometry.' Its reception everywhere augurs
well for the success of your book, which is an extension and practical
application of Fraucceur's. It has fulfilled my wishes, and I shall imme-
diately introduce it into my school.
Yours, very respectfully, -
• WILLIAM B. FOWLE.
krom Walter R. Johnson, Principal of the Philadelphia High School.
Philadelphia, Nov. 27th, 1830.
Dear Sir— The First Lessons in Plane Geometry, with a perusal of
which I have been favored, appears to me eminently calculated to lay
the foundation of a clear and comprehensive knowledge of the demon-
strative parts of that important science.
As it has obviously been the result of actual experience in teaching, it
commends itself to the attention of the profession, by the assurance that
it is really adapted to the comprehension and attainments of those for
whom it was designed. Pennit me to express the hope that it may meet
its full share of that encouragement which works in this department are
beginning to receive in every part of our country.
I remain, dear sir, very respectfully yours,
WALTER R. JOHNSON.
PREFACE.
Popular Education, and the increased study of Mathe-
matics, as the proper foundation of all useful knowledge,
seem to call especially for Elementary Treatises on Geome-
try, as has been evinced in the favorable reception of the
first edition of this work within a few months of the date of
its publication. A few changes have been made in the
present edition, which, it is hoped, will contribute to the use-
fulness of the work as a book for elementary instruction.
The author acknowledges with pleasure the valuable aid
he has received from, some of the most experienced and
distinguished instructers ; and is, in this respect, particularly
indebted to the kindness of Messrs. E. Bailey, George B.
Emerson, and Miss Elizabeth P. Peabody, of Boston, at
whose suggestion several demonstrations have been simpli-
fied, in order to adapt the work to the capacity of early
beginners.
As regards the use of it in schools and seminaries, the
teacher will find sufficient directions in the remarks inserted
in the body of the work.
The Problems, of which the third and fourth parts are
principally selected from those of Meier Hirsch, form a
section by themselves, in order to be more easily referred to.
1*
6 PREFACE.
The teacher may, according to his own judgment, use as
many of them at the end of each section, as may be solved
by the principles the pupils have become acquainted with.
Boston, September 30, 1830.
PREFACE TO THE STEREOTYPE EDITION.
The present stereotype edition differs from the previous
ones only in the typographical arrangement, to meet the
view of the publishers, whose intention it is to reduce its
price, in order to bring it within the reach of common schools
throughout the Union.
F.J.G.
Boston, March 27, 1832.
^fc'
TABLE OF CONTENTS
Introduction 9
Definitions .' 11
Questions on Definitions 15
.Notation and Significations 18
Axioms {JO
SECTION L
Of Straight Lines and Angles 22.
Recapitulation of the Truths contained in the First Sec-
tion 34
SECTION II.
PART I.
Of the Equality of Triangles 37
PART II.
Of Geometrical Proportions and Similarity of Triangles .... 53
Theory of Geometrical Proportions 53
Similarity of Triangles 67
Recapitulation of the Truths contained in Section IL,
Part I 76
Recapitulation of the Truths contained in Part II 79
Questions on Proportions 79
Questions on Similarity of Triangles 81
«*
8 CONTENTS.
SECTION III.
Of the Measurement of Surfaces 83
Recapitulation of the Truths contained in the Third Sec-
tion 99
SECTION IV.
Of the Properties of the Circle 102
What is meant by Squaring a Circle 128
Recapitulation of the Truths contained in the Fourth Sec-
tion 134
SECTION V.
Application of the foregoing Principles to the So-
lution of Problems 141
PART I.
Problems relative to the Drawing and Division of Lines
and Angles % 141
PART II.
Transformation of Geometrical Figures 159
PART III.
Partition of Figures by Drawing 171
PART IV.
Construction of Triangles t 182
Appendix, containing Exercises for the Slate 189
GEOMETRY.
INTRODUCTION.
If, without regarding the qualities of bodies, viz :
their smoothness, roughness, color, compactness, tenacity,
&c, we merely consider the space which they fill — their
extension in space — they become the special subject of
mathematical investigation, and the science which treats
of them, is called Geometry. -
The extensions of bodies are, called dimensions. Every
body has three dimensions, viz : length, breadth, and
depth. Of a wall or a house, for instance, you can form
no idea, without conceiving it to extend in length, breadth,
and depth ; and the same is the case with every other
body you can think of.
The limits or confines of bodies are called surfaces
(superfices), and may be considered independently of the
bodies themselves. So you may look at the front of a
house, and inquire how long and how high is that house,
without regarding its depth; or you may consider the
length and breadth of a field, without asking how deep it
goes into the ground, &c. In all such cases, you merely
consider two dimensions. A surface is, therefore, defined
to be an extension in length and breadth without depth.
16 GEOMETRY.
The limits or edges of surfaces are called lines, and
may again be considered independently of the surfaces
themselves. You may ask, for instance, how long is the
front of such a house, without regarding its height ; or
how far is it from Boston to Roxbury, without inquiring
how hroad is the road. Here, you consider evidently
only one dimension ; and a line, therefore, is defined to
be an extension in length without breadth or depth.
The beginning and end of lines are called points
They merely mark the positions of lines, and can, there-
fore, of themselves, have no magnitude. To give an
example : when you set out from Boston to Roxbury, you
may indicate the place you .start from, which you may
call the point of starting. If this chances to be Marl-
borough Hotel, you do not ask how long, or broad, or
deep that place is ; it suffices for you to know the spot
where you begin your journey. A point is, therefore,
denned to be mere position, without either length or
breadth.
Remark. A point is represented on paper or on a
board, by a small dot. A line is drawn on paper with a
pointed lead pencil or pen ; and on the board, with a
thin mark made with chalk. The extensions of sur-
faces are indicated by lines ; and bodies are represented
on paper or on the board, according to the rules of
perspective.
Before we begin the study of Geometry, it is neces-
sary, first, to acquaint ourselves with the meaning of some
terms, which are frequently made use of in books treating
on that science.
GEOMETRY. H
Definitions.
A line is called straight, when every part of it lies in
the same direction, thus,
Any line in which no part is straight, is called a curve
line.
A geometrical plane is a surface, in which two points
being taken at pleasure, the straight line joining them
lies entirely in that surface.* A surface in which no
part is plane, is called a curved surface. Any plane sur-
face, terminated by lines, is called a geometrical figure.
The simplest rectilinear figure, terminated by three
straight lines, is called a triangle.
A geometrical figure, terminated by four straight lines,
is called a quadrilateral — by 5, a pentagon — by 6, a hexa-
gon — by 7, a heptagon — by 8, an octagon, &c.
Any geometrical figure, terminated by more than three
straight lines, is (by some authors) called a polygon^
When two straight lines meet, they form an angle ; the
point at which they meet is called the vertex, and the
lines themselves are called the legs of the angle. When
a straight line meets another, so as to make the two
adjacent angles equal, the angles are called right an-
* The teacher can give an illustration of this definition, by taking
anywhere on a piece of pasteboard, two points and joining them
by a piece of stiff wire. Then, by bending the board, the wire,
which represents the line, will be off the board, and you have a
curved surface ; and by stretching the board, so as to make the
wire fall upon it, you have a plane.
t Legendre calls all geometrical figures polygons.
12
GEOMETRY.
gles, and the lines are said to be perpendicular to each
other.
Any angle smaller than a right angle is called acute,
and when greater than a right angle, an obtuse angle.*
Two lines which, lying in the same plane, and how-
ever far extended in both directions, never meet, are said
to be parallel to each other.
When two lines, situated in the same plane, are not par-
allel, they are either converging or diverging. Two lines
are said to be converging, if, when extended in the direc-
tion we consider, they grow nearer each other; and
diverging, if the reverse takes place.
Converging.
Diverging.
* Angles are measured by arcs of circles, described with any
radius between their legs. Here the teacher may state, that the
circle is divided into 360 equal parts, called degrees ; each degree,
again, into 60 equal parts, called minutes; a minute, again, sub-
divided into 60 equal parts, called seconds, &c. ; and that the
magnitude of an angle can thus be expressed in degrees, minutes,
seconds, &c. of an arc of a circle, contained between its legs.
GEOMETRY.
13
A triangle is called equilateral, when all its sides are
equal.
A triangle is called isosceles, when two of its sides only
are equal.
A triangle is called scalene, when none of its sides are
equal.
A triangle is also called right-angled, when it contains
a right angle ;
and oblique-angled, when it contains no right angle.
A parallelogram is a quadrilateral whose opposite sides
are parallel.
v_^
A rectangle, or oblong, is a right-angled parallelogram
14 GEOMETRY.
A square is a rectangle whose sides are all epual.
A rhombus or lozenge, is a parallelogram whose sides
are all equal.
A trapezoid is a quadrilateral in which two sides only
are parallel.
A straight line joining two vertices, which are not on
the same side of a geometrical figure, is called a diagonal.
The side which is opposite to the right angle, in a
right-angled triangle, is called the hypothenuse.
A circle is a surface terminated on all sides by a curve
line returning into itself, all points of which are at an
equal distance from one and the same point, called the
centre.
The curve line itself is called the circumference. Any
GEOMETRY. 15
part of it is called an arc. A straight line, drawn from
the centre of a circle to any point of the circumference,
is called a radius. A straight line, drawn from one point
of the circumference to the other, passing through the
centre, is called a diameter. A straight line, joining any
two points of the circumference, without passing through
the centre, is called a chord.
The plane surface included within an arc of a circle
and the chord on which it stands is called a segment.
The arc of a circle which stands on a diameter is
called a semi-circumference. The plane surface included
within a semi-circumference and a diameter is called a
semi-circle.
The plane surface included within two radii and an
arc of a circle is called a sector. (See the figure, page
14.) If the two radii are perpendicular to each other,
the sector is called a quadrant.
A straight line, which, drawn without the circle, and
however far extended in both directions, meets the cir-
cumference only in one point, is called a tangent.
QUESTIONS ON DEFINITIONS.
What is that science called, which treats of the exten-
sions of bodies, considered separately from all their other
qualities 1
What are the extensions of bodies called ?
What are the limits or confines of bodies called T
How do you define a surface 1
]5 GEOMETRY.
What are the limits of surfaces called ?
How do you define a line ?
What are the beginning and end of lines called ?
How do you define a point ?
How is a geometrical point represented?
How is a line represented ? How a surface ?
How do you define a straight line ?
What do you call a line in which no part is straight ?
What is that surface called, in which, when two points
are taken at pleasure, the straight line joining them lies
entirely in it ?
What do you call a surface in which no part is plane ?
What is a plane surface called when terminated by lines?
By how many straight lines is the simplest rectilinear
figure terminated ?
What do you call it ?
What do you call a geometrical figure terminated by
four straight lines ?
What, if terminated by five straight lines?
What, if by six ? By seven ? By eight ?
What are all geometrical figures terminated by more
than three straight lines called ?
When two straight lines meet, what do they form ?
What is the point where the lines meet called ?
What do you call the lines which form the angle ?
If one straight line meets another, so as to make the
two adjacent angles equal, what do you call these angles ?
What are the lines themselves said to be?
What is an angle which is smaller than a right angle
called?
What an angle larger than a right angle ?
What do you call two lines, which, situated in the
same plane, and however far extended both wavs never
meet?
GEOMETRY. 17
When are two lines said to be converging? When,
diverging ?
When a triangle has all its sides equal, what is it
called ?
When two of its sides only are equal, what 1
When none of its sides are equal, what ?
What is a triangle called, when it contains a right
angle ?
What, if it does not contain one ?
What is a quadrilateral, whose opposite sides are par-
allel, called?
What is a right-angled parallelogram called ?
What is an equilateral rectangle called ?
What, an equilateral parallelogram ?
What, a quadrilateral in which two sides only are
parallel ?
How is a circle terminated ?
What is the line called which terminates a circle ?
What is any part of the circumference called?
What, a straight line, drawn from the centre, to any
point in the circumference T
What, a straight line joining two points of the circum-
ference, and passing through the centre ?
What, a straight line joining two points of the circum-
ference, without passing through the centre ?
What is the plane surface, included within an arc and
the chord which joins its two extremities, called ?
What is that part of the circumference called, which is
cut off by the diameter ?
What, the plane surface within a semi-circumference
and a diameter ?
What, the surface within an arc of a circle and the
two radii drawn to its extremities ?
2* •
18 GEOMETRY.
What is the sector called, if the two radii are perpen-
dicular to each other ?
What is the name of a straight line, drawn without the
circle, which, extended both ways ever so far, touches
the circumference only in one point*
NOTATION AND SIGNIFICATIONS.
For the sake of shortening expressions, and thereby to
facilitate language, mathematicians have agreed to adopt
the following signs :
=r stands for equal ; e. g., the line AB — CD means,
that the line AB is equal to the line CD.
-}- stands for plus or more ; e. g., the lines AB -}- CD
means, that the length of the line CD is to be added to
the line AB.
— stands for minus or less; e.g.,liue AB — CD means,
that the length of the line CD is to be taken away from
the line AB.
X is the sign of multiplication.
: is the sign of division.
<^ stands for less than ; e. g., the line AB <^ CD means,
that the line AB is shorter than the line CD.
> stands for greater than; e. g., the line AB ]> CD
means, that the line AB is longer than the line CD.
A point is denoted by a single letter of the alphabet
chosen at pleasure ; e. g. t
the point B.
A line is represented by two letters placed at the be-
ginning and end of it ; e. g. t
A. B
the line AB.
GEOMETRY
19
An angle is commonly denoted by three letters, the
one that stands at the vertex always placed in the middle ;
A
8
the angle ABC or CBA. It is sometimes also repre-
sented by a single letter placed within the angle ; e. g.,
the angle a.
A triangle is denoted by three letters placed at the
three vertices; e. g.,
the triangle ABC.
A polygon is denoted by as many letteis as there are
vertices ; e. g.,
£!
£>
the pentagon ABCDE.
A quadrilateral is sometimes denoted only by two let-
ters, placed at the opposite vertices; e. g. t
the quadrilateral AB.
20 GEOMETRY.
QUESTIONS ON NOTATION AND SIGNIFICATIONS.
What is the sign of equality ?
What sign stands for plus or more ?
What for minus or less 1
What for multiplication ?
What for division 1
What for less than ?
What for more than ?
How is a point denoted ?
How a line ?
How an angle ?-
How a triangle ?
How a quadrilateral 1
How any polygon 1
Axioms.
There are certain invariable truths, which are at once
plain and evident to every mind, and which are frequently
made use of, in the course of geometrical reasoning. As
you will frequently be obliged to refer to them, it will be
well to recollect the following ones particularly :
TRUTH I.
Things which are equal to the same thing, are equal
to one another.
TRUTH II.
Things which are similar to the same thing, are similar
to one another.
- GEOMETRY. 21
TRUTH III.
If equals be added to equals, the wholes are equal.
TRUTH IV.
If equals be taken from equals, the remainders are
equal.
TRUTH V.
The whole is greater than any one of its parts.
TRUTH VI.
The sum of all the parts is equal to the whole.
TRUTH VII.
Magnitudes which coincide with one another, that is,
which exactly fill the same space, are equal to one another.
TRUTH VIH.
Between two points only one straight line can be drawn.
TRUTH IX.
The straight line is the shortest way from one point to
another.
TRUTH X.
Through one point, without a straight line, only one
line can be drawn parallel to that same straight line.
SECTION I.
OF STRAIGHT LINES AND ANGLES.
QUERY I.
In how many points can two straight lines cut each other?
Answer. In one only.
Q. But could not the two A
straight lines AB, CD, which
cut each other in the point E, ^M-
have another point common; ^,— — - -~~ ' ^NvM"
that is, could not a part of the Zs jg
line CD bend over and touch the line AB in M t
A. No.
Q. Why not 1
A. Because + here would be two straight lines drawn
between the same points E and M, which is impossible.
(Truth VIII.)
QUERY II.
If two lines have any part common, what must necessa-
rily follow ?
A. They must coincide with each other throughout,
and make hut one and the same straight line.
Q. How can you prove , j£ ^
this, for instance, of the
two lines CA, BM, which have the part MA common?
A . The common part MA belongs to the line MB as
well as to the line AC, and therefore MC and AB are, in
this case, but the continuation of the same straight line AM.
GEOMETRY.
23
M
E
QUERY III.
How great is the sum of the two adjacent angles, which
are formed by one straight line meeting another , taking
a right angle for the measure ?
A. It is equal to two right angles.
Q. How do you prove this
of the two angles ADE, CDE,
formed by the line ED, meet-
ing the line AC,at the point D?
A. Because, if at D you
erect the perpendicular DM,
the two angles, ADE and CDE, occupy exactly the same
space, as the two right angles, ADM and CDM, formed
by the meeting of the perpendicular; namely, all the
space on one side of the line AC. (See Truth VII.)
Q. Can you prove the same of the sum of the two
adjacent angles, formed by the meeting of any other two
straight lines ?
QUERY IV
Wliat is the sum of any
number of angles, a, b, c,
d, e, fyc, formed at the
same point, and on the same
side of the straight line
AC, taking again a right
angle for the measure ?
A. It is also equal to two right angles.
Q. Why?
A. Because, by erecting at the point B a perpendicular
to AC, all these angles will be found to occupy the same
space as the two right angles, made by the perpendicular
MB.
24
GEOMETRY.
QUERY V.
When two straight lines, AB, CD, cut
each other, what relation do the angles which
are opposite to each other at the vertex M,
bear to each other ?
A. They are equal to each other.
Q. How can you prove it 1
A. Because, if you add the same angle a,
first to b, and then to e, the sum will, in both
cases, be the same ; namely, equal to two right angles ;
which could not be, if the angle b were not equal to the
angle e (see Truth III) ; and in the same manner I can
prove that the two angles, a and d, are equal to each
other.
Q. If the lines CD, AB, are
perpendicular to each other,
what remark can you make in
relation to the angles d, b,
e f a?
A. That each of these an-
gles is a right angle.
Q. And what is the sum of
all the angles, a, b, c, d, e, f,
around the same point, equal
to?
A. To four right angles.
Q. Why?
A. Because if, through that
point, you draw a perpendicu-
lar to any of the lines, for in-
stance the perpendicular MN, to the line OP, all the
angles, a, b, c, d, e,f taken together, occupy the same
d
M
GEOMETRY.
25
space, which is occupied by the four right angles, formed
by the intersection of the two perpendiculars MN, OP.
QUERY VI.
If a triangle has one side, and the two adjacent angles,
equal to one side and the tivo adjacent angles of another
triangle, each to each, what relation do these triangles
bear to each other ?
A. They are equal.
Q. Supposing in. this diagram the side a b equal to
AB ; the angle at a equal to the angle at A, and the an-
gle at b equal to the angle at B ; how can you. prove that
the triangle a b c is equal to the triangle ABC ?
A. By applying the side ah to its equal AB, the side
ac will fall upon AC, and be upon BC ; because the
angles at a and A, b and B, are respectively equal ; , and
as the sides ac, be, take the same direction as the sides
AC, BC, they must also meet in the same point in which
the sides AC, BC, meet ; that is,, the point c will fall
upon C; and the two triangles abc, ABC, will coincide
throughout. ? •
Q. IVJiat relation do yon here discover between the
equal sides and angles ?
A. That the equal angles -ate and C, are opposite to
the equal sides ab, AB, respectively. ,.
26
GEOMETRY.
QUERY VII.
If two straight lines are both perpendicular to a third
line, what relation must they bear to each other ?
A. They must be parallel.
Q. Let us suppose the
;m
two lines AB, CD, to be
both perpendicular to a
third line, GH ; how can
you convince me that AB
and CD are parallel ?
A. Because, if you ex- ^
tend AB and CD, in the
directions BE, DF, making
BE and DF equal to BA
and DC respectively, every
thing will be equal on
both sides of the line GH.
A
I ^
i
O
B
D
M
F
II
jsr
Now if the l.ines AB, CD, are not parallel, they must either
be converging or diverging. If they are converging, AB
and CD wiir, when sufficiently extended, cut each other
somewhere, say in M ; but then (every thing being equal
on both sides of the line GH) the same must take place
with the lines BE, DF, on the other side of the line GH,
which must cut each other somewhere in N ; and there
would be two straight lines cutting each other in two
points, which is impossible. If the lines AB, CD, were
diverging, BE, DF, would be- the same ; but it is equally
impossible for two straight lines to diverge in two direc-
tions : consequently the two straight lines, AB, CD, can
neither be converging nor diverging, and therefore they
must be parallel..
Q. Can two straight lines which meet each other, be
perpendicular to the same straight line ?
GEOMETRY.
27
A. No.
Q. Why not?
A. Because, if they are both perpendicular to a third
line, I have just proved that they must be parallel ; and
if they are parallel, they cannot meet each other.
Q. From a point without a straight line, how many
perpendiculars can there be drawn to that same straight
line ?
A. Only one. '
Q. Why can there not more be drawn !
A. Because I have proved that two perpendiculars to
the same straight line must be parallel to each other ; and
two lines, parallel to each other, cannot be drawn from
one and the same point. .
QUERY VIII.
If a straight line, MN,
cuts two other straight
lines at equal angles ; that
is, so as to make the angles ,
CIN and AFN equal;
what relation exists be- *»'
tween these two lines 1
A. They are parallel
to each other.
Q. How can you prove
it by this diagram ? The line IF is bisected in O, and,
from that point O, a perpendicular OP is let fall upon the
line AB, and afterwards extended until, in the point R,
it strikes the line CD.
A. I should first observe that the triangles OPF and
ORI are equal ; because the triangle OPF has a side and
two adjacent angles equal to a side and two adjacent
angles of the triangle ORI, each to each. (Query 6.)
Wffft-
OS GEOMETRY.
Q. Which is that side, and which are the two adjacent
angles 1
A. The side OI, which is equal to OF ; because the
point O bisects the line IF. One of the two adjacent
angles is the angle IOR, which is equal to the angle FOP \
because these angles are opposite at the vertex : and
the other is the angle OIR, which is equal to the angle
OFP ; because the angle CIN, which, in the query, is
supposed to be equal to AFN, is also equal to the angle
OIR, to which it is opposite at the vertex. (Truth I.)
Q. But of what use is your proving that the triangle
ORI is equal to the triangle OPF ?
A. It shows that since the triangle OPF is right-an-
gled in P, the triangle ORI must be right-angled in R ;
for, in equal triangles, the equal angles are opposite to
the equal sides (remarks to Query 6, page 25) ; conse-
quently the two lines AB, CD, are both perpendicular to
the same straight line PR, and therefore parallel to each
other. (Last query.)
Q. Supposing, now, two
straight lines, AB, CD, to
be cut by a third line, MN, so
as to make the alternate an-
gles AEF and EFD, or the
angles BEF and EFC, equal,
what relation icould the lines
AB, CD, then bear to each other ?
A. They icould still be parallel.
Q. How can you prove this?'
A. If the angle AEF is equal to the angle EFD, the
angles AEF and CFN are also equal ; because EFD and
CFN are opposite angles at the vertex. And, in the same
manner, it may be proved, that if the angles BEF and
EFC are equal, MEA and EFO arc also equal ; there-
GEOMETRY. 29
fore, in both cases, there are two straight lines cut by a
third line at equal angles ; consequently they are parallel
to each other.
Q. There is one more case, and that is : If the two
straight lines AB, CD (in our last figure), are cut by a
third line MN, so as to make the sum of the two interior
angles AEF and EF<3 y equal to two right angles, how
are the straight lines AB, CD, then, situated with regard
to each other ?
A.. They are still parallel to each other. For the sum
of the two adjacent angles EFC and CFN is also equal
to two right angles ; and therefore, by taking from each
of -the equal sums the common angle EFC', the two re-
maining angles AEF and- CFN must be equal (Truth
IV.) ; and you have again the first case, viz : two straight
lines cut by a third line at equal angles.
Q. Will you now state the different cases in which two
straight lines are parallel 1
A. 1. When they are cut by a third, line at equal angles.
2. When they are cut by a third line so as to make the
alternate angles equal ; and,
3. When the sum of the two interior angles, made by
the intersection of a third line, is equal to two right ^
angles.
3*
30
GEOMETRY.
QUERY IX.
Supposing the two straight lines CD, EFj are cut by
a third line AM at unequal angles, ABC, BHE (Fig. I.
and II) ; or so as to have the alternate angles CBH and
BHF, or DBH and BHE unequal ; or in such a man-
ner, that the sum of the two interior angles CBH and
BHE (Fig. I), or DBH and BHF (Fig. II), is less
than two right angles ; what will then be the case with the
two straight lines CD, EF? . -
Fig. I. . Fig. II.
KCE. C& M
L
n
-M
Dp F p*s F
A. They will, in every one of these cases, cut each
other, if sufficiently extended.
Q. How can you prove this 1
A. By drawing, through the point B, another line NP
at equal angles with EF, and which will then also make
the alternate angles, NBH, BHF and PBH, BHE, equal,
and the sum of the two interior angles, NBH and BHE,
equal to two right angles ; this line NP will be parallel to
the line EF ; consequently the line CD cannot be parallel
to it ; because through the point B only one line can be
drawn parallel to the line EF. (Truth X.)
GEOMETRY.
31
QUERY X.
Can you now tell the relation which the eight angles }
a, b, c, d, e,f, g, h, formed by the intersection of two par-
allel lines, by a third line, bear
to each other ?
A. Yes. In the first place,
the angle a is equal to the angle
e ; the angle c equal to the angle
g ; the angle b equal to the angle
f ; and. the angle d equal to the
angle h; — 2d. the angles a, d-,
e, h, as well as the angles b, c, f, g, are respectively equal
to one another ; — and finally, the sum of either c and e, or
d and f, must make two right angles. For if either of
these cases were not true, the lines would not be parallel,
(Last query.)
QUERY XE .
From what you have, learned of the properties of paral-
lel lines, what law do you discover respecting the distance
they keep from each other?
A. Parallel lines remain throughout equidistant.
Q. When do you call two lines equidistant?
A . When all the perpendiculars, let -fall from one line
upon the other, are equal.
Q. How can you prove, that the perpendicular lines
OP, MI, RS, &c. are all equal to one another 1
M O £
A-
&
I
D
A. By joining MP, the two triangles MPO, MPI, have
the side MP common ; and the angle a is equal to. the
32 GEOMETRY.
angle b ; because a and b are alternate angles, formed by
the two parallel lines MI, OP (Query 10) ; and the angle
c is equal to the angle d; because these angles are formed
in a similar manner by the parallel Hues AB, CD : there-
fore we have a side and two adjacent angles in the trian-
gle MPO, equal to a side and two adjacent angles in the
triangle MPI ; consequently these two triangles are equal ;
and the side OP, opposite to- the angle c, in the triangle
MPO, is equal to the side MI, opposite to the equal angle
d, in the triangle MPI. In precisely the same manner I
can prove that RS is equal to, MI, and consequently also
to OP; and so I might go on, and show that every per-
pendicular, let fall from the line AB, upon the parallel
line CD, is equal to RS, MI, OP, &c- The two parallel
lines* AB and CD are therefore, throughout,- at ah equal
distance from each other ; and the same can be proved
of other parallel lines.
QUERY XII.
If two lines are parallel to a third line, what relation
do they bear to each other ? .
Fig. I.
C- i , D
a — '■ — —I 1 — : — —b
F
Fig, II.
A. 1 j -B
C 1 1 D
E ! ■ F
They are parallel to each other.
Q. How can you prove this 1
A. From the line CD being parallel to AB, it follows
that every point in the line CD is at an equal distance
from the line AB ; and because EF is also parallel to AB,
GEOMETRY.
every point in the line EF is also at an equal distance
from the line AB ; and therefore (in Fig. I.) the whole
distances between the lines CD and EF, or (in Fig. II.)
the differences between the equal distances, are equal :
that is, the lines CD, EF, are likewise equidistant ; and
consequently parallel to each other.
QUERY XIII.
What is the sum of all the angles in every triangle
equal to ?
A. To two right angles,
Q. How do you prove this 1
A. By drawing, through the
vertex of the angle b, a straight
line parallel to the basis BC, the
^(f
angle a is equal to the angle d, and the angle c is equal
to the angle c (Query 10) ; and as the sum of the three
angles a y b r c, is equal to two right angles (Query 4), the
sum of the three angles d, b f e, in the triangle, is also
equal to two right angles.**
Q. Can you now find
out the relation which the
exterior angle e bears to
the two interior angles a
and hi • '
A. The exterior angle
e is equal to the sum of
the two interior angles , a and b.
Q. How can you prove this ?
A. Because, by adding the angle c to the two angles a
* The teacher may give his pupils an ocular demonstration of
this truth, by cutting the three angles b, d, e, from a triangle, and
then placing them along side of each other ; they will be in a
straight line.
34 GEOMETRY.
and b, it makes with them two right angles ; and by add-
ing it to the angle e alone, the sum of the two angles,
c and c, is also equal to two right angles (Query 3), which
could not be, if the angle e alone were not equal to the
two angles a and b together. (Truth III.)
Q. What other truths can you derive from the two
which you have just now advanced?
A. 1. The exterior angle e is greater than either of the
interior opposite ones, a or b.
2. If two angles of a triangle are known, the third
angle is also determined.
3. Wlien .two angles of a triangle are equal to two
angles of another triangle, the third angle in the one is
equal to the third angle in the other.
4. No triangle can contain more than one right angle.
5. No triangle can contain more than one obtuse angle.
6. No triangle can contain a right and -an obtuse angle
together.
7. In a right-angled triangle, the right angle is equal
to the sum of the two other angles*
Q. How can you convince, me of the truth of each of
these assertions 1
RECAPITULATION OF THE TRUTHS CONTAINED IN
THE FIRST SECTION.
Can you now repeat the different principles of straight
lines and angles which you have learned in this section?
Ans. 1. Two straight lines can cut each other only in
one point.
• 2.. Two straight lines which have two points common,
must coincide with each other throughout, and form but
one and the same straight line.
GEOMETRY. 35
3. The sum of the two adjacent angles, which one
straight line makes with another, is equal to two right
angles.
4. The sum of all the angles, made by any number of
straight lines, meeting in the same point, and on the same
side of a straight line, is equal to two right angles.
5. Opposite angles at the vertex are equal.
6. The sum of all the angles, made by the meeting of
ever so many straight lines around the same point, is
equal to four right angles. - . ■ S ■ •
7. When a triangle has one side and the two adjacent
angles, equal to one side and the two adjacent angles
in another triangle, each to each, the two triangles are
equal.
8. In equal triangles the equal angles are opposite to
the equal sides. ' . •
9. If two straight lines are perpendicular to a third
line, they are parallel to each other.
10. If two lines are cut by a third line at equal angles,
or so as to make the alternate angles equal, or so as 10
make the sum of trie two interior angles formed by the
intersection of a third line, equal to two right angles, the
two lines are parallel.
11. If two lines are cut; bv a third line at unequal
angles ; or so as to have the alternate angles unequal ; or
in such a way as to make the sum of the two 'interior
angles less than two right angles, these two lines will,
when sufficiently extended, cut each other.
12. If two parallel lines are cut by a third line, the
alternate angles are equal.
13. Parallel lines, are throughout equidistant.
14. If two lines are parallel to a third line, they are
parallel to each other.
36 GEOMETRY.
15. The sum of the three angles in any triangle, is
equal to two right angles.
16. If one of the sides of a triangle is extended, the
exterior angle is equal to the sum of the two interior
opposite angles.
17. The exterior angle is greater than either of the
interior opposite ones.
18. If two angles of a triangle are given, the third is
determined. -.
•19. There can be but one right angle, or one obtuse
angle, and never a right angle and obtuse angle together,
in the same triangle.
20. In a right-angled triangle, the right angle is equal
to the sum of the two other angles.*
* The teacher may now ask his pupils to repeat the demonstra-
tions of these principles.
SECTION II.
OF EQUALITY AND SIMILARITY OF TRIANGLES.
PART I.
OF THE EQUALITY OF TRIANGLES.
r
Preliminary Remark. There are three kinds of equality to be
considered in triangles, viz : equality of area, without reference to
the shape; equality of shape, without reference to the area — simi-
larity ; and equality of both shape and area — coincidence. All
questions, asked in this section, will refer only to the last two kinds
of equality ; and those in the first part, only to the coincidence of
triangles.
UUERY I.
If two sides and the angle which is included by them in
one triangle, arc equal to two sides and the angle which is
included by them in another triangle, each to each, what
relation do these two triangles bear to each other ?
Ans. They are equal to each other in all their parts,
that is, they coincide with each other throughout.
Show me that this must be the case with any two tri-
angles, ABC, abc, in which we will suppose the side
AB — ab, AC = ac, and the angle at A equal to the angle
at a.
4
38 GEOMETRY.
A. By placing the line ac upon its equal AC, the angle
at a will coincide with the angle at A, because these two
angies are equal ; and the line ah will fall upon the line
AB ; and as ah = AB, the point h will fall upon B ; that
is, the three points of the triangle ahc will fall upon the
three points of the triangle ABC, thus :
The point a upon A,
" h " B,
" c " C;
consequently these two triangles must coincide.
Q. What remark can you here make with respect to the
sides and angles of equal triangles ?
A. The equal sides, cb, CB, are opposite to the equal
angles at a and A.
QUERY II.
If one side and the two adjacent angles in one triangle,
are equal to one side and the two adjacent angles in
another triangle, each to each, what relation do the two
triangles hear to each other ?
A. They are equal, and the angles opposite to the equal
sides are also equals as has been proved in the 1st Section.
(Query 6.)
QUERY III.
What remark can you make with respect to the two
angles at the basis of an isosceles triangle ?
A. They are equal to each other.
GEOMETRY. 39
Q. How can you prove it ?
A. Suppose we had two equal
isosceles triangles, ABC and
abc, or, as it were, another im-
pression, abc, of the triangle
ABC, that is,
The side ab = AB,
« ac =z AC,
" be =zBC.
The angle at a sd angle at A,
b= " B,
cz= " C.
Then the sides AB, AC, ab, ac, being all equal to one
another, and the angle at a remaining the same, which-
ever way we place it, the whole of the two triangles, abc,
and ABC, will still coincide, when abc is placed upon
ABC in such a manner that ac will fall upon AB, and ab
upon AC (for you will still have two sides and the angle
which is included by them in the one, equal to two sides
and the angle which is included by them in the other) ;
therefore the angle at c must be equal to the angle at B.
And as the angle at c is only, as it were, another impression
of the angle at C, the angles C and B must also be equal ;
that is, the two angles at the basis of the isosceles triangle
ABC are equal : and the same can be proved of the two
angles at the basis of every other isosceles triangle.
QUERY IV.
If the three sides of one triangle are equal to the three
sides of another, each to each, what relation do the two
triangles bear to each other ?
A. Tltey coincide with each other throughout ; that is,
their angles are also equal, each to each.
40
GEOMETRY.
Q. How can you prove this, for
instance, of the two triangles ABC
and abc, in which we will suppose
the side AB = ab,
AC = ac,
BC — bc?
That you may easier find out your
demonstration, I have placed the
two triangles, as you see, along
side of each other, with their bases, AB and ab } together,
and have joined their opposite vertices, C and c by the
straight line Cc. What do you now observe with regard
to the two triangles ACc and BCc ?
A. Both are isosceles ; for the sides AC and ac, BC
and be, are respectively equal ; and, therefore, the angles
* and y, o and w, must be equal, each to each ; and as
the angle x is equal to the angle y, and the angle o equal
to the angle w, the sum of the two angles x and o, that
is, the whole angle ACB, must be equal to the sum of the
two angles y and w, that is, to the whole angle acb ; and
the two triangles, ABC, and abc, having two sides, AC,
BC, and the angle which is included by them in the one,
equal to the two sides ac, be, and the angle which is in-
cluded by them in the other, each to each, must coincide
throughout, and have, consequently, all their angles re-
spectively equal to one another. (Query 1, Sect. II.)
QUERY V.
WJiich of two angles in a triangle is greater, that
which is opposite to the smaller, or that which is opposite
to the greater side ?
A. That which is opposite to the greater side.
Q. How can you prove it ?
A. Because if in any triangle, for instance in the
GEOMETRY. 41
triangle ABD, one side, AB, is j>
greater than another, AD, the side
AB will contain a part which is
equal to AD ; # and therefore, by A- {j^-^
taking upon AB the distance AC
equal to AD, and joining DC, the, triangle ACD will be
isosceles, and the angle x will be equal to the angle y,
(Query 3, Sect. II.) ; and as the exterior angle y must
be greater than the interior opposite angle CBD, in the
triangle DBC, (Query 13, Sect. J.) the angle at x will
also be greater than the angle CBD ; and the angle ADB
being greater still than the angle x, must consequently
be still more so than the angle CBD ; that is, the angle
ADB, opposite to the greater side AB, is greater than the
angle at B, opposite to the smaller side AD : and the same
can be proved of two unequal sides in any other triangle.
Q. What truth can you directly derive from this, re-
specting the three angles and sides of a triangle !
A. That the greatest of the three angles is opposite to
the greatest of the three sides. For if the side AD, for
instance, is greater than the side DB, it can be proved
that the angle at B, opposite to the side AD, is greater
than the angle at A, opposite to the side DB ; and as the
side AB is greater still than AD, the angle ADB, opposite
to AB, must be greater still than the angle at B, and is
therefore the greatest angle in the triangle ABD.
Q. From ichat you have learned of the relation which
exists between the sides and. angles of a triangle, can you
now tell which of the sides of a right-angled triangle is
the greatest ?
A. Yes. That which is apposite to the right angle.
Q. Why?
* If the magnitude A is greater than B, A must contain a part
equal to B.
4*
GEOMETRY.
A. Because, in a right-angled triangle, the right angle
is greater than either of the two other angles. (Conseq.
Query 13, Sect. I.)
QUERY VI.
It has been proved before (Query 3,
Sect. II.), that in an isosceles triangle, the A
angles at the basis are equal : can you now j \
prove the reverse; that is, that a triangle / \
must be isosceles ivhen- it contains two equal / \
angles? A B
A. Yes. Because, if either of the two
sides AC, BC, were greater than the other, the angle
opposite to that side would also be greater than the angle
which is opposite to the other side ; but the two angles
at A and B are equal, therefore the sides AC, BC, are
also equal.
Q. If the three angles in a triangle are equal to one
another, what relation do the sides bear to each other ?
A. They are also equal, and the triangle is equilateral.
Q. How can you prove this ?
A. If, in the triangle ABC, for
instance, the angle at A is equal
to the angle at B, I have just
proved that the side BC must be
equal to the side AC ; and if the
angle at B is also equal to the
angle at C, the side AC must likewise be equal to the
side AB ; that is, the three sides AB, BC, AC, are equal
to one another, and the triangle ABC is equilateral.
QUERY VH.
Can any one side of a triangle be greater than, or equal
to, the sum of the two other sides ?
GEOMETRY. 43
A. No. A straight line being
the shortest way from one point
to another, it follows that, in any
triangle, ABC for instance, the
side AB is smaller than the sides AC and BC together.
QUERY VIII.
If, from a point M, in a triangle ABC, two lines, AM,
BM, are drawn to the two extremities of any side, AB,
in that triangle, what relation does the angle AMB, made
by these two lines, bear to the angle ACB, which is opposite
to the side AB in the triangle ? And what do you ob-
serve with regard to the sum of the two lines, AM and
MB, which include the angle AMB, and that of the two
sides, AC, BC, of the triangle which include the angle
ACB?
C
DM
D
DM
A MB
A. The angle AMB, made by the lines AM, BM, is
always greater than the angle ACB, opposite to the side
AB, in the triangle ABC; but the sum of the two
lines AM, MB, is in all cases smaller than the sum of the
two sides AC, CB, of the triangle.
Q. How can you prove both your assertions I
A. The exterior angle MDB is greater than the inte-
rior opposite angle ACD, in the triangle ACD (Query
13, Sect. I.) ; and for the same reason is the exterior
angle AMB greater than the interior opposite angle
44 GEOMETRY.
MDB, in the triangle MDB; and therefore the angle
AMB is greater still than the angle ACB. Idly. The
three sides AB, AC, BC, by which the greater surface
is bound, enveloping the three sides AB, AM, MB, it
follows that their sum is greater than the sum of the
three sides AB, AM, BM, by which the smaller surface
ABM is bound ; and, taking from each of the unequal
sums the same line AB, which serves both as a common
basis, the greater will remain where the greater was
before ; that is, the sum of AC and BC will still be
greater than the sum of AM, BM.
QUERY IX.
If, from a point A , without a straight line MN, you
let fall a perpendicular, AB, upon that line; and, at the
same time, draw other lines, AD, AE, AF, fyc, obliquely
to different points, I), E, F, fyc, in the same straight
line; which is the shortest, the perpendicular, or one of
the oblique lines I
A.
A. The perpendicular is the shortest.
Q. How can you prove it 1
A. Because the triangles, ABD, ABE, ABF, ABN,
&c. are all right-angled in B ; and in every right-angled
triangle, the greatest side is opposite to the right angle.
(Page 41.)
Q. And what other truths do you derive from the one
you have just mentioned?
A. 1st. The perpendicular AB measures the distance
GEOMETRF. 45
of the point A from the line MN; for it is the shortest
line that can be drawn from that point to that line*
2dly» The angles o, p, r, t, Sfc. are all obtuse, because
they are exterior angles of the right-angled triangles,
ABD, ABE, ABF, &C;, and, therefore, greater than the
interior opposite right angle at B.
3dly. The angles o, p, r, t, 8$c. become successively
greater, and the angles u, q, s, &o. smaller, as the lines
AD, AE, AF, fyc, are drawn farther from the perpen-
dicular. For the exterior angle p is greater than the
interior opposite one o, in the triangle ADE ; the exterior
angle r is greater than the interior opposite one p, in the
triangle AEF ; the exterior angle t, again, is greater than
the interior opposite one r, in the triangle AFN ; and
so on.
4thly. The oblique lines, AD, AE, AF, fyc. become
successively greater, as they are drawn farther from the
perpendicular; that is, the line AD is greater than the
line AB ; the line AE than the line AD ; the line AF
than the line AE ; and so on. For the angles o, p, r, &,c.
are all obtuse, and become successively greater, as the
triangles ADE, AEF, &c. are more remote from the per-
pendicular ; and, therefore, the sides AE, AF, AN, &c,
which are successively opposite to these angles, in the
triangles ADE, AEF, AFN, must become greater with
them.
5thly. The straight lines, AC, AD, drawn on both
sides of, and at an equal distance from, the perpendicular
AB, are equal. For the two triangles ABC, ABD, have
the side AB common, and the side BC equal to the side
BD (because the lines AC, AD, are at an equal distance
from the perpendicular AB) ; and as the line AB is per-
pendicular to CD, the angle ABC, included by the sides
AB, BC, in the triangle ABC, is equal to the angle ABD,
46 - GEOMETRY.
included by the sides AB, BD, in the triangle ABD ; con-
sequently, these two triangles are equal ; and the third
side AC in the one triangle, is equal to the third side
AD in the other. (Query 1, Sect. II.)
6thly. There is but one point in the line MN, on each
side of the perpendicular, such, that a straight line,
drawn from it to the point A, is of a given length. This
follows from No. 4.
7thly. There is but one point in the line MN, on each
side of the perpendicidar , in which a line drawn to the
point A forms with the line MN an angle of a given
magnitude. This follows from No. 3.
QUERY X.
If two sides, and the angle which is opposite to the
greater of them, in one triangle, are equal to two sides ana
the angle which is opposite to the greater of them in
another, each to each, what relation do these two trian*
gles bear to each other ?
A. They coincide w'ith each other in all their parts;
that is, they are equal to each other.
Q. How can you prove it?
I A. Because, if, in a triangle, Fig ' L
ABC, for instance, you have the -£*•
sides AB and AC, and the angle / j \ Sn s s .
at B, which is opposite to the nZ. — ! — !^ _^w
greater side AC, given, the whole
triangle is determined. For, in the jy
first place, by the angle at B, the / 1\^"^^
direction of the sides AB, BC, is jg }"~jj > jg : ^tf
determined. %dly. By the length
of the side AB, the distance of the point A from the line
BC is determined. 2dly. If you imagine the perpendic-
GEOMETRY. 47
ular AD to be let fall upon BC (Fig. I.), or if the angle
ABC be obtuse (as in Fig. II.) on its further extension
BE, there can be but one point in the line BC, on this
side of the perpendicular, from which a line drawn to the
point A, is as long as the line AC (see consequence 6th
of the preceding query) ; therefore, by the length of the
line AC, the point C, and thereby the whole of the third
line BC, is also determined.
Q. But is it not possible for the line AC to fall on the
other side of the perpendicular ?
A. No. Because the line AC, being greater than the
line AB, would in this case be farther from the perpen-
dicular, than the line AB (conseq. 4, preceding query),
and the angle at B would then fall without the triangle ;
and because the whole triangle ABC is entirely deter-
mined, when two of its sides, and the angle which, is
opposite to the greater of them, are given : therefore, all
triangles, in which these three things are equal, must be
equal to one another.
Q. What truth can you infer from this respecting the
case where the hypothcnuse, and one side of a right-angled
triangle, are equal to the hypothcnuse and one of the
sides in another right-angled triangle ?
A. That these two right-angled triangles are equal to
each other. For, in this case, we have two sides, and
the right angle which is opposite to the greater of them,
in the one, equal to two sides, and the angle which is
opposite to the greater of them, in the other.
48 GEOMETRY.
Q. But if, in Fig. II. (page 46) the two sides AC, AB, and the
angle at C, opposite to the smaller side AB, be given, would not
this be sufficient to determine the triangle ABC ?
A. No. For the two lines, AB, AE, being equal, there would
be two triangles, ABC and AEC possible, containing the same
three things, and it would be doubtful which of the two triangles
was meant.
QUERY XI.
If you have two sides, ab, be, of a triangle, abc, equal
to two sides, AB, BC,of another triangle, ABC, each to
each ; hut the angle ABC included by the two sides, AB,
BC, in the triangle ABC, greater than the angle abc,
included by the sides ab, be, in the triangle abc ; ichat
remark can you make with regard to the two sides ac,
AC, which are respectively opposite to those angles?
A. That the side ac, opposite to the smaller angle abc,
in the triangle abc, is smaller than the side AC, opposite
to the greater angle ABC, in the triangle ABC.
Q. How do you prove this 1
A. By placing the triangle abc upon the triangle
ABC, with the side ab upon AB (its equal), the side be
will fall within the angle ABC, because the angle abc is
smaller than the angle ABC ; and the ex-tremity c, of the
line be, will either fall without the triangle ABC, as you
see in the figure before you, or within it, or it may also
fall upon the line AC itself.
Is*. If it falls without the .triangle ABC, by imagining
GEOMETRY. 49
the line Cc drawn, the triangle cBC will be isosceles ; for
we have supposed the side be equal to BC ; and because
i.he angles at the basis of an isosceles triangle are equal
(Query 3, Sect. II.), the angle z is equal to the sum of
the two angles x and y ; consequently greater than the
angle y alone; and if the angle z is greater than the
angle y, the two angles z and w together will be greater
still than the same angle y; therefore, in the triangle
ACc, the angle AeC is greater than the angle ACc;
consequently the side AC, opposite to the greater angle
AcC, must be greater than the side ac, opposite to the
smaller angle ACc. * " ♦ "
%dly. If the extremity of the line be falls within the
triangle ABC, the sum of the two sides ac, be, must be
smaller than the sum of the two
sides AC, BC (Query 8, Sect. II.) ;
therefore, by taking from each of
these sums the equal lines be, BC,
respectively, the remainder, AC, of
the greater sum (AC-f-BC) is greater than the remain-
der, ac, of the smaller sum (ac-\-bc).
Finally. If the point c falls upon the line, AC itself,
it is evident that the whole line AC must be greater than
its part Ac.
a
QUERY XII.
If, in a parallelogram, ACDB,
you draw a diagonal CB, ivhat
relation do the two triangles,
ABC, CDB, bear to each other?
5
50 GEOMETRY.
A. They are equal to each other , and the parallelo-
gram is divided into two equal parts.
Q. How can you prove this 1
A. The two triangles, ABC and CDB, have the side
CB common ; and the angle y is equal to the angle w ,
because y and w are alternate angles, formed by the in-
tersection of the two parallel lines CD, AB, by a third
line, CB; and the angle x is equal to the angle z, because
these two angles are formed in a similar manner, by the
parallel lines AC, DB (Query 10, Sect. I.) : and as the
triangle ABC has a side CB, and the two adjacent angles,
x and w, equal to the same side CB, and the two adjacent
angles, z and y, in the triangle CDB, each to each ;
therefore these two triangles are equal (Query 6, Sect. I.),
and the diagonal CB divides the parallelogram into two
equal parts.
Q. What other properties of a parallelogram can you
infer from the one just learned?
1st. The opposite sides of a parallelogram are equal;
that is, the side CD is equal to the side AB, and the side
CA to the side DB; for in the equal triangles, ABC,
CDB, the equal sides must be opposite to the equal angles.
(Conseq. of Query 1, Sect. II.)
2dly. The opposite angles in a parallelogram are
equal ; for in the two equal triangles, ABC, CDB, the
same side, CB, is opposite to each of the angles, at D and
A. (Conseq. of Query 6, Sect. I.)
3dly. By one angle of a parallelogram, all four are
determined; for the sum of the four angles in a parallelo-
gram is equal to four right angles ; because the sum of
the three angles in each of the two triangles, ABC, CDB,
is equal to two right angles. Now, if the angle at D, for
instance, is known, the angle at A is equal to it ; and
there remain but the two angles ACD and ABD, each
»>
GEOMETRY. £l
of which must be equal to half of what is wanting to
complete the sum of the four right angles.
Q. If you have a quadrilateral, in which the opposite
sides are respectively equal, does it follow that the figure
must be a parallelogram ?
A. Yes. For if, in the last figure, you have the side
CD equal to the side AB, and the side AC equal to the
side BD ; by drawing the diagonal BC, you have the
three sides of the triangle ABC, respectively, equal to the
three sides of the triangle CDB ; therefore, these two
triangles are equal ; and the angle y, opposite to the side
DB, is equal to the angle w, opposite to the equal side
AC ; and the angle x, opposite to the side AB, is equal to
the angle z, opposite to the equal side CD ; that is, the
alternate angles, y and w, x and z, are respectively equal :
therefore the side CD is parallel to the side AB, and the
side AC to the side BD, and the figure is a parallelo-
gram.
Q. If, in a quadrilateral, you know but tioo sides to be
equal and parallel, what will then be the name of the
figure 1
A. It will still be a parallelogram. For if, in the last
figure, the side CD is equal and parallel to AB, by drawing
the diagonal CB, you have the two sides, CB and CD, in
the triangle CDB, equal to the two sides, CB, AB, in the
triangle ABC, each to each ; and because the side CD is
parallel to the side AB, the included angle y is equal to
the included angle w; therefore the two triangles are
equal (Query 1, Sect. II.), and the side AC is also equal
and parallel to the side DB, as before.
52 GEOMETRY.
QUERY XIII.
If, from one of the vertices of a
rectilinear figure, diagonals . are
drawn to all the other vertices, into
how many triangles will this recti-
linear figure he divided?
A. Into as many as the figure
has sides less two. For it is evident, M D
that if, from the vertex .A, for instance, you draw the di-
agonals AF, AE, AD, AC, to the vertices F, E, D, C,
each of the two triangles AGF, ABC, will need for its
formation two sides of the figure, and. a diagonal ; but
then every one remaining side of the figure will, together
with two diagonals, form a triangle ; therefore there will
be as many triangles formed, as there are sides less the
two, which are additionally employed in the formation of
the two triangles AGF, ABC.
Q. And what is the sum of all the angles, 'BAG, AGF,
GFE, FED, EDC, DCS, CBA, equal to 1
A. To as many times two right angles as the figure
ABCDEFG has sides less two. For as every rectilinear
figure can be divided into as many triangles as there are
sides less two ; and because the sum of the three angles
in each triangle is. equal to two rigjit angles (Query 13,
Sect. I.) there will be as many times two right angles in
all the angles of your figure, as there are triangles ; that
is, as many as the figure has sides less two.
SECTION II.
PART II.
OF GEOMETRICAL PROPORTIONS,* AND SIMILARITY
OF TRIANGLES.
Whenever we compare two things with regard to
their magnitude, and inquire how many times one is
greater than the other, we determine the ratio which
these two things bear to each other. If, in this way, we
find out that the one is two, three, four, &c. times
greater than the other, we say that these things are in the
ratio of one to two, to three, to four, fyc. : e. g. If you
compare the fortunes of two persons, one of whom is
worth $10,000, and the other $20,000, you say, that
their fortunes are in the ratio of one to two. Or if you
compare two lines, one of which is two, and the other
six feet long, you say of these lines, that they are in the
* It is the design of the author to give here a perfectly element-
ary theory of geometrical proportions, and to establish every prin-
ciple geometrically, and by simple induction. Intending the
above theory for those who have not yet acquired the least knowl-
edge of Algebra, he is not allowed to identify the theory of pro-
portions with that of algebraic equations (as it is done by some
writers on Mathematics), and then to find out the principles of the
former by an analysis of the latter. There are several disadvan-
tages inseparable from the algebraic method of considering a ratio
as a fraction, besides the difficulty of making such a theory accessi-
ble to beginners. Neither can an algebraic demonstration be
made obvious to the eye like a geometrical one.
5*
54 GEOMETRY.
ratio of one to three, because the second line is three
times as long as the first.
It frequently occurs, that two things are to each other
in the same ratio in which two others are ; we then say
that these things are in proportion. This is frequently
the case in the fine arts ; but particularly in the science
of Geometry, from which these proportions are called
geometrical. To give an example : If you draw a
house, you must draw it according to a certain scale ;
that is, you must draw it one thousand, two thousand,
three thousand, &-c. times smaller than the building itself:
but then you are obliged to reduce every part of it in
proportion. If, for instance, you draw the front of the
house one thousand times smaller than the original, you
must reduce the windows, doors, and every other part,
in the same ratio. If, on the contrary, the windows
were reduced two thousand times, whilst the doors and
other parts were reduced only owe thousand times^ your
picture would be out of proportion, because the different
parts would be reduced by different ratios. In this case
your picture would be distorted ; and would not resem-
ble the original.
The same is the case with resemblance, produced in
any other kind of drawings ; but particularly in geomet-
rical figures.
C
Fig. L J\ Fig. II. Fig. HI.
a/\>
a b a e
a c a c
I o b **
GEOMETRY. 55
If the two triangles, ABC, abc, are to be similar to
each other, it is necessary that they should be construct-
ed after the same manner, and that the side AC should
be exactly as many times greater than the side ac, as the
side BC is greater than be, and the side AB than ab. If
(Fig. I. and II.) the side AB, for instance, is twice as
great as the .side ab ; that is, if. the side ab is half of the
side AB ; the side ac must also be half of the side AC,
and the side be half of the side BC ; that is, the three
sides, ab, ac, be, of the triangle abc, must be in propor-
tion to the three sides, AB, AC, BC, of the triangle ABC.
Again, if (Fig. I. and III.) the side AB is three times as
great as the side ab ; that is, if the side ab is one third
of the side AB ; the side ac must also be one third of the
side AC, and the side be one third of the side BC ; or
the triangles abc, ABC, would not be similar to each
other. The same holds true of all other geometrical
figures, composed of any number of sides. If they are
similar, their sides are proportional to each other.
There are different ways of denoting a geometrical
proportion. Some mathematicians express the propor-
tionality of the sides, ab, ac, of the triangle abc (Fig. II.),
to the sides AB, AC, of the triangle ABC (Fig. I.), in
the following manner :
AB : ab : : AC : ac;
or,
AB-^-ab : :• AC-^-ac ;
and also
AB : ab = AC : ac*
which is read thus :
AB is to ab, as AC is to ac.
* The first manner of expressing a proportion is now in general
use among the English and French mathematicians ; the second is
sometimes met with in old English writers, and the third way is
adopted in Germany.
56 GEOMETRY.
As a proportion is nothing less than the equality of two
ratios, the third way of denoting a proportion, in which
the sign of equality is put between the two ratios, seems
to be the most natural. The reason why the sign of di-
vision (see Notation and Significations), is put between
the two terms, AB, ab, of a ratio, is obvious ; for a ratio
points out how many times one term (the side ab) is con-
tained in the other (the side AB).
The first and fourth terms of a proportion, together, are
called extremes ; because one of them stands at the be-
ginning , and the other at the end, of a proportion : the
second and third terms, standing in the middle, are, to-
gether, called the means.
The following principles of geometrical proportions
ought to be well understood and remembered r
1st. It is important to observe, that in every geometrical
proportion the two ratios may be inverted; that is, in-
stead of saying,
AB : ab = AC : ac,
you may say,
ab : AB = ac : AC ;
for, the order of terms being changed in both ratios,
they are still equal to one another ; but, leaving one ratio
unaltered, if you change the order of terms in the other,
the proportion will be destroyed. You cannot say,
ab : AB = AC : ac ;
for the smaller side, ab, is contained twice in the greater
side, AB (Fig. I. and II.) ; but the greater side, AC, is
not contained once in the smaller side, ac.
2d. Another remarkable property of geometrical pro-
portions is, that you may change the order of the mcanr,,
or extremes, without destroying the proportion. Thus
you may change the proportion
Geometry. 5 7
AB : ab = AC : ac ..... (I.)
into
AB : AC = ab : ac . . . . . . (II.)
or by changing the extremes into
ac : ab = AC : AB ..... (III.)
The reason why you have a right to do this, is easily
comprehended. If, in the first proportion, the side AB
is as many times greater than ab, as AC is greater than
ac, the ratio of AB to AC will be the same as that of
ab to ac. In Fig. I. and II. (page 54), we have ab
equal to one half of AB ; consequently ac is also equal
to one half of AC ; and, therefore, let the ratio of the
two lines, AB to AC, be whatever it may, their halves,
ab and ac, must be in the same ratio. To give another ex-
ample : If A's garden is five times greater than B's, half
of A's garden is also five times greater than half of B's
garden. The second proportion (II.) would still be correct,
if, as in Fig. I. and III., the sides AB, AC, were three
times as great as the sides ab, ac ; for then the thirds
of AB and AC would still be in the same proportion as the
whole lines AB and AC. Nothing can now be easier
than to extend this mode of reasoning, and show the
generality of the principle here advanced. The correct-
ness of the third proportion might be proved precisely in
the same manner as that of the second ; for the third
proportion (III.) differs from the second (II.) only in the
order in which the two ratios are placed ; and of two
equal things, it does not matter which you put first. The
correctness of the second proportion proves, therefore,
that of the third proportion.
3d. If you have two geometrical proportions, which
have one ratio common, the two remaining ratios will,
again, make a proportion ; for if two ratios are equal to
the same ratio, they must be equal to each other. (See
Axioms, Truth I.) If you have the two proportions
58 GEOMETRY.
AB : ab = AC : ac
AB : ab — BC : be
you will also have the proportion
AC : ac = BC : 6c.
For an illustration of this principle, we may take the two
triangles ABC, abc (Fig. I. and II.) : If the sides AB
and ab are in proportion to the sides AC and ac, and
also in proportion to the sides BC and be, the three sides
of the triangle ABC will be in proportion to the three
sides of the triangle abc; therefore, any two sides of the
first triangle will be in proportion to the two correspond-
ing sides of the other triangle.
4th. Another important principle of geometrical pro-
portions is this: If you have several geometrical propor-
tions, of which the second has a ratio common with the
first, the third a ratio common with the second, the fourth
u ratio-common with the third, and so on; the sum, of all
thejiftt terms of these proportions will bear the same ratio
to the sum of all the second terms, which the sum of all the
third terms docs to the sum of all the fourth terms, that
ti>, the sums will again make a proportion.
To prove this, we will, in the first place, consider the
simplest case ; that of two proportions only ; and, the
easier to comprehend it, take the same two proportions
wnich we have just had under consideration, viz :
ac : AC = ab : AB
ab : AB = be : BC.
We know, from the two triangles, ABC and abc (Fig. I.
and II.), that, in the first proportion, ac is half of AC;
consequently ab is also half of AB, and, in the second
proportion, be is also half of BC. Thus, each of the two
first terms, ab, ac, is half of its second term ; and conse-
quently each of the third terms, be, ab, is half of its cor-
responding fourth term ; therefore, adding ab and ac
GEOMETRY. 59
together, their sum will be one half of the sum of AB
and AC ; and so will be and ab, be, together, one half of
the sum of BC and AB. For the sake of illustration,
you may measure off the length of ab and ac, upon
a
B
the line be, and the length of AB and AC on another
line BC ; and you will find that the line be is exactly one
half of the line BC. For the line be, composed of two
parts, ab, ac, each measuring exactly one half of the
corresponding two parts, AB, AC, of which the line BC
is composed, must evidently be one half of the whole line
BC. In the same way you may convince yourself that
cv
A B C
the line ac, composed of the two parts, ab and be, meas-
ures one half of the second line AC, composed of the two
parts' AB and BC : and therefore you will have
ab-\-ac : AB -f AC = be -f ab : BC -f AB.*
Although, in our example, we have chosen a proportion
in which the first and third terms are exactly one half
of the second and fourth terms, yet it is easy to perceive,
that the same course of reasoning will apply to any other
two proportions. Thus, if the first terms in the above
proportions were one third, or one fourth, or one fifth,
* The lines over ab-\-ac, AB-f-AC, &c, mark that ab-\-ac,
AB-|- AC, &c, are but single lines composed of the two parts, ab,
ac, and AB, AC.
60 GEOMETRY.
&*c, of the corresponding second terms, the sum of all
the first terms would also be one third, or one fourth, or
one fifth, &,c, of the sum of all the second terms ; and
the same would be the case with regard to the sum of the
third and fourth terms. It is also evident, that our prin-
ciple would still hold true, if, instead of two proportions,
we had three, four, or more proportions given, of which
two and two have a common ratio. If, for instance, we
had the three proportions
ac : AC = ab : AB
ab : AB = be : BC
be : BC = ac : AC,
we should, according to our principle, have
be -f- ab -f ac : BC -f- AB -f AC — ac -\- be -f ab
: AC + BC-f-AB.
Each of the three lines, be, ab, ac, would be one half of
its corresponding second term ; and in the same way
would each of the three lines, ac, be, ab, be one half of
its corresponding fourth term ; and, therefore, the sum of
the three lines, be, ab, ac, or, which is the same, a line
as great as the three lines, be, ab, ac, together, would be
one half of the sum of the three lines, BC, AB, AC, or a
line as great as the three lines, BC, AB, AC, together ;
and the same would be the case with the sum of the third
and fourth terms. And in like manner can this principle
be extended to four, five, six, and more proportions.
5th. Another principle, which it is important to recol-
lect, is, that by adding the second term of a proportion
once, or any number of times, to the first term, and the
fourth term the same number of times to the third term,
you will still have a proportion. To give an example :
In the proportion
•*■
GEOMETRY. Ql
AB : ab = AC : ac,
let there be added the second term ab, in the first place,
once to the first term AB ; and the fourth term ac also
once to the third term. AC. Our proportion is then
changed into
AB + a0 '■ a0 — AC -f- ac : ac,
in which the first term, AB -f- ab, instead of being only
twice as great as ab; is now, by the addition of the term
ab itself, three times as .great as ab ; and for the same
reason is AC -f- ac three times as gr,eat as ac: The two
new ratios, "•■'*, • - ■
AB -\-ab : ab,and
AC -f- ac : ar, ' "•
are therefore equal, and consequently make a proportion.
The same would be- the case,* if, instead of adding the
second and fourth terms once, you would add them twice
respectively to. the first and third terms ; with the only
difference, that the first term, AB -{-Hab, would then be
four times as great as the second term ab. A similar
change would lake place with regard to the third- term,
AC -}- 2ac, which would then be four time's as great as
the term ac ; and you would have the proportion
AB-\-2a6 \.ab — AC -f 2ac : ac. i
If the second term were added . three times to the first
term, the first term, AB -f- 'Sab, would be five times- as
great as ah; and the third term, AC -j- Sac, would also
be five times. as great as ac i and so on.
In precisely the same manner you may prove that, by
adding ilit first term q nee, or any number of times, to the
second term, and the third term the same, number of times
to the fourth term, the result will still be a proportion.
Thus, our proportion
6
62 GEOMETRY.
AB : ab — AC : ac,
may be changed into
AB : a6 + AB = AC rac-j-AC,
or into \ ~
AB : a6-j-2AB — AC :ac-|-2AC, &c.
It is also evident that the same principle will hold of any
other geometrical proportion.*
Gth. For the same reason that the second term of a
geometrical proportion may be added once or any number
of titties' to the first term, and the fourth term the same
number of times to the third term, without destroying the
proportion; the second term may also be subtracted once
or any number of times from the first term, provided the
fourth term, is the same number of times subtracted from
the third term, and the result will still be a proportion.
If, in the geometrical proportion
AB : ab — AC : ac ,
the first term (AB) is twice as great a.s ab, and AC twice
as great as «c, we shall, .by subtracting ab from AB, and
ac from AC, make the two terms jn each ratio equal ;
and we shall have a new proportion,
A3 — ab:ab =2 AC — ac : ac.
* The teacher had better show this to the pupil, particularly
as the above mode of demonstrating this principle admits of an
ocular demonstration by measurements. For if the "teacher uses
lines for the terms of his proportions, and not abstract numbers,
which are always more difficult to be comprehended, he can actu-
ally perform, these additions, by extending the Hne AB, for
instance, to once or twice the length of the line ab, and then show,
by measuring these lines, that the fjrst term is really as many
times greater than the second term, as th^e third" terra is greater
than the fourth term. In this manner the demonstrations will not
only be perfectly geometrical, but also have the advantage of the
'jvhictive method.
GEOMETRY. 63
If AB were three times as great as ab, AC would, of
sourse, be three times as great as ac ; and therefore, by
subtracting ab from AB, and ac from AC, the first term
(AB — ab), in the last proportion, would be twice as
great as ab; and for the same reason would AC — ac, be
twice as great as ac. In the same manner may this
principle " be applied to ever/ other., geometrical propor-
tion ; and it may also be proved, that, by subtracting the
first term of a geometrical proportion once or any number
of times from the second term^ and the third term the same
number of times- from the fourth term^ the proportion icill
not be destroyed.: ■ .
7th.. If all the ■ terms of a geometrical proportion are
multiplied or -divided by the same number, the proportion
remains the same.
For an example, we will again take the proportion
AB : ab 3= AC : ac,
in which ab is half of AB, and ac half of AC. Then it
is evident, that a line, which is, for instance, ten times
as long as ab, that is, a line which contains the line ab
ten times, is still half of a line winch contains the line
AB ten times; and in like manner is a line ten times as
long as ac still half of a line ten times as long as AC ;
consequently the proportion
10 AB : 10a&=:10AC : 10 ac
is. the same as
AB : ab — AC : ac,
because in both proportions the first term in each ratio is
double of the second term.
Neither would our proportion change, if, instead of
multiplying each term by 10, we were to multiply it by 2,
by 3, by 4, &c, or even by fractions; for the reasoning
would, in every one of these cases, be precisely the same
as in the case of our multiplying by ten.
64 GEOMETRY.
It is also easy to apply the same principle to any other
geometrical proportion.
If, instead of multiplying each term of the proportion
AB : ah sen AC :■ ac,
we divide it by ten, it is evident that the tenth part of
the line ah will still be half of the tenth part of the line
AB ; and so will the tenth part of the line ac be half
of the tenth part of the line AC ; consequently the pro-
portion
T VAB : T ^ah — T \AC : T \ac
is still the same as
AB : ah = AC : ac ;
and the same reasoning may be applied to the division
by any other number, and to any other geometrical pro-
portion.
8th. If three terms of a proportion he given, the fourth
term can easily he found. Let there be the three terms
of a proportion,
AB:«S = AC:
to which the fourth term is wanting. Then, by knowing
how many times the line ab is smaller than the line AB,
or., which is the same, whatever part of the line AB the
line ah is, you can easily take the same part of the line
AC, which will be the fourth term of your proportion.
If you know, for instance, that the line ab is one lialf of
the line AB, you would at once conclude, that the re-
quired fourth term in your proportion must be one half
of the line AC : this is, as we know, really the case with
our proportion, where the fourth term ac, which we sup-
posed here to be unknown, is really one half of AC. If
ah were one third of AB, you would conclude that your
fourth term must be one third of AC ; and so on. If,
instead of the fourth term, another, for instance the
GEOMETRY. 65
second term, were unknown, you could find it in a man-
ner similar to the one just given. For, one ratio being
expressed, you will always know the relation which £fl
term you are to find must bear to the term with which it
is to form a ratio. . -
9th. Geometrical proportions are also frequently made
use of in common- Arithmetic , and in Algebra. You can
say of the two numbers 3 and 6, that they are in propor-
tion to the numbers 4 and 8 ; because 3 are as many
times contained in 6, as 4 in 8, which may be expressed
thus : " •
3 :6 = 4: :8.
For this reason, if four lines are in a geometrical pro-
portion, their length, expressed in numbers of rods, feet,
&c, will be in the same proportion.
10th. It is to be remarked, that in every geometrical
proportion, expressed in numbers* the product obtained
by multiplying the two mean terms together, is equal
to the product obtained by multiplying the two extreme
terms. In the above proportion, 3:6=4:8, for in-
stance, we have, 3 times 8 equal to 6 times 4. For, the
first of our extreme terms, 3, being exactly asjnany times
smaller than the first of our mean terms, 6, as the second
of our extreme terms, 8, is greater than the second of
our mean terms, 4 (namely, twice) ; what the multiplier
3, in the one case, is smaller than the multiplier 6, in the
other, is made up by the multiplicand 8, which is as many
times greater than the multiplicand 4, as the multiplier 3
is smaller than the multiplier 6 ; and in a similar manner
* For wc cannot multiply lines together, but merely the abstract
numbers, which express their relative length.
6* -r
66 GEOMETRY.
we could prove the same of any other geometrical pro-
portion. To give but one more example : In the pro-
portion
2:4r=3:_6,
we have, again, twice 6 equal to 4 times 3 ; because the
first multiplier, 2, is exactly as many times smaller than the
second multiplier, 4, as the first multiplicand, 6, is greater
than the second multiplicand 3 (namely, twice).*
If both ratios of our proportion were inverted, as, for
instance, 4 : 2 — 6 : 3, our principle would still prove to be
correct. For we have again 4 times 3 equal to twice 6.
The only difference consists in the mean terms having
now become the extreme terms, and vice versa. If we
change the order of the means and extremes^ their pro-
ducts refnain still the same. For 3 times 8 are the same
as 8 times 3 , v ^nd 6 times 2 the same as. twice 6.
•When, in a geometrical proportion, the two mean
terms are equal to one another, either of t-htem is called a
mean proportional between the two extremes. Thus, in
the proportion, . '
4:6 = 6:9,
6 is a mean proportional between 4 and 9.
What you have learned of geometrical proportions
will enable you- to understand every principle in plane
Geometry ; we will therefore continue our inquiries into
the principles of geometrical figures.
,
* The teacher may illustrate this principle by a balance ; show-
ing that 2 weights, of 6 pounds each, are in equilibrium with 4
weights of 3 pounds each. The weights in this example, 6 pounds
and 3 pounds, are the multiplicands, and their number 2 and 4 are
the respective multipliers.
GEOMETRY. 07
QUERY XIV.
If you divide one side, AB~
of a triangle, ABC, into any
number of equal parts, for
instance four, and then, from
the points of division D, F, H,
draw the lines DE, FG, JIK,
parallel to the side AC, ivhat
remark can you make with regard to the other side BC?
A. That the other side, BC, is divided into as many
equal parts as the side AB.
Q. How can you prove this?
A. By drawing- the lines DL, FM, HN, parallel to the
side BC, the triangles, BDE> DFL, FHM, ilAN, are all
equal to one another. For, comparing, in the first place,
the two triangles, BDE, DFL, we see that the side BD
is equal to DF (beeause we have divided the line AB
into equal parts) ; and the angle x is equal to the angle z,
because these- angles are formed by the two parallels DL
and BC, being intersected by the straight line AB (Query
10, Sect I.) ; and the angle y is equal to the angle iv,
because y and w are formed, in a similar manner, by the
two parallels, DE, FG, being intersected by the same
straight line AB : consequently we have one side, DB,
and the two adjacent angles x and y, in the triangle BDE,
equal to one side DF, and the two adjacent angles, * and
w, in the triangle DFL ; therefore these two triangles are
equal to each other (duery 6, Sect I.) ; and the side DL,
opposite to the angle w, in the triangle DFL, is equal to
the side BE, opposite to the equal angle y, in the triangle
BDE ; and in the same manner it can be proved, that
FM and HN, are also equal to BE. Now, each of the
quadrilaterals, DELG, FGMK, HKNC, is a parallels
OS GEOMETRY.
gram (because the opposite sides are parallel) ; and as
the opposite sides of a parallelogram are equal, DL must
be equal to EG, FM to G.K, and HN to KC. But each
of the lines DL, FM, HN, is equal to BE; therefore,
each of the lines EG, GK, KC, must also be equal to
BE ■ consequently the line. BC is divided into the same
number of equal parts as. the line AB.
Q. Could you prove the same principle in the case
where the line AB is divided into five, six, or more equal
parts 1 ■
QUERY XV.
If, in a triangle j ARC, .you draw a
line,DE, parallel to one of the sides,
say AC; what relation do the parts
BD,DA; BE, EC, into which the
sides AB and BC qre divided, bear
to each other, and to the whole of the
sides AB, BC ?
A. The upper parts, BD and BE,
as well as the loivcr parts, DA and EC, arc in the same
ratio, in ichich the whole sides AB, BC, themselves are.
Q. Why?
A. Because you can imagine the side AB to be suc-
cessively divided into smaller aiid smaller parts, until one
of the points of division shall have fallen upon the point
D : then, by drawing, through all the points of division,
parallel lines to the side AC, the side BC will be
divided into as many equal parts* as the side BA (last
Query) ; and as the line DE itself will be one of these
parallels, BE will have as many of these parts marked as
BD ; and EC as many as DA : and therefore the ratio of
GEOMETRY. 69
the whole of the line BA to the whole of the line BC, must
be the same as that of BD to BE, or DA to EC.
Q. How can you express these proportions in writing ?
A. BA : BC = BD : BE
B A : BC = DA : EC ;
consequently, also,
BD:BE = DA:EC
(3d principle of proportion).
Q. Is the reverse of the same principle also true ? that
is, must the line DE be parallel to AC, when the parts
BD and BE, and DA and EC, are proportional to each
other, or to the whole of the sides BA, BC ?
A> Yes. For you need only imagine the side BA to
be again successively divided into smaller and smaller
parts, until one of the points of division shall have fallen
upon D. Then, it is evident that, by drawing, as before,
through the points of division, parallels to the side AC,
DE itself must be one of them, if BE shall again have as
many of these parts marked as BD, and EC as many as
DA ; for only in this case can BE, BD, and EC, DA, be
proportional to each other, and to the- whole of the sides
BC and BA.
Remark. It has already been stated (page 55), that two geomet-
rical figures Cannot be similar to each other, unless they are con-
structed after the same manner, and have their sides proportional.
We will now give the strictly geometrical definition of the same
principle for rectilinear figures. .
In order that two rectilinear figures may be similar to each
other, it is necessary,
1st, That both figures should be composed of the same number
of sides /* •
* This will, of course, always be the case in triangles.
70 GEOMETRY.
2dly, That the angles in one figure should be equal to the
angles in the other, each to each ;
3dly, That these angles should follow each other in precisely
the same order in both figures ; and,
4thly, That the sides, ivhich include the equal angles in both
figures (and which are therefore called the corresponding or
homologous sides*), should be in a geometrical proportion.
QUERY XVI.
If, in a triangle, ABC, you draw
a line, DE, parallel to one of the
sides, say AB, tvhat relation docs
the triangle DEC, which is cut off ,
bear to the wliole of the triangle
ABC?
A. The triangle J) EC is simi-
lar to the triangle ABC.
Q Why? .,../'.%
A. Because tli e three angles, x, y, z, of the triangle
DEC, are equal to the three angles, w, y, t, of the triangle
ABC, each to each ; for the angles x and z are respec-
tively equal to the angles w, t; because the line DE is
parallel to AB (Query 10, Sect. I.). This satisfies the
three first conditions of similarity. Moreover, we have
the proportion CD : CE rz: CA : CB (preceding Query),
and by drawing DH parallel to the side CB, also the pro-
portion CD : BH (or ED) r= AC : AB ; therefore, the
three sides of the triangle DEC are proportional to the
three sides of the triangle ABC, which is the fourth con-
dition of similarity : consequently these two triangles are
similar to each other.
* In triangles, the corresponding sides are those which are oppo-
site to the equal angles.
GEOMETRY. 71
QUERY XVII.
If the three angles in one
triangle are equal to the
three angles in another tri-
angle, each to each, ivhat
relation do these triangles
bear to each other ?
A. They arc similar.
Q. How can you prove' it 1
A. By applying the. triangle abc to the triangle ABC,
the angle at c will coincide with the angle at C, and the
side ca will fall upon CA, and cb upon CB ; and as the
angles at a and b, in the triangle abc l are respectively
equal to the angles at A and B, in .the triangle ABC, the
side ab will fall parallel to the side AB (Query 8, Sect.
I.) and we shall have the same case as in the preceding
Query: consequently, the triangles abc and ABC will
be similar to each other.
Q. Supposing you have a triangle, of which you know
only two angles, respectively, equal to two angles in another
triangle, whaf can you infer with regard to these two
A. -That tliey must still be similar to each other. For
two angles of a triangle always., determine the third one
(page 34, 2d.).
QUERY XVIII.
If you have two triangles, abc, ABC (see the last
figure), and only know that one angle at c, in the one, is
equal to one angle at C in the other,, but that the sides,
which include that angle in both triangles, are in a geo-
metrical proportion, what inference can you draw from it 1
A. That these triangles are again similar to each
72
GEOMETRY.
other. For if you imagine the triangle abc placed, as
before, upon the triangle ABC, the angle at c will again
coincide with the angle at C, and the side ca will fall
upon CA, and cb upon CB ; and as ca and cb are pro-
portional to CA and CB, the side ab will fall parallel to
the side AB (Query 15, Sect. II.); and- we shall once
more have the same case as in Query 16, Sect: II. ; con-
sequently the triangle abc will be similar to the triang\e
ABC
QUERY XIX.
L°.t us now consider the case, where all the angles of
two triangles arc unknown-, but the three 'sides -of the
one (ire in prdportion to the . three sides of the other ;
what relation will these tfiahglcs hear io each other?
A. They will still be similar to each other ?
Q. How can you prove it ? J
A. Let us suppose, for instance,
that the three sides of the triangle
ab' are in proportion to the three
sides of the' triangle ABC ; that
is, iet us have the proportions
ac : ab = AC : AB
ac : cb = AC : CB. .
Then make CD equal to ca, and
draw through the point D the
line DE, paralfel" to AB ; and the
triangle CDE will be similar to
the triangle CAB (Query 16,
Sect. II.), and we shall "have the
proportions
DC : I>E = AC : AB
DC : CE = AC : CB,
in which the two ratios, AC : AB, and AC : CB, are he
same as in the first two proportions ; consequently, eom-
GEOMETRY. 73
paring these two proportions with the two preceding
ones, we shall have
DC :DE = ac:ab
DC : CE = ac : cb
(see theory of proportions, principle 3d).
Now, as I have made DC equal to ac, I can write ac
instead of DC, in the two last proportions ; and they will
then become
ac : DE = ac : ab,
ac : CE = ac : cb.
The upper one expresses, that the line DE is as many
times greater than ac, as the line ab is greater than the
same line ac (Definition of geometrical proportions) ;
consequently the line DE is equal to the line ab. In
like manner does the lower one express, that CE is as
many times greater than ac, as cb is greater than the
same line ac ; consequently CE is also equal to cb ; and
the three sides of the triangle DEC, are equal to the three
sides of the triangle abc, each to each ; therefore these
two triangles are equal to one another (Query 4, Sect.
II.) ; and as the triangle DEC is similar to the triangle
4.BC the triangle abc will also be similar to it.
Q. Will you now briefly state the different cases, in
which two triangles are similar to one another 1
A. 1st. When the three angles in one triangle are
equal to the three angles in another, each to each ; and
also when two angles in one triangle are equal to two
angles in another, each to each ; because then the third
angle in the one is also equal to the third angle in the
other.
2dly. When an angle in one triangle is equal to an
angle in another, and the two sides which include that
7
74 GEOMETRY.
angle in the one triangle, are in proportion to the tioo
sides which include that angle in the other triangle.
3dly. When the three sides of one triangle are in pro-
portion to the three sides of another.
QUERY XX.
If you have a right-an-
gled triangle, ABC, and
from the vertex A, of the
riglit angle, let fall a per-
pendicular, AD, upon the
hypothenuse BC, what re-
lation do the two triangles,
ABD and ACD, into which the whole triangle is di-
vided, bear to each other, and to the whole triangle ABC
itself ?
A. The two triangles, ABD and ACD, are similar to
each other, and to the whole triangle ABC.
Q. How can you prove this 1
A. The triangle ABD is similar to the whole triangle
ABC, because the two triangles being both right-angled,
and having the angle at x common, have two angles in
one triangle, respectively, equal to two angles in the other
(page 73, case 1st) ; and for the same reason is the tri-
angle ACD similar to the whole triangle ABC (both
being right-angled, and having the angle y common) ;
and as each of the two triangles, ABD, ACD, is similar
to the whole triangle ABC, these two triangles must be
similar to each other. (Truth II.)
Q. IVJiat important inferences can you draw from the
principle you have just established ?
A. 1st. In the two similar triangles, ABD and ACD,
the sides which are opposite to the equal angles, must be
GEOMETRY. 75
in proportion (condition 4th of geometrical similarity,
page 70) ; and we shall therefore have the proportion
BD: AD=AD : DC;*
that is, the perpendicular AD is a mean proportional
between the two parts into which it divides the hypothec
nuse. (Theory of proportion, page 66.)
A2%. From the two similar triangles, ABC, ABD, we
shall have the proportion
BD : AB = AB : BC ;
that is, the side AB, of the right-angled triangle ABC,
is a mean proportional between the whole hypothenuse BC,
and the part BD, cut off from it by the perpendicular
AD.f
Sdly. The two similar triangles, ACD and ABC, give
the proportion
DC : AC = AC : BC ;
that is, the other side, AC, of the right-angled triangle
ACD, is also a mean proportional between the whole
hypothenuse and the other part, DC, cut off" from it by the
perpendicular AD.
Remark. The five last queries comprise one of the most im
portant parts of Geometry. The principles contained in them are
applied to the solution of almost every geometrical problem. The
beginner will therefore do well to render himself perfectly familiar
with them.
* The first ratio is formed by the two sides, BD and AD, of the
triangle ADB, of which BD is opposite to the angle z, and AD to
the angle x ; and the second ratio is formed by the two correspond-
ing sides, AD, DC, of the triangle ADC ; because the sides AD,
DC, are opposite to the angles y and u, which are respectively
equal to z and x.
t The part BD of the hypothenuse, situated between the ex-
tremity B of the side AB, and the foot D of the perpendicular AD,
is sometimes called the adjacent segment to AB. (Legendre's
Geometry, translated by Professor Farrar.)
76 GEOMETRY.
RECAPITULATION OF THE TRUTHS CONTAINED IN
THE SECOND SECTION.
Ques. Can you now repeat the different principles re-
specting the equality and similarity of triangles, which
you have learned in this section ?
Ans. 1. If, in two triangles, two sides of the one are
equal to two sides of the other, each to each, and the
angles which are included by them are also equal to one
another, the two triangles are equal in all their parts, that
is, they coincide with each other throughout.
2. In equal triangles, that is, in triangles which coin-
cide with each other, the equal sides are opposite to the
equal angles.
3. If one side and the two adjacent angles in one tri-
angle are equal to one side and the two adjacent angles
in another triangle, each to each, the two triangles are
equal, and the angles opposite to the equal sides are also
equal.*
4. The two angles at the basis of an isosceles triangle
are equal to one another.
5. If the three sides of one triangle are equal to the
three sides of another, each to each, the two triangles
coincide with each other throughout ; that is, their angles
are also equal, each to each.
6. In every triangle, the greater side is opposite to the
greater angle, and the greatest side to the greatest angle.
7. In a right-angled triangle, the greatest side is oppo-
site to the right angle.
* This principle, though already demonstrated in the first
section, is repeated here, in order to complete what is said on the
equality of triangles.
GEOMETRY. 77
8. When a triangle contains two equal angles, it also
has two equal sides, and the triangle is isosceles.
9. If the three angles in a triangle are equal to eac*.
other, the sides are also equal, and the triangle is equi-
lateral.
10. Any one side of a triangle is smaller than the sura
of the two other sides.
11. If from a point within a triangle, two lines are
drawn to the two extremities of one of the sides, the
angle mader by those lines is always greater than the
angle of the triangle which is opposite to that side ; but
the sum of the two lines, which make the interior angle,
is smaller than the sum of the two sides which include
the angle of the triangle.
12. If from a point without a straight line, a perpen-
dicular is let fall upon that line, and, at the same time,
other lines are drawn obliquely to different points in the
same straight line, the perpendicular is shorter than any
of the oblique lines, and is therefore the shortest line that
can be drawn from that point to the straight line.
13. The distance of a point from a straight line is
measured by the length of the perpendicular, let fall from
that point upon the straight line.
14. Of several oblique lines drawn from a point with-
out a straight line, to different points in that straight line^
that one is the shortest, which is nearest the perpendicu-
lar, and that one is the greatest, which is farthest from
the perpendicular.
15. If a perpendicular is drawn to a straight line, then
two oblique lines drawn from two points in the straight
line, on each side of the perpendicular, and at equal dis-
tances from it, to any one point in that perpendicular, are
equal to one another.
16. If a perpendicular is drawn to a straight line, there
7*
78 GEOxMETRY.
is but one point in the straight line, on each side of the
perpendicular, such, that a straight line drawn from it to
a given point in that perpendicular, is of a given length.
17. If a perpendicular is drawn to a straight line, there
is but one point in the straight line, on each side of the
perpendicular, from which a line drawn to a given point
in that perpendicular, makes with the straight line an
angle of a required magnitude.
i 18. If two sides and the angle which is opposite to the
greater of them, in one triangle, are equal to two sides
and the angle which is opposite to the greater of them in
another triangle, each to each, the two triangles coincide
with each other in all their parts ; that is, they are equal
to each other.
19. If the hypothenuse and one side of a right-angled
triangle, are equal to the hypothenuse and one side of
another right-angled triangle, each to each, the two right-
angled triangles are equal.
20. If in two triangles two sides of the one are equal to
two sides of the other, each to each, but the angle in-
cluded by the two sides in one triangle, is greater than
the angle included by them in the other, the side opposite
to the greater angle in the one triangle, is greater than the
side opposite to the smaller angle in the other triangle.
21. Every parallelogram is, by a diagonal, divided into
two equal triangles.
22. The opposite sides of a parallelogram are equal to
each other.
23. The opposite angles in a parallelogram are equal
to each other.
24. By one angle of a parallelogram a\\four angles are
determined.
25. A quadrilateral, in which the opposite sides are
respectively equal, is a parallelogram.
GEOMETRY. 719
26. A quadrilateral, in which two sides are equal and
parallel, is a parallelogram.
27. If from one of the vertices of a rectilinear figure,
diagonals are drawn to all the other vertices, the figure is
divided into as many triangles as it has sides less two.
28. The sum of all the angles in a rectilinear figure, is
equal to as many times two right angles as the figure has
sides less two.
RECAPITULATION OF THE TRUTHS CONTAINED IN
PART II.
1. On Proportions.
Ques. 1. How is a geometrical ratio determined ?
■Q. 2. What is the ratio of a line 3 inches in length,
to a line of 12 inches? What, the ratio of a line 2 inches
in length, to one of 10 inches, &c?
Q. 3. When two geometrical ratios are equal to one
another, what do they form ?
Q. 4. What is a geometrical proportion ?
Q. 5. What signs are used to express a geometrical
proportion ?
Q. 6. What sign is put between the two terms of a
ratio 1
Q. 7. What sign is put between the two ratios of a
proportion ?
Q. 8. What are the first and fourth terms of a geomet-
rical proportion called ?
Q. 9. What are the second and third terms of a geo-
metrical proportion called 1
Q. 10. What are the most remarkable properties of
geometrical proportions ?
80 GEOMETRY.
Ans. a. In every geometrical proportion the two ratios
may be inverted,
b. In every geometrical proportion the order of the
means or extremes may be inverted.
c. If two geometrical proportions have a ratio common,
the two remaining ratios make again a proportion.
d. If you have several geometrical proportions, of
which the second has a ratio common with the first, the
third a ratio common with the second, the fourth a ratio
common with the third, &,c.,the sum of all the first terms
will be in the same ratio to the sum of all the second
terms, as the sum of all the third terms is to the sum
of all the fourth terms ; that is, the sums make again a
proportion.
e. The second term of a proportion being added once,
or any number of times, to the first term, and the fourth
term the same number of times to the third term, they
will still be in proportion ; and in the same manner can
the first term be added a number of times to the secoud
term, and the third the same number of times to the
fourth term, without destroying the proportion.
f. The second term may also be once, or any number
of times, subtracted from the first- term, and the fourth
from the third term, without destroying the proportion ;
or the first term may also be subtracted from the second,
and the third from the fourth — and the result will still be
a geometrical proportion.
g. If all the terms of a geometrical proportion are
multiplied or divided by the same number, the proportion
remains unaltered.
h. From three terms of a geometrical proportion the
fourth term can be found.
t. If four lines are together in a geometrical propor-
GEOMETRY. gl
tion, their lengths, expressed in numbers of rods, feet,
inches, &.C., are in the same proportion.
k. In every geometrical proportion, the product of the
two mean terms is equal to that of the two extremes.
I. When the two mean terms of a geometrical proport
tion are equal to each other, either of them is called a
mean proportional between the two extremes.
QUESTIONS ON SIMILARITY OF TRIANGLES.
Ques. What other principles do you recollect in the
second part of the second section 1
Ans. 1. If one side of a triangle is divided into any
number of equal parts, and then, from the points of
division, lines are drawn parallel to one of the two other
sides, the side opposite to the one that has been divided
will, by these parallels, be divided into as many equal
parts as the first side.
2. If, in a triangle, a line is drawn parallel to one of
the sides, that parallel divides the two other sides into
such parts as are in proportion to each other and to the
whole of the two sides themselves ; and the reverse of this
principle is also tf ue ; namely, a line must be parallel to
one of the sides of a triangle, if it divides the two other
sides proportionally.
3. If, in a triangle, a line is drawn parallel to one of
the sides, the triangle which is cut off by it is similar to
the whole triangle.
4. If the three angles in one triangle are equal to the
three angles in another triangle, each to each, the two
triangles are similar to one another ; and the same is the
case if only two angles in one triangle are equal to two
angles in another, each to each.
82 GEOMETRY.
5. If an angle in one triangle is equal to an angle in
another, and the two sides which include that angle in
the one triangle are in proportion to the two sides which
include the equal angle in the other, these two triangles
are similar to each other.
6. If the three sides of one triangle are in proportion
'o the three sides of another, the two triangles are similar
to each other.*
7. If, in a right-angled triangle, a perpendicular is let
fall from the vertex of the right angle upon the hypothe-
nuse, that perpendicular divides the whole of the triangle
into two parts, which are similar to the whole triangle,
and to each other.
8. The perpendicular let fall from the vertex of a right-
angled triangle upon the hypothenuse, is a mean pro-
portional between the parts into which it divides the
hypothenuse.
9. In every right-angled triangle, each of the sides
which include the right angle is a mean proportional
between the hypothenuse and that part of it, which lies
between the extremity of that side and the foot of the
perpendicular let fall from the vertex of the right angle
upon the hypothenuse.
* The teacher will do well to let the pupil repeat the different
eases where two triangles are similar to each other. (Page 73.)
SECTION III.
OF THE MEASUREMENT OF SURFACES.
Preliminary Remarks. We determine the length of a line,
by finding how many times another line, which we take for the
measure, is contained in it. The line which we take for the
measure is chosen at pleasure ; it may be an inch, a foot, a fathom,
a mile, &c. If we have a line upon which we can take the length
of an inch 3 times, we say that line measures 3 inches, or is 3
inches long. In like manner, if we have a line upon which we
can take the length of a fathom 3 times, we call that line 3 fath-
oms, &c. To find out which of two lines is the greater, we must
measure them. If we take an inch for our measure, that line is
the greater, which contains the greater number of inches. If we
take a foot for our measure, that line is the greater, which contains
the greater number of feet, &c.
To measure the extension of a surface, we make use of another
surface, commonly a square ( □ ), and see how many times it can
be applied to it ; or, in other words, how many of those squares it
takes to cover the whole surface. The length of the square side is
arbitrary. If it is an inch, the square of it is called a square inch ;
if it is a foot, a square foot ; if it is a mile, a square mile, &c. The
extension of a surface, expressed in numbers of square miles, rods
feet, inches, &c, is called its area.
Rcinark 2. If we take one of the
sides of a triangle for the basis, the
perpendicular let fall from the vertex
of the opposite angle, upon that side,
is called the altitude or height of the
triangle.
If, in the triangle ABC, (Fig. I.)
for instance, we call AC the basis, the
perpendicular BD will be its height. If
the perpendicular BD should fall with-
out the triangle ABC (as in Fig. II.),
Fig. I.
84 GEOMETRY.
we need oiily extend the basis, and then let fall the perpendicular
BD upon its farther extension CD.
* M a * &
a*
2sT V
$
If, in a parallelogram, ABCD, we take AD for the basis, any per
pendicular, MN, CO, PQ, &c., let fall from the opposite side BC,
or its farther extension CR, upon tbat basis, or its farther extension
DS, measures the height of the parallelogram. For in a parallelo-
gram the opposite sides are parallel to each other (see Definitions),
and all the perpendiculars, let fall from one of two parallel lines to
the other, are equal (Query 11, Sect. I.). What in this respect holds
of a parallelogram is applied also to a square, a rhombus, and a
rectangle^; for these three figures are only modifications of a paral-
lelogram. (See Definitions.)
As in every rectangle, ABCD, the adjacent (*. T)
sides, AB, BD, are perpendicular to each I _
other, it is evident that if AB is taken for the
basis, the side BD itself is the height of the rectangle.
Remark 3. We call two geometrical figures equal* to one
another, when they have equal areas (see preliminary remark to
Sect. II.). Thus a triangle is said to be equal to a rectangle when
it contains the same number of square miles, rods, feet, inches, &c,
as that rectangle.
* The term equivalent would undoubtedly be better ; but as
there is no generally adopted sign in mathematics to express that
two things are equivalent without being exactly the same, we are
~° the term equal.
GEOMETRY.
85
D 1 2
4 5
L
6 C
4
12 3 4 5 6
QUERY I
If the basis, AJB, of a rectan-
gle, ABCD, measures 6 inches,
and the height, the side BC, 4
inches, how many square inches
are there in the rectangle ?
A. Twenty-four.
Q. How can you prove this ?
A. If a rectangle is four inches high, I can divide it,
like the rectangle ABCD (see the figure), into four rect-
angles, each of which ,is one inch high, and has its basis
equal to the basis of the whole rectangle. And as the
basis, AB, of the rectangle measures 6 inches, by raising
upon it, at the distance of an inch from each other, the
perpendiculars 1, 2, 3, 4, 5, each of the four rectangles
will be divided into 6 square inches ; and therefore the
whole rectangle ABCD into 24 square inches.
Q. How many square inches are
there in a rectangle, whose basis is
5, and height 3 inches ?
A. Fifteen. Because in this case
I can divide the rectangle into 3
rectangles of 5 square inches each.
Q. Supposing the measurements of the first rectangle
were given in feet, in rods, or in miles, instead of inches,
how many square feet, rods, or miles would there be in
the rectangle 1
A. If its measurements were given in feet, it would
contain 24 square feet ; if they were given in rods, it
would contain 24 square rods, &c. ; for in these cases I
need only imagine the lines, 1, 2, 3, 4, &c, to be drawn
a foot, a rod, &c, apart ; the number of divisions will
8 " .
C 1 2 3 4 5JD
3
2
1
12 3 4 5
86
GEOMETRY.
remain the same ; nothing but their size will be altered.
And the same reasoning applies to the second rectangle.*
Q. Can you now give a general rule for finding the
area of a rectangle ?
A. Yes. Multiply the length of the basis given in rods,
feet, inches, fyc, by the height expressed in units of the
same kind.
Q. Can you now tell me how to find the area of a
square ?
A. The area of a square is found by multiplying one
of its sides by itself For a square is a rectangle whose
sides are all equal (see Definitions) ; and the area of a
rectangle is found by multiplying the basis by an adja-
cent side.
QUERY II.
If a parallelogram ABEF
stands on the same basis, AB,
as a rectangle, ABCD, and
has its height equal to the
height of that rectangle, what
relation do the areas of these
two figures bear to each other ?
A. The area of the parallelogram ABEF is equal to
the area of the rectangle ABCD ; therefore I can say
that the parallelogram ABEF is equal to the rectangle
ABCD (see remark 3d, Introd. to Sect. III.).
C D F
E
/
V /
'£\
- 1
A. - 1
H
* The teacher may also give his pupils a rect-
angle whose measurements are both- given in
fractions ; for instanc*e r a rectangle of 3£ inches
in length and 2^ inches high, and then show by
the figure that this rectangle measures 6 square
inches, 2 half square inches, | and J of a square inch; in the whole
7} square inches, which is the answer to the multiplication of 3J
fa
'/*
'/+
:■>.
1
1
1
fk
1
1
1
r*
by 2£.
m
GEOMETRY. 87
Q. How can you prove it 1
A. The right-angled triangle ACF has the hypothe-
nuse AF and the side AC, equal to the hypothenuse BE
and the side BD, in the right-angled triangle BDE, each
to each (AF and BE, AC and DB, being opposite sides
of the parallelogram ABEF, and the rectangle ABCD,
respectively) ; therefore these two triangles are equal
(page 47) i and by taking from each of the two equal tri-
angles ACF, BDE, the part DGF common to both, the
remainders, AGDC, BGFE, are also equal(Truth IV.);
and then, by adding again to each of the equal remainders
the same triangle ABG, the sums, that is, the rectangle
ABCD and the parallelogram ABEF are equal to one
another. (Truth III.)
Q. What important truths can you infer from the one
you have just demonstrated ?
A. 1st. All parallelograms, which have equal bases and
heights, are equal to one another ; for each of them is
equal to a rectangle upon the same basis, and of the same
lieight. (Truth I.)
2dly. Parallelograms upon equal bases, and between the
same parallels, am equal to one another ; for if they are
between the same parallels, their heights must be equal,
(Query 11, Sect. I.)
3dly. The area of a parallelogram is found by multi-
plying the basis, given in rods,, feet, inches, fyc, by the
Jteight, expressed in units of the-same kind. Because the
area of the rectangle upon the same bas-is and of the same
height to which it is equal, is found in the same manner.
4thly. The area of a rhombus or lozenge is found like
that of a parallelogram, a lozenge being only a peculiar
hind of parallelogram.
5thly. The areas of parallelograms are to each other,
as the products obtained by multiplying the length of the
m
88 GEOMETRY.
bases of the parallelograms by their heights; because
these products are the areas of the parallelograms.
The parallelogram
ABCD, for instance, *pf- — ty E_0__F
is to the parallelogram / / / ~7
GHEF, as the product Z L/ /_ !__/
of the basis AB, by the A NB a rH
height MN, is to the product of the basis GH, by the
height OP ; because AB multiplied by MN is the area of
the parallelogram ABCD, and GH multiplied by OP is
the area of the parallelogram GHEF. This proportion
may be expressed thus :
Parallelogram ABCD : parallelogram GHEF = AB
X MN : GH x OP.
6thly. Rectangles or parol- .
hlograms, which have equal / — j j „ 1
bases, are to each other as their / / / ) ~J
heights. L U L \J
For if, in the above propor- A NB G Pir
tion, the basis AB is equal to the basis GH, you can write
AB instead of GH, and thereby change it into
Parallelograms ABCD : parallelograms GHEF =
AB X MN : AB X OP ;
that is, the parallelogram ABCD is to the parallelogram
GHEF, as AB times the height MN is to AB times the
height OP ; or, which is the same, as the height MN
alone is to the height OP alone ; which is written thus:
Parallelogram ABCD : parallelogram GHEF =
MN : OP.
7thly. In precisely the same manner it may be proved,
that if the heights MN and OP are equal, the parallelo-
grams ABCD, GHEF, are to each other as their bases;
which may be expressed thus :
Parallelogram ABCD : parallelogram GHEF =
AB : GH.
GEOMETRY. 89
QUERY III.
If two triangles, ABC, ABE, stand on the same basis
AB, and have equal heights CK, EG, what relation do
the areas of these triangles bear to each other ?
nam j2r
A. The areas of these triangles are equal.
Q. How can you prove it 1
A. Draw the line AD parallel to BC ; BF parallel to
AE ; and through the two vertices G and E, the line DF
parallel to AG (which is possible since the heights CK
and EG are equal). The area of the parallelogram
ABCD will be equal to the area of the parallelogram
ABEF (Query 2, Sect. III.) ; and as the triangle ABC
is half of the parallelogram ABCD (Query 12, Sect. II.),
and the triangle ABE half of the parallelogram ABEF,
the areas of these two triangles are also equal to one
another ; for if the wholes are equal, the halves must be
equal ; and in the same way it may be proved that trian-
gles, which have equal bases and heights, are equal to
one another.
Q. What consequences follow from the principle just
advanced 1
A. 1st. Every triangle is half of a parallelogram upon
equal basis and of the same height. (This is evident from
looking at the figure, and from Query 12, Sect. II.)
2d. The area of a triangle is half of the area of a
parallelogram upon the same basis and of the same height.
Thus the area of a triangle is found by multiplying its
■ 8*
90 GEOMETRY.
basis by its height, and dividing the product by 2 ;* for
the area of a parallelogram is equal to the whole pro-duct
of the basis by the height.f
3d. The areas of triangles upon the same basis and
between the same parallels are equal; because if they are
between the same parallels, their heights are equal ; and
we have the same case as in the last query ; namely, tri-
angles upon the same basis, and of equal heights.
4th. The areas of triangles are to each other as the
products of their bases by their heights : for the halves of
these products being the areas of the triangles, the whole
products must be in the same ratio. Thus the area of
the triangle ABC is
to the area of the
triangle EGP, as the
basis AB multiplied Ji^- jy- \ B JE7
by the height CN, is
to the basis EG, multiplied by the height PM ; which
may be expressed thus :
Triangle ABC : triangle EGP = ABx CN : EG X PM.
5th. Hie areas of triangles upon equal bases are to
each other as the heights of the triangles ; because the
areas of parallelograms upon the same bases and of the
same heights, are to each other in the ratio of the heights ;
and their halves (the areas of the triangles) must be in
* Instead of multiplying the basis by the whole height and
dividing the product by 2, you may multiply the basis by half the
height, or the height by half the basis.
t If the basis of a triangle is 8 feet and the height 4 feet, the
area of the triangle is equal to 4 times 8, divided by 2; that is,
16 square feet ; whereas the rectangle upon 8 feet basis and 4
feet high, measures 32 square feet, which is double the area of the
triangle.
1
GEOMETRY. 01
the same ratio.* Thus if the two triangles ABG, ECF,
O
/i\ F
have their bases AB, EC, equal to each other, we have
the proportion :
Triangle ABG : triangle ECF — CM : FN.
6th. The areas of triangles, which have equal heights,
are to each other as the bases of the triangles. This truth
follows like the preceding-, one from the same principle
established with regard to parallelograms, of which the
triangles are the halves. (Page 88, 7thl.y.)
QUERY IV.
How do you find the -A J? j£
3
area of a trapezoid ?
A. By multiplying .
the sum of the two par- ~ I? D
allel sides by their distance, and dividing the product
by 2.
Q. How can you prove this 1
A. By drawing the diagonal AD, the trapezoid ABCD,
will be divided into the two triangles ACD and ABD.
The area o'fthe triangle ACD is found by multiplying its
basis, CD, by its height AF, and dividing the product
by 2. (Page 89, 2d.) In the same manner we find the
area of the triangle ABD, by multiplying its basis, AB,
* This principle and the following one might have been estab-
lished immediately from the proportion :
Triangle ABC : triangle EGP = AB X CN : EG X PM,
in precisely the same manner, as it has been proved for parallelo-
grams. (Page 88, 6thly.)
92 GEOMETRY.
by its height DE, and dividing the product by 2; and
as the height, DE, of the triangle ABD, is equal to
the height AF, of the triangle ACD (because DE and
AF are perpendiculars between the same parallels), we
can find the area of the two triangles, or of the whole
trapezoid, ABCD, at once, by multiplying the sum of the
two parallel lines AB, CD, by their distance AF, and
dividing the product by 2.*
QUERY V.
Hoic do you find the area of,-
a polygon ABCDEF, or-, in
general, of any other rectilinear
figure ?
A. By dividing it by means
of -diagonals [as in the figure ~~ •&
before you), or by any other means into triangles. The
area of each of these triangles is then easily found by the
rule given (page 89, 2d.) ; and the sum of the areas of
all the triangles, into which the figure is divided, is the
area of it. .
* If you multiply two numbers successively by the same number,
and then add the products together, the answer will be the same
as the sum of the two numbers at once multiplied by that number.
Multiply each of the numbers, 6 and 5, for instance, by 4, and
then add the products, 24 and 20, together, you will have 44 ; and
adding, in the first place, 6 to 5, and then multiplying the sum,
11, by 4, you will again have 44.
Instead of multiplying the sum of the two parallel sides by their
distance, and then dividing the product by 2, you may multiply, at
once, half the sum of the two parallel sides by their distance ; or
the sum of the two parallel sides by half their distance.
GEOMETRY.
93
UUERY VI.
If, upon each of the three sides AB, AC, BC, of a
right-angled triangle ABC, you construct a square , what
relation do the squares constructed upon the sides AC y
BC, bear to the square constructed upon the hypothenuse y
AB1
A. The square ABHK, constructed upon the hypothe-
nuse AB, equals, in area, the two squares ACDE,
BCGF, constructed upon the two sides AC, BC.
Q. How can you prove it by this diagram, in which
the perpendicular CM, is let fall from the vertex C, of
the right-angled triangle ABC, upon the hypothenuse
AB, and extended until, in I, it meets the side HK,
opposite to the hypothenuse ; and DB and CH, are
joined? v : .
A. In the first place, I should remark that the two
sides AB, AD, of the triangle ABD, are equal to the two
sides AH, AC, of the triangle ACH, each to each (AH
94 GEOMETRY.
and AB, being sides of the same square, ABHK ; and,
AC and AD, being sides of the square ACDE) ; and
that the angle DAB, included by the sides AD, AB, is
also equal to the angle CAH, included by the two sides
AC, AH (for each of these angles is formed by the angle
CAB being added to the right angle of a square) ; there-
fore these two triangles are equal to each other, (Query 1,
Sect. II.)
• Q. Having proved that the triangle ABD is equal to
the triangle ACH, what can you infer from it?
A. That the area of the square ACDE, is equal to the
area of the rectangle AHIM. For the area of the triangle
ABD, is half of the area of the square ACDE ; because
the triangle ABD stands upon the same basis AD, as the
square ACDE, and has its height BQ,, equal to the height
AC of that square ; and it has been proved that the" area
of every triangle is half of the area of a rectangle or
square of equal basis and height. (Page 89, 1st.) For the
same reason is the area of the triangle ACH, equal to
half the area of the rectangle AHIM ; for the* triangle
ACH, stands on the same basis AH, as the rectangle
AHIM, and has its height CO. equal to the height AM,
of that rectangle ; and as the halves , the two triangles
ABD and ACH, are eqnal to eacrf other, the wholes, the
squares ADEC and AHIM, must also be equal to each
other, fn precisely the same, manner I can prove
from the equality of the two triangles ABG and BCK,
that the square BCFG is^equal to the rectangle 'MBIK ;
and because the area of the square ADEC, is equal to
the area of the rectangle AHIM, and the area of the
square BCFG is equal to the area of the rectangle
MBIK ; therefore the sum of the areas of the two rectan-
gles, AHIM and MBIK, that is, the area of the square
upon the hypothcnuse AB, is equal to the sum of the
GEOMETRY. 95
areas of the squares constructed upon the two sides AC,
BC.
Remark. For the discovery of this principle, we are indebted to
Pythagoras, a famous Greek mathematician. It is a very important
one, and teaches how to find one of the sides of a right-angled tri-
angle when the two others are given* if, for instance, the two
sides AC, BC, of the right-angled triangle ABC, were known to
measure,. one 5, the other 6 inches, the sum of their squares 25
(5 time3^5.),.and 36 (6 times 6), equal to 61, would be the area of
the square of the hypothenuse ; and the square root of that number
would be the hypothenuse AB, itself. If the hypothenuse and one
of the sides are given, -you need only subtract the square of the
side from the square of the hypothenuse, and then the square root
of the remainder is. the other side. If, for instance, the hypothe-
nuse of a right-angled triangle were 10 feet, and one of the sides
6 feet ; the square of the hypothenuse would be 10 times 10, or 100,
and the square of 6, which is 36, subtracted from 100, ldaves*64,
which would be the. square of the side to be found ; and taking the
square root of it, which is 8 (because 8 times "8 are 64), you will
have the side itself*
QUERY VII.
It has ' been proved (page 90, 4th), that the areas of
any two triangles are to each other as their bases, multi-
plied by their heights ; can you now find out the relation
which the bases and heights of similar triangles bear to
each other ?
A. In similar triangles the bases are in proportion to
the heights.
Q. How can you prove this ?
* We shall hereafter give the geometrical solutions of these
problems.
9(5
GEOMETRY.
A. Let there be any two similar triangles ABC, AED.
Place the smaller one ABC, upon the larger AED, in
such a way that the angle at A falls upon the angle at A,
and from the vertices, C and D, let fall the perpendicu-
lars CM," DO, upon AO. Then the two triangles BCM,
EDO, are both right-angled, and the angle CBM is equal
to the angle DEO (because in the two similar triangles
ABC, AED, the angles ABC and AED, are equal to
each other, and CBM and DEO, make with them, respec-
tively, two right angles) ; therefore the third angle BCM
in the triangle BCM, is also equal to the third angle
EDO, in the triangle EDO, and the two triangles BCM,
EDO, are similar. (Page 73, 1st.) But in similar trian-
gles the sides opposite to the equal angles are propor-
tional, consequently we have
CM :DO = CB : DE;
and in -the similar- triangles ABC, ADE, •
AB : AE^CB : DE.
These two proportions have the second ratio common ;
therefore the two first ratios must again make a propor-
tion (Theory of Proportions, Principle 3d.), namely:
AB : AE = CM : DO.
This proportion expresses that the bases AB of the
smaller triangle ABC, is to the bases AE of the larger
triangle AED, as the height CM, of the first triangle
ABC, is to the height DO, of the triangle AED.
GEOMETRY.
97
QUERY VIII.
From what you have learned in the preceding query,
can you determine the proportion zohich the areas qfsimi*
lar triangles bear to each other ?
A. The areas of similar triangles are to each other as
the squares upon the corresponding sides.
li
r
a^
^^ ,
BM E O
Q. How can you prove this, for instance, of the two
similar triangles ABC, ABD ?
A. Let us place the smaller triangle ABC upon the
larger AED, as in the last query; and upon AB and
AE, construct the squares ABST, AERP. Then the
triangles ABC, AED, have the same bases AB, AE, as
the triangles ABT, AEP, and their heights CM, DO, are
in proportion to the heights TB, PE, of the triangles
ABT, AEP (TB and PE being respectively equal to
AB, AE, which, in the last query, are proved to be pro-
portional to CM and DO) ; therefore the areas of the
triangles ABC, AED, are in proportion to the areas of
the triangles ABT, AEP. (Page 90, 5thly.) But if the
two triangles ABC, AED, are in proportion to the two
triangles ABT, AEP, which are the halves of squares
ABST, AERP, they must also be in proportion to the
squares themselves ; which may be expressed thus :
9
I
98
GEOMETRY.
Triangle ABC : triangle AED = AB X AB : AE X AE,
and is read :
The area of the triangle AED is as many times greater
than the area of the triangle ABC, as the area of the
square upon the side AE, is greater than the area of the
square upon the corresponding side AB.
Q. Can you prove that the same ratio exists also be-
tween the squares upon the sides AC and AD, and also
between the two sides CB, DE, of the similar triangles
ABC, AED 1
A. Yes. For to prove it of the two sides AC and
AD, I need only take them for the bases of the two trian-
gles ; and to prove it of the sides CB, DE, I must take
CB and DE for the bases ; the reasoning would be the
same as that I just went through.
QUERY DC,
From the ratio which you have proved to exist between
the areas of similar triangles, can you now find out the
ratio which exists between the areas of similar polygons ?
(See Definitions.)
A. Yes. The areas of similar polygons are to each
other, as the areas of the squares constructed upon the cor-
responding sides. The areas of the two similar polygons
ABCDEF, abcdef for instance, are to each other as the
GEOMETRY. 99
areas of the squares constructed upon the sides AB, «6,
or as the areas of the squares upon the sides BC, be, &c.
For, by drawing in the polygon ABCDEF, the diagonals
AC, AD, AE, and in the polygon abedef, the correspond-
ing diagonals, ac, ad, ae, the triangle ACB, is similar to
the triangle abc, the triangle ACD, to the triangle acd,
&c. ; because, if the whole polygons ABCDEF, abedef \
are similar, their similarly disposed parts must also be
similar ; and the same proportion which exists between
their parts, must necessarily exist also between the whole
polygons; consequently, as the areas of the triangles
ABC, abc, ACD, acd, &c, are in the ratio of the areas
of the squares constructed upon their corresponding
sides, the whole polygons must be in the same ratio,
which may be expressed thus :
Polygon ABCDEF : polygon abedef
= AB X AB : ab X ab-
RECAPITULATION OF THE TRUTHS IN THE THIRD
SECTION.
Ques. |. How do you determine the length of a line?
2. How do you find out which of two lines is the
greater 1
3. How can you measure a surface ?
4. What do you call the area of a surface ?
5. If you take one of the sides of a triangle for the
basis, how do you determine the height of the triangle ?
6. How is the height of a parallelogram determined ?
How that of a rectangle ? A rhombus ? A square ?
7. When do you call a triangle equal to a square?
to a parallelogram ? to a rectangle, &>c. ?
100 GEOMETRY.
8. When can you call two geometrical figures equal to
one another, though these figures do not coincide with
each other ?
9. Can you repeat the different principles respecting
the areas of geometrical figures, which you have learned
in this section 1
Ans. 1. The area of a rectangle is found by multiply-
ing its basis, given in miles, rods, feet, inches, &c, by
its height expressed in units of the same kind.
2. The area of a square is found by multiplying one
of its sides by itself.
3. If a parallelogram stands on the same basis as a
rectangle, and has its height equal to the height of that
rectangle, the area of the parallelogram is equal to the
area of the rectangle 1
4. The areas of all parallelograms, which have equal
bases and heights, are equal to one another.
5. Parallelograms upon equal bases, and between the
same parallels, are equal to one another.
6. The area of a parallelogram is found by multiplying
the basis given in rods, feet, inches, &c, by the height,
expressed in units of the same kind.
7. The area of a rhombus or lozenge is found like
that of a parallelogram.
8. The areas of parallelograms are to each other, as
the products obtained by multiplying the bases of the
parallelograms by their heights.
9. Rectangles, or parallelograms which have equal
bases, are to each other as their heights.
10. Rectangles, or parallelograms which have equal
heights, are to each other as their bases.
11. If two triangles stand on the same basis, and have
equal heights, their areas are equal to one another.
GEOMETRY. jqj
12. Every triangle is half of a parallelogram upon equal
basis and of the same height.
13. The area of a triangle is half of the area of a par-
allelogram upon equal basis and of the same height ; and,
therefore, the area of a triangle is found by multiplying
the length of its basis by its height, and dividing the
product by 2.
14. The areas of triangles upon the same basis, and
between the same parallels, are equal.
15. The areas of triangles are to each other, as the
products of their bases by their heights.
16. The areas of triangles upon equal bases are to
each other, as the heights of the triangles.
17. The areas of triangles, which have equal heights,
are to each other, as their bases.
18. The area of a trapezoid is found by multiplying
half the sum of the two parallel sides, by their distance.
19. The area of any rectilinear figure, terminated by
any number of sides, is found by dividing that figure,
either by diagonals or by any other means, into triangles,,
and then adding the areas of these triangles together.
20. If, upon each of the three sides of a right-angled
triangle, a square is constructed, the square upon the
hypothenuse equals, in area, the two squares constructed
upon the two sides, which include the right angle.
21. The bases of similar triangles are to each other,
as the heights of the triangles.
22. The areas of similar triangles are to each other,
as the areas of the squares upon the corresponding sides.
23. The areas of similar polygons are to each other, as
the squares constructed upon the corresponding sides.
9*
SECTION IV.
OF THE PROPERTIES OF THE CIRCLE *
QUERY I.
In how many points can a straight line, CD, meet the
circumference of a circle 1
A. In two points,
C :
M, N, only. For,
letting fall, from the
centre of the cir-
cle, the perpendicu-
lar OP upon the
straight line CD,
there is but one point in the line CD, on each side of
the perpendicular, such, that a line, drawn from it to
the point O of the perpendicular, has the length of the
radius ON. (Page 46, 6thly.)
QUERY IT.
In what cases do the
circumferences of two
circles cut each other ?
A. When the distance,
OP, between their cen-
tres, O and P, is less
than the sum of their radii, OM, PM.
* Before entering on this section, the teacher ought to recapitu-
late with his pupils the definitions of a circle, of an arc, of a chord,
a segment, &c.
GEOMETRY.
103
QUERY III
When do two circles
touch each other exteriorly ?
A. Wlien the distance,
OP, between their centres,
O and P, is equal to the
sum of their radii OM,
PM.
QUERY IV.
WJicn do two circles touch each
other interiorly 1
A. When the distance, OP, be-
tween their centres, O and P, is
equal to the difference between
their radii, OM and PM.
QUERY V.
When are the circumferences of
two circles parallel to each other 7
A. When they are concentric,
that is, when they are described
from the same point,. C, as the
centre.
QUERY VI,
If from the centre, C, of a cir-
cle, a perpendicular, CD, is let
fall upon a chord, AB, in that
circle, what relation do the two
parts, AD, BD, into which the
chord AB is divided, bear to each
other 1
104 * GEOMETRY.
A. Tlie two parts AD, BD, are equal to each other ;
that is, the chord AB is bisected in the point D.
Q. How can you prove this 1
A. By drawing the two radii AC, BC, the right-angled
triangle ACD has the hypothenuse AC, and the side CD,
equal to the hypothenuse BC, and the side CD, in the
right-angled triangle BCD, each to each ; therefore these
two triangles are equal (page 47) ; and the side AD, in
the triangle ACD, is equal to the side BD in the equal
triangle BCD.
Q. Wliat other truths can you infer from the one you
have just established ?
A. 1. A straight line, drawn from the centre of a circle
to the middle of a chord, is perpendicular to that chord.
2. A perpendicular, drawn through the middle of a
chord, passes, tvhen sufficiently far extended, through the
centre of the circle.
3. Two perpendiculars, each drawn through the middle
of a chord in the same circle, intersect each other at the
centre ; for each of them must go through the centre.
4. The two angles, x and y, which the radii AC, BC,
drawn to the extremities of the chord AB, make with the
perpendicular CD, are equal to one another ; for they are
opposite to the equal sides AD, BD, in the equal triangles
ADC, BDC.
QUERY VH.
If the two chords, AD, AB,
are equal to each other, what re
mark can you make with regard
to the arcs AD, AB, subtended*
by these chords ?
A. The two arcs, AB, AD,
* The arcs AD, AB, standing on the chords AB, AD, are said to
be subtended by these chords.
GEOMETRY. 105
subtended by the equal chords, AB, AD, are equal to
one another.
Q. Why?
A. This follows from the perfect uniformity with
which a circle is constructed. For, if the chord AB is
placed upon its equal, the chord AD, the arcs, AB and
AD, must coincide with each other ; because every point
in both these arcs is at the same distance from the centre,
C, of the circle.
Remark. It is to be observed that each chord subtends two
arcs, one of which is smaller, and the other greater than the semi-
circumference, both together completing the whole circumference.
In speaking of an arc, subtended by a chord, we always mean that
which is smaller than the semi-circumference.
Q. TVIiat other truths can you infer from the one you
have just proved 7
A. 1. That equal arcs stand on equal chords ; for, by
placing one of the equal arcs AB, AD, upon the other,
the beginning and end of the two chords AB, AD, and
therefore the whole chords themselves, coincide with each
other.
2. The greater arc stands on the greater chord, and
the greater- chord subtends the greater arc. The chord
AF, for instance, is greater than the chord AD ; because
the arc AF, belonging to the greater chord AF, is greater
than the arc AD, belonging to the smaller chord AD.
3. Among all the chords, AD, AF, AM, AN, AB,
fyc, tohich can be drawn in a circle, the diameter AM is
the greatest ; because the greatest arc, the semi-circum-
ference, stands on it.
Remark. All that has been said of chords and arcs in the same
circle, holds true also of chords and arcs in equal circles.
106 GEOMETRY.
QUERY VIII.
What relation do you discover
between the angles ACB, BCD, at
the centre, C, of a circle, and the
arcs AB, BD, intercepted between
their legs ?
A. The angles ACB, BCD, at
the centre, are to each other in the
same ratio as the arcs AB, BD, of the circumference.
Q. How can you show this ?
A. I divide the whole of the arc AD successively into
smaller and smaller parts, until one of the points of
division shall have fallen upon B. Then, it is evident
that by drawing to the points of division the radii Cm,
Cn, Co, &c, the angles ACB and BCD are divided into
as many equal parts as the arcs AB, BD (for the sectors
ACm, mCn, raCB, &c, will all coincide with each other,
when they are placed upon one another); and therefore
the same ratio which exists between the arcs AB, BD,
exists also between the angles ACB, BCD. In our figure,
we have the ratio of the arc AB to the arc BD as 3 to 6 ;
and the same ratio (as 3 to 6) exists also between the
angles ACB and BCD at the centre of the circle ; that
is, the arc BD is as many times greater than the arc AB,
as the angle BCD is greater than the angle ACB (Def.
of Geom. Proportions).
JVJiat inference can you draw from the truth you have
just advanced?
Ans 1. If the arcs AB, BD, are equal to one another,
the angles ACB, BCD, at the centre, are also equal to
one another ; for they are in the same ratio as the arcs
AB, BD (namely, then, in the ratio of equality).
GEOMETRY. 107
2. If the angles ACS, BCD, at the centre, are equal
to one another, the arcs AB, BD, are also equal to one
another ; because they are to each other in the same ratio
as the angles at the centre.
Remark 1. It has already been stated (note to page 12), that
angles are measured by arcs of circles, described with any radius
between their legs. The reason is now apparent; for the arcs
intercepted between their legs are in proportion to the angles at
the centre.
Remark 2. If the circumference of a circle is divided into 360
equal parts, called degrees ; each degree again into 60 equal parts,
called minutes; each minute again into 60 equal parts, called
seconds, &,c; it is easy to perceive, that the magnitude of an angle
does not depend upon the length of the arc intercepted between
its legs ; but merely upon the number of degrees, minutes, sec-
onds, &c, it measures of the circumference of the circle of which
it is a part.
Thus, if the angle BAC meas-
ures 3 degrees by the arc MN,
it measures the same number of
degrees by the arc OP, the same
number of degrees by the arc CB,
&c, although the degrees them-
selves vary in size.
Remark 3. As the sum of all the angles around the same point
is equal to 4 right angles (page 24), the sum of all the angles
around the centre of a circle is also equal to 4 right angles ; there-
fore the circumference of a circle is the measure of 4 right angles ;
the semi-circumference that of 2 right angles, and the arc of a
quadrant, that of one right angle. If the circumference of a circle
is divided into 360 degrees, 90 of them are the measure of 1 right
angle, 180 that of 2 right angles, and 360 that of 4 right angles.
108
GEOMETRY.
QUERY DC.
If a straight line is drawn per-
pendicular to the extremity A, of
the radius AC, in how many
points ivill that line meet the cir-
cumference of the circle ?
A. In one only {namely, the
point A) ; and therefore the line
DE is a tangent to the circle.
(See Def. page 15.)
Q. But why can the line ED
have no other point common with the circumference 1
A. Because the perpendicular AC is the shortest line
which can be drawn from the point C, the centre of the
circle, to the straight line ED (page 44) ; therefore every
other line, CG, CF, CD, drawn from the centre, C, to the
straight line ED, will be greater than the radius AC ;
consequently every point in the line ED, except the point
A itself, is without the circle.
Q. And what other truths can you infer from the one
last established ?
A. 1. A radius or diameter, drawn to the point of
tangent, is perpendicular to the tangent.
2. A line drawn through the point of tangent perpen-
dicular to the tangent, passes, when sufficiently far ex-
tended, through the centre of the circle.
GEOMETRY. 100
What relation does the angle ACB, measured by the
arc AB bear to the angle y, formed by the tangent BD
and the chord AB, which subtends the arc AB ?
A. The angle ACB, at the centre of the circle, is
twice as great as the angle y, formed by the tangent BD,
and the chord AB.
Q. How can you prove this ?
A. From the centre of the circle, let fall the perpen-
dicular CI upon the chord AB, and extend it until it
meets the circumference in K. Then the angles w and
z, and consequently the arcs AK, KB, are equal to one
another (page 104, 4thly). We have further the' triangle
BIC right-angled, and therefore the two angles x and z,
together, equal to a right angle (page 34, Tthly) ; and
because the tangent DB is perpendicular to the radius
CB, the angles x and y are together also equal to a right
angle ; therefore the angle z is equal to the angle y
(Truth III, page 21) : and as the angle z is half of the
angle ACB, the angle y (its equal) is also half of the
angle ACB. *&&*
Q. And what remark can you make with regard to the
arc BK1
A. That the arc BK, which measures the angle z, may
be taken also for the measure of the angle y (its equal);
10
HO GEOMETRY.
and as the arc BK is half of the arc AB, the angle y,
made by the tangent BD and the chord AB, may likewise
be measured by half the arc AB.
Q. What do you mean by saying that half the arc AB
measures the angle y ?
A. That if the arc AB is given in degrees, minutes,
seconds, &c, the angle y measures half as many degrees,
minutes, seconds, &c, as the arc AB. Thus if the arc
AB were 12 degrees and 30 minutes, the angle y would
measure 6 degrees and 15 minutes.
*
>> hat relation does the angle w, formed by the two
radii CA, CF, bear to the angle y, formed by the Uco
chords AB, FB, if the legs of both these angles stand on
the extremities of the same arc AF ?
A. The angle w, formed by the two radii CA, CF, is
twice as great as the angle y, formed by the two chords
AB, FB.
Q. How can you prove it ?
A. Drawing in the point B a tangent, DE, to the
circle, the angles x, y, z, being together equal to two
right angles (duery 4, Sect. I.), will have for their
measure half the circumference of the circle (page 107,
remark 3d). Now, the angle x, formed by the tangent
DB and the chord AB, is measured by half the arc AB,
as has been proved in the last query ; and for the same
GEOMETRY.
Ill
reason is the angle z measured by half the arc BF ; and
therefore the remaining angle y is measured by half the
arc AF ; because half of the arc AF makes with half of
the arcs AB and BF, half the circumference. But the
angle, w, at the centre is measured by the whole arc AF;
therefore the angle w is twice as great as the angle y.
Q. What important truths can you infer from the one
you have just learned?
A. That every angle made by two chords at the cir-
cumference of a circle, measures half as many degrees,
minutes, seconds, fyc, as the arc on the extremity of which
these chords stand.
2. The angles x, y, z, at the
circumference, having their legs
standing on the extremities of the
same arc, ACB, are all equal to
one another ; because each of
them is measured by half the arc
ACB.*
QUERY XII.
If two chords, AH, CD, in the
same circle, are parallel to each
other, what relation do the arcs,
AC, BD, intercepted by them, on
both sides of the circumference,
bear to each other ?
A. The arcs AC, BD, are equal to each other.
Q. How can you prove it ?
A. Joining AD, the alternate angles x andy are equal
to one another (Query 10, Sect. I.) ;. therefore the arcs
* The arc ACB is designated by three letters, in order to dis-
tinguish it from the upper arc AB.
112 GEOMETRY.
AC and BD, measured by the halves of these angles, are
also equal to one another.
QUERY XIII.
If from the same point, A,
without a circle, you draw a
tangent, AB, to the circle, and,
at the same time, another line,
AC, cutting the circle, what
relation exists between the tan-
gent AB, and the line AC, which cuts the circle ?
A. The tangent AB is a mean proportional (Theory
©f Prop., page 66), between the whole line AC and the
part AD, which is without the circle.
Q. How can you prove it 1
A. By joining BD and BC, the triangle ABD is simi-
lar to the whole triangle ABC ; because the angle at A
is common to both triangles, and the angle y in the tri-
angle ABD, is equal to the angle x, in the triangle ABC
(both angles being measured by half the arc BD*) ;
therefore we have the proportion
AD : AB = AB : AC,
where the tangent AB is a mean proportional between
the whole line AC and the part AD without the circle.
(The sides AD and AB, in the triangle ABD, are
opposite to the angles y and z in the same triangle, and
the sides AB and AC, ih the triangle ABC, are opposite
to the angles x and CBA, which are respectively equal to
the angles y and z.)
* The angle x is formed at the circumference by the two chords
BC and DC, whose extremities stand on the arc BD (Query 11,
Sect. IV.) ; and the angle y is formed by the tangent BA, and tho
chord BD, which subtends the arc BD. (Query 10, Sect IV.)
GEOMETRY.
113
QUERY XIV.
If two chords, AD, BC, cut each
other within the circle, what relation
exists between the parts AE, ED,
BE, EC, into which they mutually
divide each other ?
A. The two parts AE, ED, are
in the inverse ratio of the two parts
BE, EC ; that is, we shall have the proportion
EC : EA — ED : EB,*
Q. How can you prove it ?
A. Joining AC and BD, the angle iv is equal to the
angle z ; because each of these two angles, w, z, measures
half as many degrees as the arc AB ; for the same reason
is the angle x equal to the angle y ; because each of these
angles measures half as many degrees as the arc CD
(Query 11, Sect. IV.); and the angles AEC, BED, are
also equal to each other, being opposite angles at the ver-
tex (Q,uery 5, Sect. I.); therefore the three angles of the
triangle AEC are equal to the three angles of the triangle
BED, each to each ; consequently these two triangles
are similar to one another ; and the sides opposite to the
equal angles, in both triangles, are in the proportion
EC : EA — ED : EB
(EC and EA are opposite to the angles x and z in the
triangle AEC ; and ED and EB are opposite to the an-
* The ratio ED to EB, is called inverse or inverted, because the
two parts ED, EB, are not in direct proportion to the two parts
EC, EA; that is, the part EC of the chord BC, is to the part EA
of the chord AD, not as the other part EB of the first chord BC, is
to the other part ED of the chord AD, but as the part ED of the
second chord is to the part EB of the first one.
10*
114
GEOMETRY.
gles y and w, which are equal to the angles x and z, each
to each).
QUERY XV.
If from a point, A, without a circle,
two lines, AB, AC, arc drawn, cut-
ting the circle ; what relation exists
between the lines AB, AC, and their
parts AD, AE, without the circle?
A. The whole lines, AB, AC, are
to each other in the inverse ratio of
their parts, AD, AE, without the cir-
cle ; that is, we have the proportion,
AB : AC = AE : AD (see the note to page 113).
Q. Why is this so ?
A. If you join BE and DC, the two triangles ABE
and ADC are similar to each other ; because two angles
in the one are equal to two angles in the other, each to
each (page 73, 1st) ; the angle, at A, namely, is common
to both, and the angles at B and C are equal ; because
they have the same measure (half the arc DE) ; and as
in similar triangles the sides opposite to .the equal angles
are in proportion, we have
AB : AE = AC : AD,*
or, by changing the order of the mean terms (principle 2d
of proportion),
AB : AC r= AE : AD 3
as above.
* The teacher will do well to show his pupils again, that the
sides AB and AE are the corresponding sides to AC and AD ;
because they are opposite to the equal angles in the triangles.
GEOMETRY. 115
Remark 1. A regular polygon is a rectilinear figure which has
all its angles and all its sides equal to one another.
Remark 2. A rectilinear figure is said to be inscribed in a
circle, when the vertices of all the angles of that figure are at the
circumference of the circle.
Remark 3. A rectilinear figure is said to be circumscribed
about a circle, when every side of that figure is a tangent to the
circle.
QUERY XYI.
If you divide the circumference
of a circle into any number of
equal parts, for instance into 6
parts, and then join the points
of division by the chords AB,
BC, CD, BE, EF, FA, what
remark can you make respecting
the rectilinear figure, ABCDEF, which will be inscribed
in the circle ?
A. The figure thus inscribed in the circle is a regular
polygon.
Q. How can you prove this ?
A. The circumference of the circle being divided into
equal parts, it follows that the arcs AB, BC, CD, &c,
and consequently also the chords AB, BC, CD, &c, which
form the sides of the inscribed figure, are equal to one
another (page 105, 1st) ; and as each of the angles ABC,
BCD, CDE, &,c, has its legs standing on the whole cir-
cumference less two of the equal arcs, into which the cir-
cumference is divided, they all measure the same number
of degrees, and consequently the angles of the inscribed
figure are also equal to one another ;* therefore the in-
scribed figure ABCDEF is a regular polygon.
* The angle ABC, for instance, has its legs standing on the whole
circumference less the two arcs AB, BC ; and the angle BCD has
its legs standing on the whole circumference less the two equa!
arcs BC, CD, &c.
116 GEOMETRY.
Q. If in this manner you divide the circumference of a
circle into 3, 4, 5, 6, fyc, equal parts, what will be the
magnitude of each of the arcs AB, BC, CD, fyc. ?
A. Each of the arcs AB, BC, CD, fyc, will then be
"&> 4 > i" > i> 4* c -> °f ^ le whole circumference, that is, •£, £,
-i, £, Sfc, o/"360 degrees, according as the circumference
has been divided into 3, 4, 5, 6, <^c, parts.
Q. And what do you observe with regard to the angles
x, y, z, &fc, at the centre of the circle, which the radii
OA, OB, OC, Sfc, drawn to the points of division A,
B, C, D, Sfc, make with each other?
A. That these angles, x, y, z, Sfc, are all equal to one
another; because they are measured by the equal arcs
AB, BC, CD, &-c. They will therefore measure £, £, -£,
foe, of 360 degrees, according as the circumference of
the circle is divided into 3, 4, 5, &c, equal parts.
QUERY XVII.
Can you find the relation which one of the sides of a
regular inscribed hexagon bears to the radius of that
circle? (See the figure belonging to the last Query.)
A. The side of a regular hexagon inscribed in a circle,
is equal to the radius of that circle.
Q. Why?
A. Because each of the triangles ABO, BCO, CDO,
&c, is in the first place isosceles, two of its sides being
radii of the same circle ; and as each of the angles x,
y, z, &c, at the centre of the circle, measures £ of 360,
that is, 60 degrees (last Query), it follows that the two
angles at the basis of each of the isosceles triangles ABO,
BCO, CDO, &,c. (for instance, the two angles w and u,
at the basis of the isosceles triangle ABO), measure to-
gether 120 degrees ; because the sum of the three angles
in every triangle is equal to two right angles, or 180 de-
GEOMETRY. J 17
grees, and 60 from 180 leave 120 degrees. Now, as
the two angles at the basis of every isosceles triangle are
equal to each other (Query 3, Sect, II.) ; each of the
two angles at the basis of one of the isosceles triangles
ABO, BCO, DCO, &c, will measure half of 120, that
is, 60 degrees. But, each of the angles at the centre
measuring also 60 degrees, the three angles in each of
the triangles ABO, BCO, CDO, &c, are equal to one
another ; and therefore these triangles are not only isos-
celes, but also equilateral; consequently each of the sides
AB, BC, CD, &c, of the hexagon is equal to the radius
of the circle.
QUERY XVIH.
If, in a regular inscribed
polygon, you draw from the cen-
tre of the circle the radii Oi,
Ok, Ol, Om, fyc, perpendicular
to the chords AB,BC, CD, 8?c;
and at the extremities of these
radii, the tangents MN, NP,
PQ, fyc; what do you observe
with regard to the figure MNPQRS, circumscribed about
the circle ?
A. The figure MNPQRS, circumscribed about the
circle, is a regular polygon, of the same number of sides
as the inscribed polygon, ABCDEF.
Q. How can you prove this ?
A. The chords AB, BC, CD, &c, are perpendicular
to the same radii, to which the tangents MN, NP, PQ,,
&c.?%.re perpendicular ; consequently the chords AB, BC,
CD, &c, are parallel to the tangents MN, NP, PQ, &c.
(for two straight lines, which are both perpendicular to a
third line are parallel to each other ; Query 7, Sect. I.) ;
118 GEOMETRY.
and therefore the triangles ABO, BCO, CDO, &c, are
all similar to the triangles MNO, NPO, PQO, &c, from
which they may be considered as cut off, by the lines
AB, BC, CD, &c. being drawn parallel to the sides MN,
NP, Pa, &c. (auery 16, Sect. II.) Now, as the trian-
gles ABO, BCO, CDO, &c, are all equal to one another,
the triangles MNO, NPO, PQO, &c, are all equal to
one another. And therefore the circumscribed figure
MNPQJIS is a regular polygon, similar to the one in-
scribed in the circle.
QUERY XIX.
It has been proved (Query 16, Jg , g
Sect. IV.), that a regular poly- -''/z\ i /^\\
gon, of any number of sides, y ''-\ !/--•'' v
may be inscribed in a circle, by &r~fi} L A~ y, D
ui 'riding the circumference of the \\ / \ //
circle into as many equal parts H4 \£''
as the polygon shall have sides,
and then joining the points of division by straight lines :
can you now prove the reverse, that is, that around every
regular polygon, a circle can be drawn in such a manner,
that all the vertices of the polygon shall be at the circum-
ference ?
A. Yes. For I need only bisect two adjacent sides of
a regular polygon ; for instance, the two sides, AB, BC,
of the regular polygon ABCDEF ; and in the points of
bisection, erect the two perpendiculars g-O, kO, which
will necessarily cut each other in a point, O. Then it is
evident, that by drawing the lines OB, OC, OA, these
three lines are equal to each other ; for the line OB is
equal to OC, because the two points B and C are at an
equal distance from the perpendicular gO (page 45,
5thly) ; and for the same reason is OB also equal to OA j
GEOMETRY. UQ
because the points B and A are at an equal distance from
the perpendicular M) Thus we have in the two trian-
gles ABO, BCO, the three sides in the one, equal to the
three sides in the other ; therefore these two triangles are
both isosceles and equal to each other.
Q. But of what use is your proving that the triangle
ABO is equal to the triangle BCO?
A. It shows that each of the angles in the polygon is
bisected by one of the lines OA, OB, OC. For, in the
first place, we have in the two equal triangles BCO,
ABO, the angle o equal to the angle z ; therefore the an-
gle ABC is bisected ; and the angle o is further equal to
the angle y, and the angle z to the angle v ; therefore
the angles BCD, and FAB, are also bisected. And
now I can show that, by drawing from the point O the
lines OF, OE, OD, to the remaining vertices F, E, D,
the whole polygon is divided into equal isosceles tri-
angles. Taking, in the first place, the two triangles,
AFO and ABO, they have two sides, OA, FA, in the
one, equal to two sides, OA, AB, in the other, each to
each ; and as the angle FAB is bisected by the line OA,
the two angles v and w are also equal ; consequently, the
two triangles AFO, ABO, are equal to each other, and
the angle u is equal to the angle w (Query 3, Sect. II.).
In precisely the same manner it may be proved that the
triangles FEO, EDO, are isosceles, and equal to the tri-
angle AFO. And as the whole polygon ABCDEF is
thus divided into equal isosceles triangles, the lines OA,
OB, OC, OD, OE, OF, are all equal to one another ; and
therefore, by describing from the point O, as a centre,
with a radius OA, a circle around the polygon ABCDEF,
each of the vertices A, B, C, D, E, F, will be in the cir-
cumference of the circle.
120
GEOMETRY.
Q. What other important consequence follows from the
principle you have just proved ?
A. That in every regular polygon a circle may be
inscribed in such a manner, that every side of the polygon
is a tangent to the circle. For if, in the regular polygon
ABCDEF, you describe with a radius Og the circumfer-
ence of a circle, that circumference will touch the middle
of the sides AB, BC, CD, DE, EF, FA, of the polygon
ABCDEF ; because the lines Ok, Og, Ol, &c. are all
equal to one another, and will therefore be radii of the
inscribed circle ; and the sides AB, BC, CD, &c, being
perpendicular to the radii Ok, Og, Ol, &c, will all be
tangents to that circle. (Page 108.)
QUERY XX.
What relation do you ob'serve to exist between two regu-
lar polygons, abcdef, ABCDEF, of the same number of
sides ?
A. They are similar to one another.
Q. How can you prove it '?
A. By describing a circle around each of the regular
polygons abcdef ABCDEF, and drawing the radii Oa,
Ob, Oc, Od, Oe, Of, OA, OB, OC, OD, OE, OF, each
GEOMETRY. 12!
of these polygons is divided into as many equal triangles
as there are sides in the polygon ; and as all the angles,
x, y, z, &lc, formed at the centre of a regular polygon,
are equal to one another, I can place the centre, O, of
the polygon abcdef, upon the centre, O, of the polygon
ABCDEF, in such a manner, that the angles at the cen-
tre shall all coincide with each other ; namely, so that
the radius 0« shall fall upon the radius OA, Ob upon
OB, Oc upon OC, &c. Then, it is evident that the
sides, ab, be, cd, de, &c, of the smaller polygon, abcdef,
are parallel to the sides, AB, BC, CD, DE, &,c, of the
greater polygon, ABCDEF ; for the points a, b, c, d, e,f
and A, B, C, D, E, F, are in the circumferences of con-
centric circles (Query 5, Sect. IV.) ; therefore the trian-
gles Oab, Obc, Ocd, &c, in the smaller polygon, are all
similar to the triangles OAB, OBC, OCD, &,c, in the
greater polygon (Query 16, Sect. II.) ; consequently the
whole polygon abcdef is similar to the whole polygon
ABCDEF.
Q. What other truths can you infer from the one you
have just learned?
A. 1. The sums of all the sides of two regular polygons
of the same number of sides are to each other in the same
ratio as the radii of the inscribed or circumscribed circles.
For in the two triangles ABO and abo, for instance, we
have the proportion
AB : ab ==. AO : ao ; that is,
the side AB is as many times greater than the side ab,
as the radius AO is greater than the radius ao ; and there-
fore 6 or any other number of times the side AB, is as
many times greater than the same number of times the
side ab, as the radius OA is greater than the radius oa;
that is, the sum of all the sides of the regular polygon
ABCDEF, is as many times greater than the sum of all
11
122 GEOMETRY.
the sides of the regular polygon abcdef, as the radius OA
of the circle, circumscribed about the regular polygon
ABCDEF, is greater than the radius, oa, of the circle cir-
cumscribed about the regular polygon abcdef. In the
same manner I can prove that the sum of all the sides
of the regular polygon ABCDEF, is as many times greater
than the sum of all the sides of the regular polygon abcdef,
as the radius of the circle, inscribed in the regular poly-
gon ABCDEF, is greater than the radius of the circle
inscribed in the regular polygon abcdef.
Remark. The sum of all the sides of a geometrical figure, that
is, a line as long as all its sides together, is called the perimeter of
that figure. The above proportion may therefore be expressed in
shorter terms ; namely, the perimeters of two regular polygons of
the same number of sides, are to each other in the proportion of the
radii of the inscribed or circumscribed circles.
2. The areas of two regular polygons of the same
number of sides, are in the same ratio as the squares
constructed upon the radii of the inscribed or circum-
scribed circles. Thus the area of the regular polygon
ABCDEF, is as many times greater than the area of the
regular polygon abcdef, as the area of the square upon
the radius OA is greater than the area of the square upon
the radius oa. For the areas of the similar triangles
ABO, abo, are to each other as the squares upon the
corresponding sides (the radii OA, oa) ; therefore, any
number of times (in our figure 6 timos) the areas of
these triangles, that is, the areas of the regular polygons
ABCDEF, abcdef themselves, are to each other in the
same ratio. In the same manner I can prove that the
area of the polygon ABCDEF is as many times greater
than the area of the polygon abcdef as the square upon
GEOMETRY. J 23
the circle inscribed in the regular polygon ABCDEF, is
greater than the square upon the radius of the circle in-
scribed in the regular polygon abcdef.
QUERY XXI.
From what you have learn-
ed of the properties of regular
polygons, can you give a rule
for finding the area of a reg-
ular polygon 1
A. Yes. Multiply the sum
of all the sides (the perimeter)
by the radius of the inscribed
circle ; the product, divided by 2, will be the area of the
regular polygon.
Q. Why?
A. Because every, regular polygon, the polygon
ABCDEF, for instance, can be divided into as many
equal triangles, as there are sides in the polygon ; and
the area of each of these triangles is found by multiply-
ing the basis, that is, one of the sides of the polygon
by the height (which, in every one of these triangles,
is equal to the radius, om, of the inscribed circle), and
dividing the product by 2 ; therefore the area of the
whole polygon ABCDEF may at once be found by multi-
plying the sum of all the sides by the radius of the in-
scribed circle, and dividing the product by 2.*
* Instead of multiplying the perimeter by the whole radius, and
then dividing the product by 2, you may at once multiply the
perimeter by half the radius, or the radius by half the perimeter.
124 GEOMETRY.
If you bisect each of the arcs AB, BC, CD, fyc,
subtended by the sides AB, BC, CD, fyc, of a regular
polygon inscribed in a circle ; and then to the points of
division, m, n, o, p, q, r, draw the lines Am, mB, Bn, nC,
Co, Sfc; what do you observe with regard to the regular
polygon, AmBnCoDpEqFr, thus inscribed in the circle ?
A. The regular polygon AmBnCoDp, Sfc, has twice
as many sides as the regular polygon ABCDEF; for
the circumference of the circle is now divided into twice
as many equal parts as before. Thus if the regular
polygon ABCDEF has 6 sides, the regular polygon
AmBnCoDp, &>c, has 12 sides; and by bisecting again
the arcs Am, mB, Bn, &c, I can inscribe a regular poly-
gon of 24 sides, and so on, by continuing to bisect the
arcs, a regular polygon of 48, 96, 192, &c, sides.
Q. And what do you observe with regard to the arcs
which are subtended by the sides of the polygons,
ABCDEF and AmBwCoDpE^Fr, inscribed in the circle 1
A. The arcs, AB, BC, CD, &c, subtended by the
sides of the regular polygon ABCDEF, first inscribed in
GEOMETRY. * 125
the circle, stand farther off the sides AB, BC, CD, &c,
than the arcs Am, mB, Bra, 6lc, from the sides Am, mB,
Bw, &.c, of the regular polygon of twice the number of
sides ; consequently, if the arcs AB, BC, CD, &c, were
drawn out into straight lines, they would differ more
from the sides AB, BC, CD, &,c., of the regular polygon
ABCDEF, first inscribed in the circle, than the arcs Am,
mB, Bn, nC, &c, would, in this case, differ from the
sides Am, mB, Bra, &c, of the regular polygon AmBnCo,
&c, of twice the number of sides.
Q. Now, if, continuing to bisect the arcs, you inscribe
regular polygons of 24, 48, 96, 192, &c. sides, what
further remark can you make with regard to the arcs
subtended by the sides of these polygons?
A. These arcs differ less in length from the sides
which subtend them, in proportion as the polygon con-
sists of a greater number of sides ; because, by continuing
to bisect the arcs, and thereby to increase the number of
sides of the inscribed polygons, the arcs subtended by
these sides grow nearer and nearer to the sides them-
selves, and finally the difference between them will be-
come imperceptible.
Q. And what conclusion can you now draw respecting
the whole circumference of a circle ?
A. Thai the circumference of a circle differs very little
from the sum of all the sides of a regular inscribed poly-
gon of a great number of sides ; therefore, if the number
of sides of the inscribed polygon is very great (several
thousand for instance), the polygon will differ so little
from the circle itself, that, without perceptible error, the
one may be taken for the other.
11*
126 GEOMETRY.
QUERY XXIII.
It has been shown in the last query, that a circle may
be considered as a regular polygon of a very great num-
ber of sides ; what inferences can you now draw with
regard to the circumferences and areas of circles ?
A. 1. The circumferences of two circles are in propor-
tion to the radii of these circles ; that is, a straight line
as long as the circumference ofthefrst circle, is as many
times greater than a straight line as long as the circum-
ference of the second circle, as the radius AB ofthefrst
circle, is greater than the radius ab of the second circle.
For if, in each of the two circles, a regular polygon of a
very great number of sides is inscribed, the sums of all
the sides of the two polygons are to each other in pro-
portion to the radii, AB, ab, of the circumscribed circles
(page 122, 1st) ; and as the difference between the cir-
cumference of a circle and the sum of all the sides of an
inscribed polygon of a great number of sides, is imper-
ceptible (last Query), we may say that the circumferences
themselves are in the same ratio.*
* The teacher may give an ocular demonstration of this princi-
ple, by taking two circles, cut out of pasteboard or wood ; and
measuring their circumferences by passing a string around them.
The measure of the one will be as many times greater than the
measure of the other, as the radius of the first circle is greater than
the radius of the second circle.
GEOMETRY. f$ft
2. The areas of two circles are in proportion to the
squares constructed upon their radii ; that is, the area of
the greater circle is as many times greater than the area
of the smaller circle, as the area of the square upon the
radius of the greater circle, is greater than the area of
the square constructed upon the radius of the smaller circle.
For if in each of these circles a regular polygon of a great
number of sides is inscribed, the difference between the
areas of the polygons and the areas of the circles them-
selves will be imperceptible ; and because the areas of
two regular polygons of the same number of sides are in
the same ratio as the areas of the squares upon the radii
of the circles in which they are inscribed (page 122,
2dly), the areas of the circles themselves are in the ratio
of the squares upon their radii.
3. The area of a circle is found by multiplying the
circumference of the circle, given in rods, feet, inches, fyc,
by half the radius, given in units of the same kind. Be-
cause a circle differs so little from a regular inscribed
polygon of a great number of sides, that the area of the
polygon may, without perceptible error, be taken for the
area of the circle. Now the area of a regular polygon
inscribed in a circle, is found by multiplying the sum of
all the sides by the radius of the inscribed circle, and
dividing the product by 2 (page 123) ; therefore the area
of the circle itself is found by multiplying the circumfer-
ence (instead of the sums of all the sides of the inscribed'
polygon) by the radius, and dividing the product by 2.
For it lias been shown in the last duery, that the sides
of a regular inscribed polygon grow nearer and nearer
the circumference of the circumscribed circle, in propor-
tion as these sides increase in number ; consequently, the
circumference of a circle inscribed in a regular polygon
of a great number of sides, will also grow nearer and
128 GEOMETRY.
nearer the circumference of the circumscribed circle ;
until finally the two circumferences will differ so little
from each other, that the radius of the one may, without
perceptible error, be taken for the radius of the other.
Remark. Finding the area of a circle is sometimes called
squaring the circle. The problem to construct a rectilinear fig-
ure, for instance a rectangle, whose area shall exactly equal the
area of a given circle, is that which is meant by finding the quad-
rature of the circle. For the area of any geometrical figure, termi-
nated by straight lines only, can easily be found by the rule given
in Query 5, Sect. Ill; or, in other words, we can always construct
a square, which shall measure exactly as many square rods, feet,
inches, &c, as a rectilinear figure of any number of sides.
Now it is easy to show, that there is nothing ab9urd in the idea
of constructing a rectilinear figure, for instance a rectangle, whose
area shall be equal to the area of a given circle. For let us take a
semicircle ABM, and let us for a mo-
ment imagine the diameter AB to
move parallel to itself between the
two perpendiculars AI, BK k It is
evident that when the diameter AB is
very near it* original position, for in-
stance in CD, the area of the rectangle ABCD is smaller than the
area of the semicircle ABM ; but the diameter continuing to move
parallel to itself in the direction from A to I, there will be a point
in the line AI, where the area of the rectangle ABIK is greater
than the area of the semicircle ABM. Now as there is a point in
the line AI, below the point I, in which the area of the rectangle
ABCD is smaller than the area of the semicircle ABM, and as the
diameter, by continuing to move in the same direction, makes in
different points C, E, G, &c, of that same line, the rectangles
ABCD, ABEF, ABGH, &c, whose areas become greater and
greater, until finally they become greater than the area of the
semicircle itself; there mu3t evidently be a point in the line AI,
in which a line drawn parallel to the diameter AB, makes with it
and the perpendiculars AI, BK, a rectangle, which, in area, is
equal to the semicircle ABM ; and as there is a rectangle which,
in area, is equal to the semicircle ABM, by doubling it, we shall
have a rectangle which, in area, is equal to the whole circle.
GEOMETRY. 129
Neither is it difficult to find the area of a circle mechanically.
For the arpa of a circle being found by multiplying the circum-
ference by the length of the radius, and dividing the product by 2
(page 127, 3dly), we need only pass a string around the circumfer-
ence of a circle, and then multiply the length of that string by the
length of the radius; the product divided by 2 will be the area of
the circle. Having thus found the comparative length of the
radius and circumference of one circle, we might determine the
circumference, and thereby the area of any other circle, when
knowing its radius. For the circumferences of two circles being
in proportion to the radii of the two circles, we should have three
terms of a geometrical proportion given ; viz. the radii of the two
circles, and the circumference of the one ; from which we might
easily find the fourth term (Theory of Proportions, page 64, 8thly),
which would be the circumference of the other circle.
But the expressions of the circumference and area of a circle,
thus obtained by measurement, are never so correct as is required
for very nice and accurate mathematical calculations ; we must
therefore resort to other means, such as geometry itself furnishes,
to calculate the ratio of the radius or diameter to the circumfer-
ence ; and herein consists the difficulty of the quadrature of the
circle. For if the ratio of the radius to the circumference is once
determined, we can easily find the circumference of any circle,
when its radius is given ; and knowing the circumference and the
radius, we can find the area of the circle.
To calculate the ratio of the diameter to the circumference y
mathematicians have compared the circumference of a circle to
the sum of all the sides of a regular inscribed polygon of a great
number of sides; for it has been shown (page 125), that the cir-
cumference of a circle differs very little from the sum of all the
sides of such a polygon.
For this purpose they took a regular inscribed hexagon, each
of the sides of which is equal to the radius of the circumscribed
circle. (Query 17, Sect. IV.) For the sake of convenience they
supposed the diameter of the circle equal to unity ; the radius and
therefore the side of a regular inscribed hexagon is then ^, and
the sum of all the sides (6 times ^) equal to 3. This is the first
approximation to the circumference of a circle.
From the side of a regular inscribed hexagon, it is easy to find
that of a regular inscribed polygon of 12 sides. Supposing, for
130 GEOMETRY.
instance, the chord CD to be the side of a regular inscribed hex-
agon, by bisecting the arc CD in B, the chords BC, BD, will be
two sides of a regular inscribed poly-
gon of 12 sides, the length of which
can easily be calculated when the
chord CD and the radius AC are once
known. For the radius AB which
bisects the arc CD, makes the angles
x and y, which are measured by the
arcs DB, BC, equal to each other ;
and therefore AE is perpendicular to
the chord DC, and bisects it in E.
(Page 104, 4thly.)
Now the radius, AC and EC (half
of CD), being known, the hypothenuse and one of the sides of the
right-angled triangle AEC are given, whence it is easy to find the
other side AE, by the rule given in the remark, page 95. Thus if
the radius is supposed to be £, the side CD of the inscribed hexa-
gon is also equal to £ ; and EC (half of CD) is \. Taking the
square of ^ from that of £, and extracting the square root of the
remainder, we obtain the length of the side AE, which, subtracted
from the radius AB, leaves the length of BE. Now we can find
the side BC in the right-angled triangle BCE, by extracting the
square root of the sum of the squares of BE and EC (see the re-
mark, page 95) ; and one of the sides of the regular inscribed polygon
of 12 sides being once determined, we need only multiply it by 12,
in order to obtain the sum of all its sides, which is the second
approximation to the circumference of the circle. In precisely the
same manner can the side, and consequently also the sum of all
the sides of a regular inscribed polygon of 24 sides be obtained,
when that of a regular inscribed polygon of 12 sides is once
known ; which is the third approximation to the circumference.
Thus we might go on finding the sum of all the sides of a regular
inscribed polygon of 48, 96, 192, &c, sides, until the inscribed
polygon should consist of several thousand sides: the sum of all the
sides would then differ so little from the circumference of the cir-
cle, that, without perceptible error, we might take the one for the
other. »>r
In this manner the approximation to the circumference of the
circle has been carried further than is ever required in the minutest
and most accurate mathematical calculations.
GEOMETRY. 13]
The beginning of this extremely tedious calculation gives the
following results :
Parts of the cir-
cumference.
Sides of the inscribed
polygon.
Sum of all the sides of th«
inscribed polygon.
6
0,5
3
12
0,258819
3,105828
24
0,130526
3,132628
48
0,065403
3,139348
96
0,032719
3,141033
192
0,016361
3,141446
It is not necessary to carry this calculation any further, since
analysis furnishes us with means to obtain the same results in a
much easier manner.
In nearly the same manner has Loudolph van Ceulen
found the ratio of the diameter to the circumference of a circle
to 32 decimals. (See his * Arithmetische en Geom. Fundamenten,
page 163. Leiden. 1616 ;' also his work « De Circulo et Adscriptis,
c. 10. Leiden. 1619.'*)
Archimedes found the ratio of the diameter to the circumfer-
ence as near 7 to 22.
Franciscus Vieta found it as 1 to 3,1415926535.
Adrianus Romanus added the following decimals
89793.
Lotjdolph van Ceulen added further
23846264338327950288.
Sharp added again
41971693993751058209749445923078,
To which Machin further added
164062862089986280348253421170679,
And lastly Lagny increased them by
821480865132823066470938446.
In a manuscript in the library at Oxford, this number is still
further extended by 29 decimals, namely,
460955051822317253594081284802.
* The first work which Loudolph van Ceulen- published on
this subject, bears the title ' Van den Circkel, daer in Gheleert
wird te vinden de naeste proportie des Cirkels-Diameter tegen
synen Omloop. Leiden, d. 20 Sept. 1596.' The work is dedicated
to Prince Moriz of Orange
13£ GEOMETRY.
So that the most accurate ratio of the diameter to the circum-
ference is at present as
I to 3,1415926535897932384626433832795028841971693993751
0582097494459230781640628620899862803482534211706
7982148086513282306647093844646095505182231725359
4081284802.
The last ratio is so near the truth, that in a circle, whose diame-
ter is one hundred million times greater than that of the sun, the
error would not amount to the one hundred millionth part of the
breadth of a hair.
In general, when the calculations need not be very minute and
accurate, 7 decimals will suffice. Thus we may consider the ratio
of the diameter to the circumference to be
as 1 to 3,1415926 ; that is,
if the diameter of a circle is 1, its circumference is 3,1415926;*
consequently if the diameter is 2, -or the radius 1, the circum-
ference will be twice 3,1415926, equal to 6,2831852. Dividing
this number by 360, we obtain the length of a degree ; dividing
the length of a degree by 60, we obtain the length of a minute ;
and that again divided by 60, gives the length of a second, and so
on. In this manner we obtain the length of
1 degree equal to 0,0174533t
1 minute " " 0,0002909
1 second " " 0,0000048
1 third " " 0,0000001
Having once determined the circumference of the circle whose
radius is 1, we can easily find the circumference of any circle,
when its radius is given ; for we need only multiply the number
6,2831852 (that is, the circumference of a circle whose radius is 1),
by the radius of the circle whose circumference is to be found;
the product will be the circumference sought. Thus if it is required
to find the circumference of a circle whose radius is 6 inches, we
need only multiply the number 6,2831852 by 6 ; the product
37,6991112 is the circumference of that circle.
If it be required to find the length of an arc of a given number
* The number 3,1415926 is sometimes represented by the Greek
letter n. Thus the circumference of a circle, whose radius is 1,
may be represented by 2*.
t The last figure in these expressions has been corrected,
GEOMETRY. 133
of degrees, minutes, seconds, &c, in a given circle, we need only
multiply
the degrees by 0,0174533
the minutes by 0,0002909
the seconds by 0,0000048, &c ;
the different products added together give the length of an arc of
the same number of degrees, minutes, seconds, &c, in the circle
whose radius is 1 ; and multiplying this product by the radius of
the given circle, we shall have the length of the arc sought. If,
for instance, it be required to find the length of an arc of 6 degrees
and 2 minutes, in a circle whose radius is 5 inches, we in the first
place multiply 0,0174533 by 6, and
0,0002909 by 2 ; the products of
these multiplications, 0,1047198 }
. _ nnn ~ i > added together,
and 0,000o818 $ b
give 0,1053016, which is the length of an arc of
6 degrees and 2 minutes in the circle whose radius is 1, and this
last product (0,1053016), multiplied by 5, gives 0,5265080, which is
the length of an arc of the same number of degrees and minutes of
a circle whose radius is 5 inches.
Now that we are able to find the circumference, or an arc of the
circumference of any circle, when knowing its radius, nothing
can be easier than to calculate the area of a circle, of a sector, a
segment, &c.
The area of a circle being found by multiplying the circumfer-
ence by half the radius, or by multiplying half the circumference
by the whole radius (page 127, 3dly), we need only take the number
3,1415926, which is half the circumference of the circle whose
radius is 1, and multiply it by the radius of the given circle ; the
product will be half the circumference of the given circle, which
multiplied again by the radius, gives us the area of it. Thus if it
is required to find the area of a circle whose radius is 5 inches, we
multiply the number 3,1415926 twicein succession by 5, that is,
we multiply it by the square of 5 ;* the product 78,5398150 is the
area sought. Hence follows the general rule :
In order to find the area of a circle, multiply the number
3,1415926 by the square of the radius.
* Multiplying a number twice in succession by 5, is the same as
multiplying that number by 25 ; which is the square of 5 ; because
5 times 5 are 25.
12 .n«*
134 GEOMETRY.
If the radius is given in rods, the answer will be square rods ; if
given in feet, the answer will be square feet, if in seconds, square
seconds, and so on. The area of a semicircle is found by dividing
the area of the whole circle by 2. In the same manner we find
the area of a quadrant by dividing the area of the whole circle
by 4, &c.
The area of a sector BCAD is found by
multiplying the length of the arc CD A by
half the radius, or we may first find what
part of the circumference the arc CDA is ;
whether a third, a fourth, a fifth, &c, and
then divide the area of the whole circle
whose radius is BC, by 3, 4, 5, &.c, according
as the arc CDA is |, \, &c, of the whole
circumference. If we are to find (he area
of the segment CDA, we must first find the
area of the sector BCDA ; then the area of the triangle ABC ;
which, subtracted from the area of the sector BCDA, will leave
the area of the segment CDA.
RECAPITULATION OF THE TRUTHS CONTAINED IN
THE FOURTH SECTION.
Q. Can you now repeat the different relations which
exist between the different parts of a circle and the
straight lines, which cut or touch the circumference ?
A. 1. A straight line can touch the circumference
only in one point.
2. When the distance between the centres of two cir-
cles is less than the sum of their radii, the two circles
cut each other.
3. When the distance between the centres of two cir-
cles is equal to the sum of their radii, the two circles
touch each other exteriorly.
4. When the distance between the centres of two
GEOMETRY. 135
circles is equal to the difference between their radii, the
two circles touch each other interiorly.
5. When two circles are concentric, that is, when they
are both described from the same point as a centre, the
circumferences of the two circles are parallel to each
other.
6. A perpendicular, let fall from the centre of a circle,
upon one of the chords in that circle, divides that chord
into two equal parts.
7. A straight line, drawn from the centre of a circle
to the middle of a chord, is perpendicular to that chord.
8. A perpendicular, drawn through the middle of a
chord, passes, when sufficiently far extended, through the
centre of the circle.
9. Two perpendiculars, each drawn through the mid-
dle of a chord in the same circle, intersect each other at
the centre.
i
10. The two angles which two radii, drawn to the
extremities of a chord, make with the perpendicular let
fall from the centre of the circle to that chord, are equal
to one another.
11. If two chords in the same circle, or in equal cir-
cles, are equal to one another, the arcs subtended by
them are also equal ; and the reverse is also true ; that
is, if the arcs are equal to one another, the chords which
subtend them are also equal.
12. The greater arc stands on the greater chord, and
the greater chord subtends the greater arc.
13. The angles at the centre of a circle are to each
other in the same ratio, as the arcs of the circumference
intercepted by their legs.
14. If two angles at the centre of a circle are equal to
one another, the arcs of the circumference, intercepted
by their legs, are also equal ; and the reverse is also true ;
136 GEOMETRY'
that is, if the two arcs intercepted by the legs of the two
angles at the centre of a circle, are equal to one another,
these angles are also equal.
15. Angles are measured by arcs of circles, described
with any radius between their legs. The circumference
is, for this purpose, divided into 360 equal parts, called
degrees; each degree into 60 equal parts, called min-
utes ; each minute, again, into 60 equal parts, called sec-
onds, &c.
16. The magnitude of an angle does not depend on
the length of the arc intercepted by its legs; but merely
on the number of degrees, minutes, seconds, &c, it
measures of the circumference.
17. The circumference of a circle is the measure of 4
right angles; the semi-circumference that of 2 right an-
gles ; and a quadrant that of 1 right angle.
1$. A straight line drawn at the extremity of the
diameter or radius, perpendicular to it, touches the cir-
cumference only in one point, and is therefore a tangent
to the circle.
19. A radius or diameter drawn to the point of tangent,
is perpendicular to the tangent.
20. A line drawn through the point of tangent, per-
pendicular to the tangent, passes, when sufficiently far
extended, through the centre of the circle.
21. The angle, formed by a tangent and a chord, is
half of the angle at the centre, which is measured by the
arc subtended by that chord ; therefore the angle, formed
by the tangent and the chord, measures half as many
degrees, minutes, seconds, &c, as the angle at the
centre.
22. The angle which two chords make at the cir-
cumference of a circle, is half of the angle made by
two radii at the centre, having its legs stand on the
GEOMETRY. 137
extremities of the same arc ; therefore every angle, made
by two chords at the circumference of a circle, measures
half as many degrees, minutes, seconds, &c, as the arc
intercepted by its legs.
23. If several angles at the circumference have their
legs stand on the extremities of the same arc these
angles are all equal to one another.
24. Parallel chords intercept equal arcs of the circum-
ference.
25. If, from a point without the circle, you draw a tan-
gent to the circle, and, at the same time, a straight line
cutting the circle, the tangent is a mean proportional be-
tween that whole line, and that part of it which is without
the circle.
26. If a chord cuts another loitldn the circle, the two
parts, into which the one is divided, are in the inverse
ratio of the two parts, into which the other is divided.
27. If, from a point without a circle, two straight lines
are drawn, cutting the circle, these lines are to each
other in the inverse ratio of their parts without the circle.
28. If the circumference of a circle is divided into
3, 4, 5, &-c, equal parts, and then the points of division
are joined by straight lines, the rectilinear figure, thus
inscribed in the circle, is a regular polygon of as many
sides, as there are parts into which the circumference is
divided.
29. If, from the centre of a regular polygon, inscribed
in a circle, radii are drawn to all the vertices at the cir-
cumference, the angles which these radii make with each
other at the centre, are all equal to one another.
30. The side of a regular hexagon inscribed in a circle,
is equal to the radius of the circle.
31. If, from the centre of a circle, radii are drawn,
bisecting the sides of a regular inscribed polygon, and
to *
138 GEOMETRY.
then, at the extremities of these radii, tangents are drawn
to the circle, these tangents form with each other a regu-
lar ci?'cwnscribed polygon of the same number of sides as
the regular inscribed polygon.
32. Around every regular polygon a circle can be
drawn in such a manner, that all the vertices of the poly-
gon shall be at the circumference of the circle.
33. Two regular polygons of the same number of sides
are similar figures.
34. The sums of all the sides of two regular polygons
of the same number of sides, are to each other in the
same ratio, as the radii of the inscribed or circumscribed
circles.
35. The areas of two regular polygons of the same
number of sides, are to each other as the areas of the
squares constructed upon the radii of the inscribed or
circumscribed circles.
36. The area of a regular polygon is found by multiply-
ing the sum of all its sides by the radius of the inscribed
circle, and dividing the product by 2 ; or we may at once
multiply half the sum of all the sides by the radius of the
inscribed circle, or half that radius by the sum of all the
sides.
37. If the arcs subtended by the sides of a regular
polygon, inscribed in a circle, are bisected, and chords
drawn from the extremities of these arcs to the points of
division, the new figure thus inscribed in the circle, is a
regular polygon of twice the number of sides as the one
first inscribed.
38. The circumference of a circle differs so little from
the sum of all the sides of a regular inscribed polygon of
a great number of sides, that, without perceptible error,
the one may be taken for the other.
39. The circumferences of two circles are in proportion
GEOMETRY. * 139
to the radii of these circles ; that is, a straight line, as
long as the circumference of the first circle, is as many
times greater than a straight line as long as the circum-
ference of the second circle, as the radius of the one is
greater than the radius of the other.
40. The areas of two circles are in proportion to the
squares constructed upon their radii ; that is, the area of
the greater circle is as many times greater than the area
of the smaller circle, as the area of the square upon the
radius of the one is greater than the area of the square
upon the radius of the other.
41. The area of a circle is found by multiplying the
circumference, given in rods, feet, inches, &c, by half
the radius, given in units of the same kind.
42. The circumference of a circle, whose radius is 1,
is equal to the number 6,2831852 ; and the circumference
of any other circle is found by multiplying the number
6,2831852 by the length of the radius.
43. The length of 1 degree in a circle, whose radius
is 1, is equal to the number 0,0174533
The length of 1 minute 0,0002909
" " " 1 second 0,0000048
" " " 1 third 0,0000001
44. The length of an arc, given in degrees, minutes,
seconds, &c, is found by multiplying the degrees by
0,0174533, the minutes by 0,0002909, the seconds by
0,0000048, &c, then adding these products together,
and multiplying their sum by the radius of the circle.
45. The area of a circle, whose radius is 1, is equal to
3,1415926 square units; and the area of any other circle
is found by multiplying the number 3,1415926 by the
square of the radius.
46. The area of a semicircle is found by dividing the
area of the whole circle bv 2.
140 ♦ GEOMETRY.
47. The area of a quadrant is found by dividing the
area of the whole circle by 4.
48. The area of a sector is found by multiplying the
length of the arc by half the radius.
49. In order to find the area of a segment, we first
draw two radii to the extremities of the arc of that seg-
ment ; then calculate the area of the sector, formed by
the two radii and that arc, and subtract from it the area
of the triangle formed by the two radii and the chord of
the segment: the remainder is the area of the segment.
SECTION V,
£*>-
APPLICATION OF THE FOREGOING PRINCIPLES TO
THE SOLUTION OF GEOMETRICAL PROBLEMS-
PART I.
Problems relative to the drawing and division of
lines and angles.
Problem I. To construct an equilateral triangle upon
a given straight line, AB.
Solution. Let AB be the
given straight line.
1. From the point A, as a
centre, with the radius AB,
describe an arc of a circle, and
from the point B, with the £ r
same radius, AB, another arc cutting the first.
2. From the point of intersection, C, draw the lines
AC, BC ; the triangle ABC will be equilateral.
Demonstration. The three sides, AB, AC, BC, of the trian-
gle ABC, are all equal to each other ; because they are radii of
equal circles.
Remark. In a similar manner can an isosceles triangle be con-
structed upon a given basis.
a
142
GEOMETRY.
Problem II. From a given point in a straight line, to
erect a perpendicular upon that line.
M
B J)
N
I. Solution. Let MN be the given straight line, and
D the point in which the perpendicular is to be erected*
1. Take any distance, BD, on one side of the point D,
and make DA equal to it.
2. From the point B, with any radius greater than
BD, describe an arc of a circle, and from the point A,
with the same radius, another arc, cutting the first.
3. Through the point of intersection, C, and the point
D, draw a straight line, CD, which will be perpendiculai
to the line MN.
Demojv. The three sides of the triangle BCD, are equal to th6
three sides of the triangle ACD, each to each, viz. •
the side BC equal to AC
m « uj) « « da
« « CD " "CD;
therefore the three angles in the triangle BCD are also equal to
the three angles of the triangle ADC, each to each (page 40) ;
and the angle x opposite to the side BC in the triangle BCD, is
equal to the angle y opposite to the equal side AC in the triangle
ACD ; and as the two adjacent angles, which the line CD makes
with the line MN, are equal to one another, the line CD is per-
pendicular to MN. (Definitions of perpendicular lines, page 12.)
143
II. Solution. Let MN be the given straight line,
and A the point in which the perpendicular is to be
drawn to it.
1. From a point, O, as a centre, with a radius, OA,
greater than the distance O from the straight line MN,
describe the circumference of a circle.
2. Through the point B and the centre O, of the cir-
cle, draw the diameter BC.
3. Through C and A draw a straight line, which will
be perpendicular to the line MN.
Demon. The angle BAC, at the circumference, measures half
as many degrees as the arc BPC intercepted by its legs (page 111,
1st). But the arc BPC is a semi-circumference ; therefore the angle
BAC, measures a quadrant ; consequently the angle BAC is a right
angle (page 107, Remark 3), and the line AC is perpendicular to
MN.
Problem III. To bisect a given angle.
Solution. Let BAC be the given
angle.
1. From the vertex, A, of the angle
BAC, with a radius, AE, taken at
pleasure, describe an arc of a circle ;
and from the two points D and E,
where this arc cuts the legs of the
given angle, with the same radius
describe two other arcs, cutting each
other in the point m.
144 GEOMETRY.
2. Through the point m, and the vertex of the given
angle, draw a straight line, Am, which will bisect the
given angle BAC.
Demon. The two triangles AmD, AmE, have the three sides
in the one equal to the three sides in the other, viz.
the side AD = to the side AE
« « ?»D=" " " mE
" " Am=" " " Am;
consequently these two triangles are equal to each other ; and the
angle x, opposite to the side mD, in the triangle AmD, is equal to
the angle y, opposite to the equal side mE, in the triangle AmE;
therefore the angle BAC is bisected.
Problem IV. From a given point without a straight
line, to let fail a perpendicular upon that line.
Solution. Let A be the
given point, from which
a perpendicular is to be
drawn to the line MN.
1. With any radius suf-
ficiently great describe an
arc of a circle.
2. From the two points B and C, where this arc cuts
the line MN, draw the straight lines BA, CA.
3. Bisect the angle BAC (see the last Problem), the
line AD is perpendicular to the line MN.
Demon. The two triangles ABD, ACD, have two sides, AB,
AD, in the one, equal to two sides, AC, AD, in the other, each to
each (AC, AB, being radii of the same circle, and the side AD
being common to both) ; and have the angles included by these
sides also equal (because the angle BAC is bisected); therefore
these two triangles are equal to one another (Query 1, Sect. II.) ;
and the angle y, opposite to the side AB, in the triangle ABD, is
equal to the angle x, opposite to the equal side AC, in the triangle
ACD. Now, as the two adjacent angles x and y, which the straight
..*
GEOMETRY.
145
line AD makes with the straight line MN, are equal to each other,
the line AD must be perpendicular to MN. (Def. of perpendicu-
lar lines.)
Problem V. To bisect a given straight line.
Solution. Let AB be the
given straight line.
1* From A, with a radius
greater than half of AB, de-
scribe an arc of a circle ; and
from B } with the same radius,
another, cutting the first in
the point C.
2. From the point C draw '
the perpendicular CM, and the line AB is bisected in M.
Demon. The two right-angled triangles AMC, BMC, are
equal, because the hypothenuse AC and the side CM in the one,
are equal to the hypothenuse and the side CM in the other
(page 47) ; and therefore the third side AM in the one, is also
equal to the third side BM in the other; consequently the line
AB is bisected in the point M.
Problem VI. To transfer a given angle.
Solution. Let x be the
given angle, and A the
point to which it is to be
transferred.
1. From the vertex of the
given angle, as a centre,
with a radius taken at pleasure, describe an arc of a cir-
cle between the legs AB, AC.
2. From the point a, as a centre, with the same radius,
describe another arc, cb.
3. Upon trie last arc take a distance, 6c, equal to the
chord BC.
146 GEOMETRY.
4. Through a and c draw a straight line ; the angle y
is equal to the angle x.
Demon. The arcs BC, 6c, are, by construction, equal to one
another ; therefore the angles x and y, at the centre, being meas-
ured by these arcs, are also equal to one another (page 106, 1st).
Problem VII. Through a given point draw a line
parallel to a given straight line.
'y
jp-
Solution. Let E be the point, through which a line
is to be drawn parallel to the straight line AB.
1. Take any point, F, in the straight line AB, and join
/ EF.
2. *In E make the angle y equal to the angle % ; the
line EG, extended, is parallel to the line AB.
Demon. The two straight lines CG, AB, are cut by a third
line EF, so as to make the alternate angles x and y equal ; there-
fore these two lines are parallel to each other (page 29, 2dly).
Mechanical Solution. Take a ruler, MN, and put
it in such a position that a right-angled triangle, passing
along its edge, as you see in the figure, will make with
it, in different points, A, C, &c, the lines AB, CD, &c.
GEOMETRY.
147
These lines are parallel to each other, because they are
cut by the edge of the ruler at equal angles.*
Problem VIII. Two adjacent sides and the angle
included by them being given, to construct a parallelo-
gram.
A
c
A—
1>
/
B
Solution. Let AB and AC be the two sides of true
parallelogram, and x the angle included by them.
1. Make an angle equal to x.
2. Make the leg AB of that angle equal to AB, and
the leg AC equal to AC.
3. Through the point C draw CD parallel to AB, and
through B, the line BD parallel to AC; the quadrilateral
ABCD is the required parallelogram.
Demon. The opposite sides of the quadrilateral ABCD, are
parallel to each other ; therefore the figure is a parallelogram.
(See Def. page 13.)
* This is a hetter way of drawing parallel lines than the common
method by a parallel ruler, which is seldom very accurate, on
account of the instrument being frequently out of order, and the
great steadiness, of hand required in the use of it
148
GEOMETRY.
Problem IX. To divide a given line into any number
of equal parts.
I. Solution. Let AB be the
given line, and let it be required to
divide it into five equal parts.
1. From the point A, draw an
indefinite straight line AN, making
any angle you please with the line
AB.
2. Take any distance Am, and
measure it off 5 times upon the line
AN.
3. Join the last point of division q,
and the extremity B of the line AB.
4. Through m, n, o, p, q, draw the straight limes bm,
en, do, ep, parallel to Bq; the line AB is divided into five
equal parts.
The demonstration follows immediately from Query 14, Sect. H.
II. Solution. Let AB be tfte —
given straight line, which is to be
divided into 5 equal parts.
1. Draw a straight line MN,
greater than AB, parallel to AB.
2. Take any distance M?i, and
measure it off 5 times upon the
line MN.
3. Join the extremities of both
the lines Mr and AB, by the straight
lines MA, rB, which will cut each
other, when sufficiently extended,
in a point I.
4. Join In, \o, Ip, \q, the line AB is divided into 6
equal parts, viz. A6, 6c, cd, de, cB.
GEOMETRY.
149
Demon. The triangles A6I, bcl, cdl, del, eBI, are similar to
the triangles Mnl, nol, opl,pqI, qrl, each to each; because the
line AB is drawn parallel to Mr (Query 16, Sect. II.) ; and as the
bases Mn, no, op, pq, qr, of the latter triangles are all equal to one
another, the bases Ab, be, cd, de, eB of the former triangles must
also be equal to one another.
Remark. If it were required to divide a line into two parts
which shall be in a given ratio, for instance, as 2 to 3, you need
only, as before, take 5 equal distances upon the line MN, and then
join the point I to the second and last point of division ; the line
AB will, in the point c, be divided in the ratio of 2 to 3. In a simi-
lar manner can any given straight line be divided into 3, 4, *>, &c.
parts, which shall be to each other in a given ratio.
Problem X. Three lines being given, to find a fourth
one, which shall be in a geometrical proportion with them.
AAA
D
Solution. Let AB, AC,
AD, be the given straight
lines, which are three terms
of a geometrical proportion,
to which the fourth term is
wanting. (See Theory of
Proportions, Principle 8th,
page 64.)
1. Draw two indefinite
straight lines AP, AQ,, mak-
ing with one another any angle you please.
2. Upon one of these lines measure off the two dis-
tances AB, AC, and on the other the distance AD.
3. Join BD, and through C draw CE parallel to BD ;
the line AE is the fourth term in the" geometrical pro-
portion
AB: AC = AD: AE.
Demon. The triangle ABD is similar to the triangle ACE,
from which it may be considered as cut off by the line BD being
13*
150 GEOMETRY.
drawn parallel to CE (Query 16, Sect. II.) ; and as in similar tri-
angles the corresponding sides are in a geometrical proportion
(page 70, 4thly), we have
AB: AC = AD: AE.
Problem XI. Two angles of a triangle being given,
to find the third one.
Solution. Let x and y be
the two given angles of the
triangle, and let it be requir-
ed to find the third angle z.
In any point O of an in-
definite straight line AB,
make two angles x and y,
equal to the two given angles of the triangle ; the remain-
ing angle z is equal to the angle z in the triangle.
Demon. The sum of the- three angles x, z, y, in the triangle,
is equal to two right angles (Query 13, Sect. I.), and the sum of
the three angles x, y, z, made in the same point O, and on the
same side of the straight line AB, is also equal to two right angles
(page 23)*; and as the angles x and y are made equal to the angle
x and y in the triangle, the remaining angle z is also equal to the
remaining angle z in the triangle.
Remark. If, instead of the angles themselves, their measure
were given in degrees, minutes, seconds, &c, you need only sub-
tract the sum of the two angles from 180 degrees, which is the
measure of two right angles ; the remainder is the angle sought
Problem XII. Through three given points, which are
not in the same straight line, to describe the circumference
of a circle.
Solution. Let A, B, C, be the three points, through
which it is required to pass the circumference of a circle.
1. Join the three points A, B, C, by the straight lines
AB, BC.
GEOMETRY. 151
2. Bisect the lines AB, BC.
3. In the points of bisection E
and F, erect the perpendiculars
EO, FO, which will cut each other
in a point O.
4. From the point O as a cen-
tre, with a radius equal to the dis-
tance AO, describe the circumference of a circle, and it
will pass through the three points A> B, C.
Demon. The two points A and B are at an equal distance
from the foot of the 'perpendicular EO ; therefore AO and BO are
equal to one another (page 45, 5thly) ; for the same reason is BO
equal to OC ; because the points B and C are at an equal distance
from the foot of the perpendicular FO ; and as the three lines AO,
BO, CO are equal to one another, the three points A, B, C, must
necessarily lie in the circumference of the circle described with
the radius AO.
Problem XIII. To find the centre of a circle, or of a
given arc.
Solution. Let the circle in the last figure be the
given one.
1. Take any three points A, B, C, in the circumfer-
ence, and join them by the chords AB, BC.
2. Bisect each of these chords, and in the points of
bisection erect the perpendiculars EO, FO ; the point O,
in which these perpendiculars meet each other, is the
centre of the circle.
In precisely the same manner can the centre of an arc
be found.
The demonstration is exactly the same as in the last problem.
152
GEOMETRY.
Problem XIV. In a given point in the circumference
of a circle, to draw a tangent to that circle.
Solution. Let A be the given
point in the circumference of the
circle.
Draw the radius AO, and at the
extremity A, perpendicular to it, the
line MN ; and it is a tangent to the
given circle.
Demon. The line MN being drawn at
the extremity AO of the radius, and per-
pendicular to it, touches the circumference
in only one point (page 108)
Problem XV. From a given point without a circle,
to draw a tangent to the circle.
Solution. Let A be the given
point, from which a tangent is to
be drawn to the circle.
1. Join the point A and* the
centre C, of the given circle.
2. From the middle of the line
AC as a centre, with a radius equal
to BC = AB, describe the circum-
ference of a circle.
3. Through the points E and D,
where this circumference cuts the
circumference of the given circle, draw the lines AD,
AE ; and they are tangents to the given circle.
Demon. Join DC, EC. The angles x and y, being both an-
gles at the circumference of the circle whose centre is B, measure
each half as many degrees as the arc on which their legs stand.
Both the angles, x and y, have their legs standing on the diameter
AC, of the circle B ; therefore each of these angles measures half
GEOMETRY.
1S3
&s many degrees as the semi-circumference (page 107, Rem. 3d) ;
consequently, they are both right angles, and the lines AE and
DA, being perpendicular to the radii CE, DC, are both tangents
to the circle C
Remark. From a point without a circle, you can always draw
two tangents to the same circle.
Problem XVI. To draw a tangent common to two
given circles.
Solution. Let A and B be the
centres of the given circles, and let
it be required to draw a tangent,
which shall touch the two circles on
the same side.
1. Join the centres of the two
given circles by the straight line
AB.
2. From B, as a centre, with a
radius equal to the difference be-
tween the radii of the given circles,
describe a third circle.
3. From A draw a tangent AE
to that circle (see the last problem).
4. Draw the radius BE, and extend it to D.
5. Draw the radius AC parallel to BD.
6. Through C and D draw a straight line, and
be a tangent common to the two given circles.
it will
Demon. The radius AC being equal and parallel to ED, it
follows that ACED is a parallelogram ; and because the tangent
AE is perpendicular to the radius BE (page 108, 1st), CD is per-
pendicular to BD ; consequently also to AC (because AC is parallel
to BD) ; and the line CD, being perpendicular to both the radii
AC, BD, is a tangent common to the two given circles.
154
GEOMETRY.
If it be required to draw a tangent
common to two given circles, which
shall touch them on opposite sides, then
1. From B as a centre, with a radius
equal to the sum of the radii of the given
circles, describe a third circle.
2. From A draw a tangent AE to that
circle.
3. Join BE, cutting the given circle
in D.
4. Draw AC parallel to BE.
5. Through C and D, draw a straight
line, and it is the required tangent, touching the circles on opposite
sides.
The demonstration is the same as the last.
Problem XVII. Upon a given straight line to describe
a segment of a circle, ichich shall contain a given angle ;
that is, a segment, such that the inscribed angles, having
their vertices in the arc of the segment and their legs
standing on its extremities, shall each be equal to a given
Solution. Let AB be the
given line, and x the given
angle.
1. Extend AB towards C.
2. Transfer the angle x to
the point A.
3. Bisect AB in E.
4. From the points A and & "£"
E, draw the lines AO and EO,
respectively, perpendicular to FG and CB.
5. From the point O, the intersection of these perpen-
diculars, as a centre, with a radius equal to OA, describe
a circle ; AMNB is the required segment.
GEOMETRY.
155
Demon; The line FG being, by construction, perpendicular to
the radius AO, is a tangent to the circle (page 108) ; and the angle
GAB, formed by that tangent and the chord AB, is equal to either
of the angles AMB, ANB, &c, that can be inscribed in the seg-
ment AMNB ; because the angle GAB measures half as many de-
grees as the arc ALB (page 109), and each of the angles AMB,
ANB, &c, at the circumference, having its legs standing on the
extremities of the chord AB, measures also half as many degrees as
the arc ALB (page 111) ; and as the angle GAB is equal to the angle
x, GAB and x being opposite angles at the vertex (Query 5,
Sect. I.), each of the angles AMB, ANB, &c, is also equal to the
given angle x.
Remark. If the angle a: is a right
angle, the segment AMB is a semi-
circle, and the chord AB a diameter.
To finish the construction, you need
only from the middle of the line AB as
a centre, with a radius equal to OA,
describe a semicircle, and it is the re-
quired segment ; for the angle AMB
at the circumference measuring half as many degrees as the semi-
circumference AB, on which its legs stand, is a right angle.
D
A
BM
a
Problem XVIII. To find a mean proportional (see
page 66) to two given straight lines.
Solution. LetAB,BC, *
be the two given lines.
1. Upon an indefinite
straight line, take the two B
distances AB, BC.
2. Bisect the whole dis-
tance AC, and from M, the middle of AC, with a radius
equal to AM, describe a semi-circumference.
3. In B erect a perpendicular to the diameter AC, and
extend it until it meets the semi-circumference in D ; the
line DB is a mean proportional between the lines AB
and BC.
156 GEOMETRY.
Demon. The triangle ADC is right-angled in D ; because the
angle ADC is inscribed in a semicircle (see the remark to the last
problem) ; and the perpendicular DB let fall from the vertex D, of
the right angle upon the hypothenuse, is a mean proportional be-
tween the two parts AB, BC,into which it divides the hypothenuse
(page 75, 1st) ; therefore we have the proportion
AB:BD = BD:BC.
Problem XIX. To divide a given straight line into
two such parts, that the greater of them shall be a mean
proportional between the smaller part and the whole of the
given line.
Solution. Let AB be the
given straight line.
1. At the extremity B of the
given line, erect a perpendicu-
lar, and make it equal to half of
the line AB.
2. From O, as a centre, with a radius equal to OB,
describe a circle.
3. Join the centre O of that circle, and the extremity
A of the given line, by the straight line AO.
4. From AB cut off a distance AD equal to AE ; then
AD is a mean proportional between the remaining part
BD, and the whole line AB ; that is, you have the pro-
portion
AB : AD = AD:BD.
Demon - . Extend the line AO until it meets the circumference
in C. Then the radius OB, being perpendicular to the line AB,
we have from the same point A, a tangent AB, and another line
AC drawn cutting the circle ; therefore we have the proportion
AC : AB = AB : AE ;
for the tangent AB is a mean proportional between the whole line
AC, and the part AE without the circle. (Query 13, Sect. IV.)
Now, in every geometrical proportion, you can add or subtract
the second term once or any number of times from the first term,
GEOMETRY. 157
and the fourth term the same number of times from the third term,
•without destroying the proportion (page 62, 6th). According to
this principle you have
AC — AB : AB = AB — AE : AE;
that is, the line AC less the line AB, is to the line AB, as the line
AB less EA, is to the line AE. But AC less AB is the same as
the line AC less the diameter CE (because the radius of the circle
is, by construction, equal to half the line AB) ; and AB less AE,
is the same as AB less AD (because AD is made equal to AE);
therefore you may write the above proportion also
AE : AB = BD : AE,* or also
AD : AB = BD : AD;
and because in every geometrical proportion the order of the terms
may be changed in both ratios (Principle 1, of Geom. Prop.), you
can change the last proportion into
AB : AD = AD : BD;
that is, the part AD of the line AB, is a mean proportional between
the whole line AB and the remaining part BD.
Problem XX. To inscribe a circle in a given triangle.
Solution. Let the given
triangle be ABC.
1. Bisect two of the angles " 2 //CT^N? 1 '
of the given triangle; for in-
stance the angles at C and B, a*
by the lines CO, BO.
2. From the point O, where these lines cut each other,
let fall a perpendicular upon any of the sides of the given
triangle.
3. From O, as a centre, with the radius OP, equal to
the length of that perpendicular, describe a circle, and it
will be inscribed in the triangle ABC.
Demon. From let fall the perpendiculars OM, ON, upon
the two sides BC, AC, of the given triangle. The angle OCM is,
* AC less the diameter CE, being equal to AE ; and BA less AD,
equal to BD.
14
158 GEOMETRY.
by construction, equal to the angle OCN (because the angle ACB
is bisected by the line CO) ; and CMO, CNO, being right angles,
the angles COM and CON are also equal to one another (because
when two angles in one triangle are equal to two angles in another,
the third angles in these triangles are also equal) ; therefore the
two triangles CMO, CNO, have a side CO, and the two adjacent
angles in the one, equal to the same side CO, and the two adjacent
angles in the other ; consequently these two triangles are equal to
one another ; and the side OM, opposite to the angle OCM in the
one, is equal to the side ON, opposite to the equal angle OCN in
the other. In the same manner it may be proved that the perpen-
dicular OM is also equal to OP ; and as the three perpendiculars
OM, ON, OP, are equal to one another, the circumference of a
circle described from the point as a centre, with a radius equal
to OP, passes through the three points M, N, P ; and the sides
AB, BC, AC of the given triangle, being perpendicular to the radii
OP, OM, ON, are tangents to the inscribed circle (page 108).
Problem XXI. To circumscribe a circle about a tri~
angle.
This problem is the same as to make the circumfer-
ence of a circle pass through three given points. (See
Problem XII.)
Problem XXII. To trisect a right angle.
Solution. Let BAC be the
right angle which is to be divid-
ed into three equal parts.
1. Upon AB take any dis-
tance AD, and construct upon
it the equilateral triangle ADE.
(Problem I.)
2. Bisect the angle DAE by the line AM (Problem
III.) ; and the right angle BAC is divided into the three
equal angles CAE, EAM, MAB.
GEOMETRY.
159
Demon-. The angle BAE being one of the angles of an equi-
lateral triangle, is one third of two right angles (page 33), and
therefore two thirds of one right angle ; consequently CAE is one
third of the right angle BAC ; and since the angle BAE is bisected
by the line AM, the angles EAM, MAB, are each of them also
equal to one third of a right angle ; and are therefore equal to the
angle CAE and to each other.
PART II.
Problems relative to the transformations of geomet-
rical figures.
Problem XXIII. To transform a giv-en quadrilateral
figure into a triangle of equal area, whose vertex shall
be in a given angle of the figure, and whose base in one
of the sides of the figure.
B
Fig. I.
^E
Fig. IL
B
A
I)
A
JE
Solution. Let ABCD (Fig. 1. and II.), be the given
quadrilateral ; the figure I. has all its angles outwards,
and the figure II. has one angle, BCD, inwards ; let the
vertex of the triangle, which shall be equal to it, fall in B,
1. Draw the diagonal BD (Fig. I. and II. ), and from
C, parallel to it, the line CE.
2. From E, where the line CE cuts AD (Fig. II.), or
its further extension (Fig. I.), draw the line EB; the
triangle ABE is equal to the quadrilateral ABCD.
160
GEOMETRY.
Demon. The area of the triangle BCD (Fig. I. and II.) w
equal to the area of the triangle BDE ; because these two triangles
are upon the same basis, BD, and between the same parallels, BD,
CE (page 90, 3dly) ; consequently (Fig. I.), the sum of the areas
of the two triangles ABD and BDC, is equal to the sum of the
areas of the two triangles ABD, BDE ; that is, the area of the
quadrilateral ABCD is equal to the sum of the areas of the two tri-
angles ABD, BDE, which is the area of the triangle ABE.
And in figure II. the difference between the areas of the two
triangles ABD, BCD, that is, the quadrilateral ABCD, is equal
to the difference between the triangles ABD, EBD, which is the
triangle ABE.
Problem XXIV. To transform a given pentagon into
a triangle, whose vertex shall be in a given angle of the
pentagon, and whose base upon one of its sides.
Solution. Let ABCDE (Fig. I. and II.), be the given
pentagon ; let the vertex of the triangle, which is to be
equal to it, be in C.
Fig. I. Fig. II.
1. From C draw the diagonals CA, CE.
2. From B draw BF parallel to CA, and from D draw
DG parallel to CE.
3. From F and G, where these parallels cut AE or its
further extension, draw the lines CF, CG ; CFG is the
triangle required.
Demon. In both figures, we have the area of the triangle
CBA equal to the area of the triangle CFA ; because these two
triangles are upon the same basis, CA, and between the same par-
GEOMETRY.
161
allels, AC, FB ; and for the same reason is the area of the triangle
CDE equal to the area of the triangle CGE ; therefore in figure I.
the sum of the areas of the three triangles CAE, CBA, CDE, is
equal to the sum of the areas of the triangles CAE, CFA, CGE ; that
is, the area of the pentagon ABCDE is equal to the area of the
triangle CFG; and in figure II. the difference between the area
of the triangle CAE and the areas of the two triangles CBA, CDE,
is equal to the difference between the area of the same triangle
CAE* and the areas of the two triangles CFA, CGE ; that is, the
area of the pentagon ABCDE is equal to the area of the triangle
CFG.
Problem XXV. To convert any given figure into a
triangle, whose vertex shall be in a given angle of the
figure, and whose basis shall fall upon one of its sides.
Fig. II.
Let ABCDEF (Fig. I. and II.) be the given figure
(in this case a hexagon), and A the angle in which the
vertex of the required triangle shall be situated. For the
sake of perspicuity, I shall enumerate the angles and sides
of the figure from A, and call the first angle A, the sec-
ond B, the third C, and so on ; further, AB the first side,
BC the second, DE the third, and so on. We shall then
have the following general solution.
1. From A to all the angles of the figure, draw the
diagonals AC, AD, AE, which, according to the order
in which they stand here, call the first, second, and third
diagonal.
14*
162 GEOMETRY.
2. Draw from the second angle, B, a line, Ba, parallel
to the first diagonal, AC ; from the point where the par-
allel meets the third side, CD (Fig. II.), or its further
extension (Fig. I.), draw a line, ab, parallel to the second
diagonal, AD ; and from the point b, where this meets the
fourth side DE (Fig. II.) or its further extension (Fig. I.),
draw another line, be, parallel to the third diagonal.
3. When, in this way, you have drawn a parallel to
every diagonal, then, from the last point of section of the
parallels and sides (in this case c), draw the line cA ;
AcF is the required triangle, whose vertex is in A, and
whose basis is in the side EF.
The demonstration is similar to the one given in the two last
problems. First, each of the hexagons is converted into the pen-
tagon AaDEF ; then the pentagon AaDEF into a quadrilateral,
A6EF ; and finally this quadrilateral into the triangle AcF. The
areas of these figures are evidently equal to one another ; for the
areas of the triangles, which, by the above construction, are suc-
cessively cut off, are equal to the areas of the new triangles which
are successively added on. (See the demonstration of the last
problem.)
Remark. Although the solution given here is only intended for
a hexagon, yet it may easily be applied to every other rectilinear
figure. All depends upon the substitution of one triangle for
another, by means of parallel lines. It is not absolutely necessary
actually to draw the parallels ; it is Only requisite to denote the
points in which they cut the sides, or their further extension; be-
cause all depends upon the determination of these points.
Problem XXVI. To transform any given figure into
a triangle ivhose vertex shall be in a certain point, in one
of the sides of the figure, or within it, and whose base
shall fall upon a given side of the figure.
GEOMETRY,
Solution. 1st Case. Let ABCDEF be a hexagon,
which is to be transformed into a triangle ; let the vertex
of the triangle be in the point M in the side CD, and the
base in AF.
1. In the first place, get rid of the angle ABC, by
drawing Ba parallel to CA, and joining Ca; the triangle
CBa, substituted for its equal the triangle ABa (for these
two triangles are upon the same basis, aB, and between
the same parallels, CA, Ba), transforms the hexagon
ABCDEF into the pentagon aCDEF.
2. Draw the lines Ma, MF, and the pentagon aCDEF
is divided into three figures, viz. the triangle MaF, the
quadrilateral MDEF on the right, and the triangle MCa
on the left.
3. Transform the quadrilateral MDEF and the triangle
MCa into the triangles MoT, M6a, so that the basis may
be in AF (see the last problem) ; the triangle Mbd is
equal to the given hexagon.
164
2d Case. Let ABCDEF be the given figure ; let the
vertex of the required triangle be situated in the point M
within the figure, and let the base fall upon AF.
1. From M to any angle of the figure, say D, draw the
line MD, and draw the lines MA, MF, by which means
the figure ABCDEF is divided into the triangle MAF,
and the figures MDCBA , MDEF.
2. Then transform MDCBA and MDEF into the tri-
angles McA, MeF, whose bases are in the continuation
of AF 5 the triangle cMe is equal to the figure ABCDEF.
The demonstration follows from those of the last three problems.
Problem XXVII. To transform a given rectangle
into a square of equal area.
A.
E
y ,„"'*
D\\
B
M G
Solution. Let ABCD be the given rectangle.
1. Extend the greater side, AB, of the rectangle, making
BM equal to BD.
2. Bisect AM in O, and, from the point O as a centre,
with a radius AO, equal to OM, describe a semicircle.
GEOMETRY. 165
3. Extend the side BD of the rectangle, until it meets
the circle in E.
4. Upon BE construct the square BEFG, which is the
square sought.
Demon. The perpendicular BE is a mean proportional be-
tween AB and BM (see Problem XVIII.) ; therefore we have the
proportion
AB : BE = BE : BM ;
and as, in every geometrical proportion, the product of the means
equals that of the extremes (Theory of Prop., Principle 10, page
65), we have the product of the side BE multiplied by itself, equal
to the product of the side AB of the parallelogram, multiplied by
the adjacent side BD (or BM). But the first of these products
is the area of the square BEFG, and the other is the area of the
rectangle ABCD.; therefore these two figures are, in area, equal
to one another.
Problem XXVIII. To transform a given triangle
into a square of equal area.
K
D JB M JP
Solution. Let ABC be the given triangle, AB its
base, and CD its height.
1. Extend AB by half the height CD.
2. Upon AM as a diameter, describe a semicircle.
3. From B draw the perpendicular BN, which is the
side of the square sought.
Demon. From the demonstration in the last problem, it follows,
that the square upon BN is equal to the rectangle, whose base is
AB, and whose height is BM (half the height of the triangle
ABC). But the triangle ABC is equal to a rectangle upon the
same base AB, and of half the height CD (page 89, 1st) ; therefor©
166
GEOMETRY.
the area of the square BNOP is equal to the area of the triangle
ABC.
Remark. It appears from this problem, that every rectilinear
figure can be converted into a square of equal area. It is only
necessary to convert the figure into a triangle (according to the
rules given in the problems 23, 24, 25), and then that triangle into
a square.
Problem XXIX. To convert any given triangle into
an isosceles triangle of equal area.
Solution. Let ABC be
the given triangle, which is
to be converted into an isos-
celes one.
1. Bisect the base AC in
D, and from D draw the perpendicular DE.
2. From the vertex, B, of the given triangle, draw BE,
parallel to the base, AC.
3. From the point E, where this parallel meets the
perpendicular, draw the straight lines EA, EC ; EAC is
the isosceles triangle sought.
Demon. The triangles AEC and ABC are upon the same basis,
AC, and between the same parallels (page 90, 3dly).
Problem XXX. To convert a given isosceles triangle
into an equilateral one of equal area. (This problem is
intended for more advanced and elder pupils.)
Solution. Let ABC be the
given isosceles triangle.
1. Upon the base, AC, of the
given triangle, draw the equilate-
ral triangle AEC (problem T.) ;
and through the vertices, E, B, of
the two triangles, draw the straight
GEOMETRY. 107
line EB, which evidently is perpendicular to AC, and
bisects the last line in D (ABC, AEC, being isosceles
triangles).
2. Upon ED describe the semicircle EFD, and from
B draw the perpendicular BF, which meets the semi-
circle in F.
3. From D, with the radius DF, describe an arc, FG,
cutting the line DE in G.
4. From G, draw the lines GH, GI, parallel to the
sides of the equilateral triangle AEC ; HGI is the equi-
lateral triangle sought.
Demon. Since the line GH is parallel to AE, and GI parallel
to EC, the angle GHI is equal to the angle EAI, and the angle
GIH to the angle ECH (page 31). Thus the two triangles GHI,
AEC, have two angles, GHI, GIH, in the one, equal to two an-
gles, EAC, EGA, in the other, each to each ; consequently they
are similar to each other (page 73, 1st) ; and the triangle GHI
must also be equilateral.
Suppose the lines DF and EF drawn ; then DF is a mean propor-
tional between DE and DB ; for the triangle EDF is right-angled
(see the Remark, page 155) in F, and if from the vertex of the
right angle, the perpendicular FB is let fall upon the hypothenuse,
the side DF is a mean proportional between the hypothenuse, ED,
and the part, BD, of it, between the foot of the perpendicular, and
the extremity, D, of the line FD (see page 75, 2dly) ; consequently
we shall have the proportion
ED : DF = DF : BD ;
and as DG is, by construction, made equal to DF,
ED : DG = DG : BD (I.)
Moreover, in the two similar triangles, ADE, HDG, the corre-
sponding sides are proportional (page 70, 4thly) ; therefore we have
the proportion
ED : DG=AD : HD (II.)
This last proportion has the first ratio common with the first pro-
portion ; consequently the two remaining ratios are in a geometri-
cal proportion (Theory of Prop., Prin. 3d) ; that is, we have
AD : HD = DG: BD;
and as, in every geometrical proportion, the product of the means is
168
GEOMETRY.
equal to that of the extremes (Theory of Prop., Principle 10th), we
have HD multiplied by DG, equal to AD multiplied by BD ; con-
sequently, also, half the product of the line HD, multiplied by the
line DG, equal to half the product of the line AD, multiplied by
BD. But half the product of the line HD, multiplied by DG, is
the area of the triangle HDG ; because the triangle HDG is right-
angled in D, therefore if HD is taken for the basis, DG is its
height ; and for the same reason is half the product of the line AD
by BD, the area of the triangle ADB ; consequently the areas of
the two triangles, ADB and HDG, are equal to one another; and
because the triangle HDG is equal to the triangle IDG, and the
triangle ABD to the triangle CBD, the area of the whole triangle
HIG is equal to the area of the whole triangle ABC ; therefore the
triangle HIG is the required equilateral triangle, which is equal,
in area, to the given isosceles triangle, ABC.
Remark 1. If BD is greater than ED, then the perpendicular,
BF, does not meet the semicircle. In this case, it is necessary to
describe the semicircle on BD, and from E to draw the perpen-
dicular. In this case, the points H, I, will not be situated in the
line AC ; but in its further extension.
Remark 2. From this and the preceding problems, it appears
how any figure may be converted into an equilateral triangle ; for
it is only necessary first to convert the figure into a triangle, this
triangle into an isosceles triangle, and the isosceles triangle into
an equilateral one.
Problem XXXI. To describe a square, which in area
shall be equal to the sum of several given squares.
E
A B C B
JE
D
A B B
Solution. Let AB, BC, CD, DE, be the sides of four
squares; it is required to find a square which shall be
equal to the sum of these four squares.
GEOMETRY. 169
1. At the extremity, B, of the line AB, draw a perpen-
dicular equal to BC, and join AC.
2. At the extremity, C, of the line AC, draw a perpen-
dicular equal to CD, and join AD.
3. At the extremity, D, of the line AD, draw a perpen-
dicular equal to DE, and join AE ; the square upon AE
is, in area, equal to the sum of the four squares upon the
lines AB, BC, CD, DE.
Demon. The square upon the hypothenuse, AC, of the right-
angled triangle ABC, is equal to the sum of the squares upon the
two sides AB, BC (Query 6, Sect. III.) ; and for the same reason
is the square upon AD equal to the sum of the squares upon CD
and AC ; consequently, also, to the squares upon CD, CB, and AB
(the square upon AC being equal to the squares upon CB and AB) ;
and finally the square upon AE is equal to the sum of the squares
upon ED and AD ; or, which is the same, to the sum of the squares
upon DE, CD, CB, and AB
Problem XXXII. To describe a square which s?ialll>e
equal to the difference of two given squares.
Solution. Let AB, AC
be the sides of two squares.
1. Upon the greater side,
AB, as a diameter, describe
a semicircle.
2. From A, within the
semicircle, draw the line
AC, equal to the given line AC, and join BC; then CB
is the side of the square sought.
Demon. The triangle ABC is right-angled in C, and in every
right-angled triangle, the square upon one of the sides, which
include the right angle, is equal to the difference between the
squares upon the hypothenuse and the other side.
15
170
GEOMETRY.
Problem XXXIII. To transform a given figure
in such a way, that it may be similar to another figure.
n c L
Solution. Let X be the given figure, and ABCDEF
the one to which it is to be similar.
1. Convert the figure ABCDEF into a square (seethe
remark, page 166), and let its side bemw, so that the area
of the square upon mn is equal to the area of the figure
ABCDEF ; convert, also, the figure X into a square, and
let its side be pq, so that the area of the square upon pq
shall be equal to the area of the figure X.
2. Take any side of the figure ABCDEF, say AF ;
and to the three lines, mn, pq, AF, find a fourth propor-
tional (Problem X), which you cut off from AF. Let
Af be this fourth proportional, so that we have the pro-
portion
mn : pq = AF : Af.
3. Then draw the diagonals AE, AD, AC, and the lines
fe, ed, dc, cb, parallel to the lines FE, ED, DC, CB ;
then Abcdef will be the required figure, which in area is
equal to the figure X, and is similar to the figure ABCDEF.
Demon. It is easily proved, that the figure Abcdef is similar
to ABCDEF. Further, we know that the areas of the two similar
figures, ABCDEF, Abcdef, are to each other, as the areas of the
squares upon the corresponding sides AF, A/, (see page 198);
Which may be expressed,
ABCDEF : Abcdef = AF X AF : A/x Af;
GEOMETRY. 171
and as the sides AF and A/ are (by construction 2) in proportion
to the lines mn, pq, the squares upon these sides, and therefore
the figures ABCDEF, Abcdef, themselves, are in proportion to the
squares upon mn and pq ; that is, we shall have the proportion
ABCDEF : Abcdef —mn X mn : pq X pq-
This proportion expresses, that the area of the figure ABCDEF
is as many times greater than the area of the figure Abcdef, as the
area of the square upon the line mn is greater than the area of
the square upon the line pq ; therefore, as the area of the figure
ABCDEF is, by construction, equal to that of the square upon the
line mn, the area of the figure Abcdef is equal to that of the
square upon the line pq. But the square upon pq is made equal
to that of the figure X ; therefore the area of the figure Abcdef
is also equal to that of the figure X ; and the figure Abcdef is the
one required.
PART III.
Partition of figures by drawing.
Problem XXXIV. To divide a triangle from one of
the vertices Jinto a given number of parts.
Solution. Let ABC be the given triangle, which is
to be divided, say, into six equal parts ; let A be the ver-
tex, from which the lines of division are to be drawn.
1. Divide the side BC, .
opposite the vertex A, into
six equal parts, BD, DE, /
EF, FG, GH, HC. //
2. From A to the points / /
of division, D, E, F, G, H, BJDHL
draw the lines AD, AE,
AF, AG, AH; the triangle ABC is divided into the six
equal triangles, ABD, ADE, AEF, AFG, AGH, AHC.
172
GEOMETRY.
Demon. The triangles ABD, ADE, AEF, AFG, AGH, AHC,
are, in area, equal to one another, because they have equal bases
and the same height, Am (page 89).
Remark. If it is required to divide the triangle ABC according
to a given proportion, it will only be necessary to divide the line
BC in this proportion, and from A to draw lines to the points of di-
vision.
Problem XXXV. From a given point in one of the
sides of a triangle, to divide it into a given number of
equal parts.
Solution. Let ABC be
the given triangle, which is
to be divided into eight equal
parts ; the lines of division
are to be drawn from T.
1. Make Aa and Bb equal
to J of AB, and from T draw .
the line TC to the vertex, C,
of the triangle.
2. From a and b draw the
lines aD, bK, parallel to TC, meeting the sides AC, BC,
in D and K.
3. Upon AC, from A towards C, measure off the dis*
tance AD as many times as possible (in this case four
times) ; and thus determine the points E, F, G ; upon
BC, in the direction from B towards C, also measure off
the distance BK, as many times as is possible (here three
times), and determine the points I, H.
4. From T draw the lines TD, TE, TF, TG, TH,
TI, TK ; then ATD, DTE, ETF, FTG, GTHC, HTI,
ITK, KTB, are the eight equal parts of the triangle ABC.
GEOMETRY.
173
Demon. Draw Ca; then the triangle AaC is \ of the triangle
ABC ; because, if AB is taken for the base of the triangle ABC, the
base Aa of the triangle AaC is, by construction, § of the base of
the triangle ABC (see page 171). Now, the triangle aDC is equal
to the triangle aDT ; because these two triangles are upon the
same base, aD, and between the same parallels, aD, TC ; therefore
(by adding to each of them the triangle aAD) the two triangles
ADT ancfaAC are also equal ; that is, ADT is also J of the trian-
gle-ABC. In the same manner (by drawing the line 6C) it may
be proved that the triangle BKT is also J of the triangle ABC.
Further, the triangles ATD a DTE, ETF, FTG, are, by construc-
tion, all equal to one another, having equal bases and heights (see
the demonstration to the last problem) ; and for the same reason are
the triangles BTK, KTI, ITH equal to one another ; therefore each
of the seven triangles ATD, DTE, ETF, FTG, BTK, KTI, ITH,is
£ of the triangle ABC ; consequently the quadrilateral GTHC
must be the remaining one eighth of the triangle ABC; and the
area of the triangle ABC is divided into eight equal parts.
Problem XXXVI. To divide a triangle, from a
given point within it, into a given number of equal parts.
[This problem is intended for elder pupils.]
A.
Solution. Let ABC be the given triangfe, which is
to be divided, say, into five equal parts ; T the point
from which the lines of division are to be drawn.
1. Through the point T and the vertex A of the trian-
gle, draw the line AT.
15*
174 GEOMETRY.
2. Take any side of the triangle, say BC, and make,
when, as here, the triangle is to be divided into five equal
parts, BE and CF equal to £ of BC, and draw the lines
Ee, 'Ff, parallel to the sides AB, AC ; these lines will
meet the line AT in the points e and/".
3. From T draw the lines TB, TC, to the vertices B
and C of the triangle ABC^ and from e and/, the lines
eI,fG parallel to TB, TC/
4. Join TI, TG ; then each of the triangles ATI,
ATG, is i of the given triangle ABC.
5. In order to determine the other points of division,
it is only necessary to cut off from the sides AB, AC, as
many distances, equal to AI, AG, respectively, as is possible
(see the solution of the last problem), and in the case where
this can no longer be effected, or in which, as in the figure,
this is impossible, proceed in the following manner :
a. Extend the two sides AB, AC, and then make IM
equal to AI, and GN equal to AG.
b. From M and N draw the lines, MH, NP, parallel to
BT and CT, and determine thereby the points H and P.
6. Draw TH, TP ; each of the quadrilaterals IBHT,
GCPT, is £ of the triangle ABC ; consequently the trian-
gle HTP is the remaining fifth of it. (If HTB were not
the last part, then it would merely be necessary to divide
this triangle by the rule given in problem XXXIV, into
as many equal parts as necessary.)
DfiMON. Draw the auxiliary lines AE, Be ; then the triangle
ABE is one fifth of the triangle ABC ; because BE is one fifth of
the basis BC (problem XXXIV) ; further, the triangle ABE is
equal to the triangle ABe; because these two triangles are upon
the same basis, AB, and, by construction 2, between the same paral-
teh, AB, Ee; and the last triangle, ABe, is also equal to the triangle
ATI ; because the triangle ABe consists of the two triangles Ale
and IeB, which are equal to the two triangles Ale and ITe (the
GEOMETRY. 175
two triangles ITe and IeB being upon the same base, Je, and, by-
construction 3, between the same parallels, Ie, BT) ; therefore the
triangle AIT is also one fifth of tbe triangle ABC ; and in the same
manner it can be proved that ATG is one fifth of the triangle
ABC.
Further, the triangle GNT is equal to the triangle A GT (the
basis GN being made equal to the basis AG, and the vertical point
T being common to both triangles) ; and the triangle GNT is equal
to the quadrilateral CGTP ; because the triangle CTN is equal to
the triangle CTP (these two triangles being, by construction, upon
the same base, TC, and between the same parallels, TC, PN) ; there-
fore the area of the quadrilateral CGTP is also one fifth of the tri-
angle ABC ; and in the same manner it may be proved that the
area of the quadrilateral IBHT is one fifth of the triangle ABC ;
and as the two triangles AGT, AIT, together with the two quad-
rilaterals CGTP, IBHT, make four fifths of the triangle ABC, the
triangle HPT must be the, remaining one fifth of it.
Problem XXXVII. To divide a given triangle into
a given number of equal parts, and in such a way, that
the lines of division shall be parallel to a given side of the
triangle.
d
Solution. Let ABC be the given triangle ; let the
number of the parts, into which it is required to be di-
vided, be five, and BC the side to which the lines of
division are to be parallel.
1. Upon one of the other two sides, say AC, describe
a semicircle, and divide the side AC into as many equal
176 GEOMETRY.
parts as the triangle is to be divided into; consequently,
in the present case, into five ; the points of division are
D, E, F, G.
2. From these points of division draw the perpendicu-
lars Dd, Ee, Ff, Gg, meeting the semicircle in the points
3. From A draw Ad, Ae, Af, Ag ; then make Am
equal to Ad, An equal to Ae, and so on, and by these
means determine the points m, n, o, p.
4. From these points draw the lines mM, wN, oO, pV,
parallel to the side BC ; then AMm, MroNn, NwO<?,
OoVp, P/)BC are the five equal parts of the triangle ABC,
which were sought.
Demon. Imagine the line dC drawn ; the triangle AdC, in-
scribed in the semicircle, is right-angled in d; consequently we
have the proportion
AD : Ad = Ad : AC ;
and as, in every geometrical proportion, the product of the mean
terms is equal to that of the extremes,
AdxAd = ADxAC;
consequently, also,
Am X Am = AD X AC
(because Am is made equal to Ad).
Further, the triangles AMm, ABC, are similar, because the line
Mm is drawn parallel to the side BC in the triangle ABC (Query
16, Sect. II.) ; and as the areas- of similar triangles are to each
other as the areas of the squares upon the corresponding sides
(Query 8, page 97), we have the proportion
triangle ABC : triangle AMm = AC X AC : Am X Am ;
therefore, also,
triangle ABC : triangle AMm = AC X AC : AC X AD
(because Am X Am is equal to AC X AD).
The last proportion expresses, that the area of the triangle ABC is
as many times greater than the area of the triangle AMm, as AC
times the side AC itself is greater than AC times the side AD; or,
which is the same, as AC is greater than AD (Prin. 7th of Geom.
GEOMETRY. 177
l>rop. page 63). But the side AD is, by construction, one fifth of
AC ; therefore the area of the triangle AMm is also one fifth of the
area of the triangle ABC. In like manner it may be proved, that
the triangle ANra is two fifths of the triangle ABC ; the triangle
AOo three fifths, and the triangle APp four fifths of it, from wbich
the rest follows of course.
Remark. If the triangle ABC is not to be divided into -<ual
parts, but according to a given proportion, it will merely be d pes-
sary, as may be readily seen from the above, to divide the lint \C
according to this proportion, and then proceed as has been ah>*uy
shown.
Problem XXXVIII. To divide a parallelogram *nto
a given number of equal parts, and in such a way, t%at
the lines of division may be parallel to two opposite sides
of the parallelogi^am.
Solution. Let ABCD B C f </ h i C
be the given parallelogram ; /
let the number of parts be / /
six ; ami let AS, CD, be the I I I I I I I
sides, to which the lines of A EE GUI B
division shall be parallel.
Divide one of the two other sides, say AD, into six
equal parts, in E, F, G, H, I, and from these points draw
the lines Ee, Yf, Gg, H//, I*, parallel to the sides AB,
CD ; then the division is done.
Remark. If it is required to divide the parallelogram according
to a given proportion, it will merely be necessary, instead of divid-
ing the line AD into equal parts, to divide it according to the given
proportion, and then proceed as before.
Problem XXXIX. To divide a parallelogram, ac-
cording to a given proportion, by a line which shall be
parallel to a line given in position.
178
GEOMETRY.
Aa * E
D
Solution. Let ABCD be the parallelogram to be
divided.
1. Divide one of its sides, say AD, according to the
given proportion ; let the point of division be in z.
2. Make zE equal to the distance Az, and draw BE.
Now, if the line BE has the required position, the tri-
angle ABE and the quadrilateral BCDE are the parts
sought.
3. But if the line of division is required to be parallel
to the line xy, bisect the line BE in t, and through this
point draw the line GH parallel to xy ; then the two
quadrilaterals ABHG, HCDG, will be the required parts.
Demox. Draw EK parallel to AB. Then the two parallel-
ograms ABEK, ABCD, having the same height, their areas are
in proportion to their hases, AE, AD (see page 88, 7th) ; that is,
we have
parallel. ABEK : parallel. ABCD = AE : AD ;
therefore
- £ of parallel. ABEK : parallel. ABCD = £ of AE : AD ;
and because the triangle AEB is equal to half the parallelogram
ABEK, and half of AE is, by construction, equal to Az, we have
triangle AEB : parallel. ABCD = Az : AD.
This last proportion expresses that the area of the parallelogram
ABCD is as many times greater than the area of the triangle
ABE, as the line AD is greater than Az ; consequently if BE has
the required position, the triangle ABE is one of the required
parts, and therefore the trapezoid BEDC the other.
Further, the line BE is (by construction 3) bisected ; the an-
GEOMETRY,
179
gles u and n are opposite angles at the vertex, and w and s
are alternate angles (page 31, 2d) ; therefore the triangle BtH,
having the side Bt, and the two adjacent angles, to and u, equal
to the side tE, and the two adjacent angles, n and s, in the
triangle GtE, these two triangles are equal to one another ; conse-
quently the area of the trapezoid ABGH (composed of the quadri-
lateral ABGt, and the triangle B/H) is equal to the area of the
triangle ABE (composed of the same quadrilateral ABGt and
the equal triangle GtE), which proves the correctness of con-
struction 3.
Problem XL. To divide a trapezoid into a given
number of equal parts, so that the lines of division may
be parallel to the parallel sides of that trapezoid.
[This problem may be omitted by the younger pupils.]
Solution. Let ABCD be the given trapezoid which
is to be divided into three equal parts.
1. Upon AB, the greater of the two parallel sides, de-
scribe a semicircle ; draw DE parallel to CB ; and from
B, with the radius BE, describe the arc of a circle, EF,
cutting the semicircle in F.
2. From F draw FG perpendicular to AB, and divide
180 GEOMETRY.
the part AG of the line AB, into three equal parts
in K and I ; from these points draw the perpendiculars
Kk, li.
3. Upon AB, from B towards A, take the distances
Bm, Bn, equal to Bk> Bi ; from the points m and n, draw
the lines mO, wM, parallel to BC ; and from the points
O, M, in which these parallels meet the side AD, the
lines MN, OP, parallel to AB : then ABNM, MNPO,
OPCD, are the three required parts of the trapezoid
ABCD
Demon. Extend the lines AD, BC, until they meet in Z. Then
the triangles DCZ, OFZ, MNZ, ABZ, are all similar to each other
(page 70) ; further, we have (by construction 3)
DC equal to BE and to BF,
OP « « Bra " " Bk,
MN « " Bn " « Bi.
The areas of the two similar triangles OPZ, CDZ, are in the ratio
of the squares upon the corresponding sides; that is, we have the
proportion
triangle OPZ : triangle DCZ = OP X OP : CD X CD ;
and since OP is equal to Bk, and CD to BF, also
triangle OPZ : triangle DCZ = Bk X B/e : BF X BF.
Imagine AF and FB joined ; the triangle AFB would be right-
angled in F, and we should have the proportion
BG : BF = BF : AB ;
and for the same reason we have
BK : Bk = Bk: AB.
Taking the product of the mean and extreme terms of the two
lastpropoitions, we have
BG X AB equal to BF X BF, and
BK x AB « « BA X Bk
Let us now take our first proportion,
triangle OPZ : triangle DCZ = Bk X Bk : BF X BF ;
and let us write BG X AB, instead of BF X BF (its equal), and
BK X AB, instead ofBk X B/c, and we shall have
triangle OPZ : triangle DCZ = AB X BK : AB X BG,
whence
triangle OPZ : triangle DCZ = BK : BG;
GEOMETRY. 181
consequently, also,
triangle OPZ — triangle DCZ : triangle DCZ = BK — BG : UB;
(Principle 6th of Geom. Prop, page 62) ; which is read thus :
triangle OPZ, less the triangle DCZ, is to the triangle DCZ,
as the line BK, less the line BG, is to the line BG;
that is, trapezoid DOPC : triangle DCZ= GK : BG;
and as GK is (by construction 2) equal to J of AG,
trapezoid DOPC : triangle DCZ = £ AG : BG.
In like manner it may be proved that
trapezoid DMNC : triangle DCZ = § AG : BG and
trapezoid DABC : triangle DCZ = AG : BG.
These proportions express that the three trapezoids DOPC,
DMNC, DABC, are to each other in the same proportion as one
third is to two thirds to three thirds ; or, which is the same, as one
is to two, to three ; whence the rest of the demonstration follows
of course.
Remark. If it is required to divide the trapezoid ABCD not
into equal parts, but according to a given proportion, it will only
be necessary to divide the line AG in this proportion, and then
proceed as before.
Problem XLI. To divide a given figure into tie*
parts according to' a given proportion, and in such a way,
that one of the parts may be similar to the zohole figure.
D
Solution. Let ABCDE be the given figure.
1. Divide one side of the figure, say AB, according to
the given proportion ; let the point of division be Z.
2. Upon AB, as a diameter, describe a semicircle, and
16
183 GEOMETRY.
from Z draw the perpendicular ZM, meeting the semi-
circle in M.
3 Make A6 ±= AM, and upon A& describe a figure,
Abcde, which is similar to the given one, ABCDE (see
Problem XXXIII) ; the line bcde divides the figure in
the manner required.
Demon. The areas of the two similar figures Abcde, ABCDE,
are to each other, as the squares upon their corresponding sides
(page 98) ; therefore we have the proportion
ABCDE : Abcde = AB X AB : A6 X A&.
Draw AM and BM ; then AM is a mean proportional between
AZ and AB ; that is, we have
AZ : AM = AM : AB ;
and as A6 is, by construction, equal to AM,
AZ: Aft — A&: AB;
consequently the product A6 X A6 is equal to AZ X AB.
Writing AZ X AB, instead of Aft X Aft (its equal), in the first
proportion, we have
ABCDE : Abcde = AB X AB : AB X AZ.
Hence ABCDE : Abcde = AB : AZ ; and therefore
ABCDE — Abcde : Abcde = AB — AZ : AZ ;
which is read thus :
ABCDE, less Abcde, is to Abcde as AB, less AZ, is to AZ ;
that is,
BCDEedcb is to Abcde as ZB is to AZ ;
consequently the figure ABCDE is divided according to the given
proportion in which the line AB is divided.
PART IV.
Construction of triangfe$.
Problem XLII. The three sides of a triangle being
given, to construct the triangle.
GEOMETRY.
183
AAB
Solution. Let AB, AC,
BC, be the three given sides
of the triangle.
1. Take any side, say
AB. and from A as a centre,
with the radius AC, describe
an arc of a circle.
2. From B, as a centre, with the radius BC, describe
another arc, cutting the first.
4. From the point of intersection C, draw the straight
lines CA, CB ; the triangle ABC is the one required.
The demonstration follows immediately from Query 4th, Sect. II.
B
Problem XLIII. Two sides, and the angle included
by them, being given, to construct the triangle.
A A
Solution. Let AB, AC, be the two given sides, and
x the angle included by them.
1. Construct an angle equal to the angle x (Problem
VI) ; make one of the legs equal to the side AB, and the
other to the side AC.
2. Join BC ; the triangle ABC is the one required.
The demonstration follows from Query 1, Sect II.
184
GEOMETRY.
Problem XLI V. One side and the two atfyaesnt
angles being given, to construct the triangle
Solution. Let AB be the given side, and x and y
the two adjacent angles.
1. At the two extremities of the line AB, construct
the angles x and y, and extend their legs, AC, BC, until
they meet in the point C ; the triangle ABC is the one
required.
The demonstration follows from Query 2, Sect. II.
Problem XLV. Two sides, and the angle opposite to
the greater of them, being given, to construct the triangle.
Solution. Let AC, BC (see the figure to Problem
XLIII) be two given sides, and x the angle, which is
opposite to the greater of them (the side BC).
1. Upon an indefinite straight line construct an angle
equal to the angle x.
2. Make the leg AC of this angle equal to the smaller
side AC, and from C as a centre, with the radius GB
equal to the greater side, describe an arc of a circle,
cutting the line AB m the point B.
3. Join BC ; the triangle ABC is the one required.
The demonstration follows from Query 10th, Sect. 11.
GEOMETRY.
185
Problem XLVI. The basis of a triangle, one of the
adjacent angles, and the height being given, to construct
the triangle.
c
M EC
BB
Solution. Let AB be the given basis, x one of the
adjacent angles, and cd the height.
1. In any point of the line AB, draw a perpendicular,
CD, equal to cd ; and through C a line parallel to AB.
2. In A make an angle equal to the given angle x,
and extend the leg AE until it meets the line MC.
3. Join EB ; the triangle AEB is the one required.
The demonstration is sufficiently evident from the construction.
Problem XLVII. The basis, the angle opposite to it 9
and the height of a triangle being given, to construct the
triangle.
As^
m
Solution. Let AB be the given base, x the angle
opposite to it, and ran the height of the triangle.
1 . Upon the base AB describe a segment of a circle
containing a given angle x (see Problem XVII).
2. In A draw a perpendicular, AD, equal to the given
height mn, and through D draw DE parallel to AB.
16*
186
GEOMETRY.
3. From C and E, where this parallel cuts the segment,
draw the straight lines CA, CB, EA, BE ; either of the
two triangles ACB, AEB, will be the one required.
The demonstration follows from the construction.
Problem XLVIII. The basis of a triangle, the angle
opposite to it, and the ratio of the two other sides being
given, to construct the triangle.
Solution. Let AB be the given basis, x the angle
opposite to it ; and let the two remaining sides bear to
each other the same ratio which exists between the two
lines mn and rq.
1. Upon AB describe a segment of a circle capable
of the given angle x (see Problem XVII).
2. In B make an angle, ABE, equal to the angle x ;
make BE equal to the line rq, BD equal to mn, and join
DE.
3. From A draw the line AC parallel to DE, and from
the point C, where it meets the segment, draw the line
CB ; the triangle ABC is the one required.
Demon. The triangle ABC is similar to the triangle DBE ;
because the two angles CAB and ACB, in the one, are equal to the
two angles BDE, DBE, in the other, each to each* (page 73, 1st) ;
therefore we have the proportion
* CAB and EDB being alternate angles, and each of the angles,
ACB, DBE, being made equal to the given angle x.
\M.
ft
GEOMETRY.
187
AC : BC = BE : BD,
which expresses that the two sides, AC, BC, of the triangle are in
the same ratio as the sides BE, BD, of the triangle DEB ; conse-
quently they are also as the lines rq, mn ; because BE and BD
are, by construction, equal to mn, rq. The rest of the demonstra-
tion is evident from the construction.
Problem XLIX. The basis of a triangle, the angle
opposite to it, and the square, which, in area, is equal to
the rectangle of the two remaining sides, being given, to
construct the triangle.*
[Let the younger pupils omit this problem.]
Solution. Let AB be the given base, x the angle
opposite to it, and ad the side of the square, equal to the
rectangle of the two remaining sides.
1. Upon AB construct the segment, AKHB, of a cir-
cle, capable of the given angle x.
2. Extend AB towards D, and in A draw the perpen-
dicular AF.
3. Make AC equal to the radius AO, AD to the side
ad of the given square, and AE to half of AD ; join EC,
and from D draw DF parallel to EC.
4. Through the point F, where this parallel meets the
perpendicular, draw FH parallel to AB ; and from the
points K and H, where this meets the segment, the lines
* By the rectangle of the two remaining sides is meant a rectan-
gle, whose base is one of these sides, and whose height is the other.
188 GEOMETRY.
AK, KB, AH, HB; then either of the two triangles AKB,
AHB, is the one required.
Demojv. Draw the diameter AL, and from either of the points
K, H, say H, let fall the perpendicular HM upon AB. The trian-
gle ALH is similar to the triangle MBH ; for the triangle ALH
being inscribed in a semicircle, each of these triangles is right-
angled, and the two angles ALH, ABH, are equal ; because both
of them measure half as many degrees as the arc AKH (page 111,
1st); therefore the remaining angles, HAL and MHB, are also
equal (page 73, 1st) ; and the corresponding sides of the two trian-
gles ALH, MBH, are in the geometrical proportion
AH : AL = HM : HB ;
consequently we have
ALX HM=AHxHB.
This proportion expresses, that the area of the rectangle, which
has for its base the diameter AL, and its height equal to the height
HM of the right-angled triangle AHB, is equal to the area of the
rectangle, which has the side AH for its base, aid the side HB
for its height.* Further, it is easy to perceive that, from the similar
triangles ACE, ADF, we have the proportion
AD : AF = AC : AE ;
consequently, also,
AD : AF = 2AC : 2AE ,
therefore,
2AC X AF = 2AE X AD ; or
diam. AL X AF = adX ad,
(because AC is equal to the radius Ao of the circle, and AE is half
of AD, and AD is equal to ad).
From this proportion it follows, that the area of the square upon
ad, is equal to that of the rectangle of AL by AF, or MH its equal
(see the figure) ; and as the rectangle AH by HB is equal to that
of AL by HM, as we have proved above, it must also be equal to
the square upon ad. The same may be proved of the rectangle of
the two sides AK, KB, of the triangle AKB.
The rest of the demonstration is sufficiently evident from the
construction.
* For the area of a rectangle is found by multiplying the base by
the height.
GEOMETRY. igg
APPENDIX.
Containing Exercises for the Slate.
1. The side of a square being 12 feet, what is its area!
2. What, if the side is 12 rods, miles, &c?
3. What is the side of a square, whose area is one
square foot?
4. What, that of a square, whose area is one square
yard, rod, mile, &c. ?
5. What, that of a square of 4, 9, 16, 25, 36, 49, 64,
81, lOOsqare feet?
6. What is the area of a rectangle, whose base is 50
feet 3 inches, and whose height 10 feet 4 inches ?
7. What, that of a rectangle, whose base is 40 feet 3
inches, and whose height is 12£ feet ?
8. If the area of a rectangle is 240 square feet 19
square inches, and its basis measures 30 feet, what is its
height ?
9. What is the basis of a rectangle, whose height is
10 feet, and whose area is 40 square feet ?
10. What is the area of a rectangle, whose basis is 4
feet, and whose height is 3 inches ?
11. What is the area of a parallelogram of 10 feet
basis, and 3 feet 4 inches high ?
12. The height of a parallelogram is 5 feet, and the
area 40 square feet : what is its basis ?
13. The sum of the two parallel sides of a trapezoid is
12 feet, and their distance 3 feet 4 inches : what is the
area of the trapezoid ?
14. The area of a trapezoid is 24 square feet, and its
height is 4 inches, 3 seconds : what is the sum of its bases 1
190 GEOMETRY.
15. What is the difference between a triangle whose
basis is 10 feet 3 inches, and height 9 feet, and a trian*
gle of 3 feet basis, and 11 inches height?
16. What is the difference between a trapezoid, the
sum of the two parallel sides of which is 14 feet 3 inches,
and height 9 inches, and a square upon 9 inches'?
17. What is the sum of the areas of a triangle of 3 feet
basis, and 9 inches height; a square upon 14 feet 3
inches, and a rectangle whose basis is 3 feet 2 inches,
and height 1 foot 4 inches ?
18. What is the area of a circle, whose radius is 9
inches 1
19. What that of a circle, whose radius is 10 feet?
20. What that of a circle, whose radius is 9 feet 6
inches ?
21. The area of a circle is 240 square feet : what is its
radius or diameter?*
22. The radius of a circle is 5 feet 8 inches : what is
its circumfereuce ?
23. What is the length of an arc of 14 degrees 29
minutes 24 seconds, in a circle whose radius is 14 inches ?
24. What that of an arc of 6 degrees 9 seconds, in a
circle whose radius is 1 foot ?
25. What that of an arc of 9 seconds, in a circle whose
radius is 1 mile ?
20. What is the area of a sector of 15 degrees, in a
circle whose radius is 3 feet ?
27. What that of a sector of 19 degrees 45 minutes, in
a circle whose radius is 1 foot 3 inches ?
The teacher may now vary and multiply these questions.
* Divide the area by n (see the note to page 132), and extract
the square root of the quotient, the answer is the radius of the
circle.
END.
*
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