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ELEMENTARY SYNTHETIC GEOMETRY.
ELEMENTARY
SYNTHETIC GEOMETRY
OF THE
POINT, LINE AND CIRCLE IN
THE PLANE
BY
N. F. DUPUIS, M.A, F.R.S.C.
PROFESSOR OF PURE MATHEMATICS IN THE UNIVERSITY OF QUEEN's
COLLEGE, KINGSTON, CANADA.
MACMILLAN AND CO.
AND NEW YORK.
1889.
[Aii rights reset fed]
CAJORl
PREFACE.
The present work is a result of the Author's experience
in teaching Geometry to Junior Classes in the University
for a series of years. It is not an edition of " Euclid's
Elements," and has in fact little relation to that cele
brated ancient work except in the subject matter.
The work differs also from the majority of modern
treatises on Geometry in several respects.
The point, the line, and the curve lying in a common
plane are taken as the geometric elements of Plane
Geometry, and any one of these or any combination of
them is defined as a geometric plane figure. Thus a
triangle is not the threecornered portion of the plane
inclosed within its sides, but the combination of the
three points and three lines forming what are usually
termed its vertices and its sides and sides produced.
This mode of considering geometric figures leads
911298
VI PREFACE.
naturally to the idea of a figure as a locus, and con
sequently prepares the way for the study of Cartesian
Geometry. It requires, however, that a careful distinction
be drawn between figures which are capable of super
position and those which are equal merely in area.
The properties of congruence and equality are accord
ingly carefully distinguished.
The principle of motion in the transformation of
geometric figures, as recommended by Dr. Sylvester,
and as a consequence the principle of continuity are
freely employed, and an attempt is made to generalize
all theorems which admit of generalization.
An endeavour is made to connect Geometry with
Algebraic forms and symbols., (i) by an elementary
study of the modes of representing geometric ideas in
the symbols of Algebra, and (2) by determining the
consequent geometric interpretation which is to be given
to each interpretable algebraic form. The use of such
forms and symbols not only shortens the statements
of geometric relations but also conduces to greater
generality.
In dealing with proportion the method of measures
is employed in preference to that of multiples as being
PREFACE. Vll
equally accurate, easier of comprehension, and more
in line with elementary mathematical study. In dealing
with ratio I have ventured, when comparing two finite
lines, to introduce Hamilton's word tensor as seeming to
me to express most clearly what is meant.
After treating of proportion I have not hesitated to
employ those special ratios known as trigonometric func
tions in deducing geometric relations.
In the earlier parts of the work Constructive Geometry
is separated from Descriptive Geometry, and short
descriptions are given of the more important geometric
drawinginstruments, having special reference to the
geometric principles of their actions.
Parts IV. and V. contain a synthetic treatment of the
theories of the mean centre, of inverse figures, of pole
and polar, of harmonic division, etc., as applied to the
line and circle ; and it is believed that a student who
becomes acquainted with these geometric extensions in
this their simpler form will be greatly assisted in the
wider discussion of them in analytical conies. Through
out the whole work modern terminology and modern
processes have been used with the greatest freedom,
regard being had in all cases to perspicuity.
viii PREFACE.
As is evident from what has been said, the whole
intention in preparing the work has been to furnish the
student with that kind of geometric knowledge which
may enable him to take up most successfully the modern
works on Analytical Geometry.
N. F. D
Queen's Coli.k;ge,
Kingston, Canada.
CONTENTS.
PART I.
I'AGE
Section L— The Line and Point. Section II.— Two
Lines — Angles. Section III. — Three or more
Lines and Determined Points — The Triangle.
Section IV.— Parallels. Section V.— The Circle.
Section VI. — Constructive Geometry, . . . i
PART II.
Section I.— Comparison of Areas. Section II.
iVIeasurement of Lengths and Areas. Section
111. — Geometric Interpretation of Algebraic Forms.
Section IV. — Areal Relations — Squares and Rect
angles. Section V. — Constructive Geometry, . 91
PART III.
Section I. — Proportion amongst Line  Segments.
Section II. — Functions of Angles and their
Applications in Geometry, 147
X CONTENTS.
PART IV.
PAGE
Section I. — Geometric Extensions. Section II. —
Centre of Mean Position. Section III.— Col
linearity and Concurrence. Section IV.— Inver
sion and Inverse Figures. Section V. — Pole and
Polar. Section VI.— The Radical Axis. Sec
tion VII. — Centres and Axes of Perspective or
Similitude, 178
PART V.
Section I.— Anharmonic Division. Section II. —
Harmonic Ratio. Section III. — Anharmonic
Properties. Section IV.— Polar Reciprocals and
Reciprocation. Section V. — Homography and
Involution, 252
ELEMENTARY SYNTHETIC GEOMETRY.
PART I
GENERAL CONSIDERATIONS.
1°. A statement which explains the sense in which some
word or phrase is employed is 2i definition.
A definition may select some one meaning out of several
attached to a common word, or it may introduce some
technical term to be used in a particular sense.
Some terms, such as space, straight, direction, etc., which
express elementary ideas cannot be defined.
2°. Def. — A Theorem is the formal statement of some
mathematical relation.
A theorem may be stated for the purpose of being sub
sequently proved, or it may be deduced from some previous
course of reasoning.
In the former case it is called a Proposition^ that is, some
thing proposed, and consists of {a) the statement or enuncia
tion of the theorem, and {b) the argument or proof. The
purpose of the argument is to show that the truth of the
theorem depends upon that of some preceding theorem
whose truth has already been established or admitted.
Ex. " The sum of two odd numbers is an even number "
is a theorem.
3°. A theorem so elementary as to be generally accepted as
true without any formal proof, is an axiom.
« A
2 SYNTHETIC GEOMETRY.
Mathematical axioms are general or particular, that is,
they apply to the whole science of mathematics, or have
special applications to some department.
The principal general axioms are : —
i. The whole is equal to the sum of all its parts, and
therefore greater than any one of its parts.
ii. Things equal to the same thing are equal to one
another,
iii. If equals be added to equals the sums are equal,
iv. If equals be taken from equals the remainders are
equal.
V. If equals be added to or taken from unequals the
results are unequal,
vi. If unequals be taken from equals the remainders are
unequal,
vii. Equal multiples of equals are equal ; so also equal
submultiples of equals are equal.
The axioms which belong particularly to geometry will
occur in the sequel.
4°. The statement of any theorem may be put into the
hypothetical form, of which the type is—
If A is B then C is D.
The first part " if A is B " is called the hypothesis^ and the
second part " then C is D " is the conclusion.
Ex. The theorem " The product of two odd numbers is
an odd number " can be arranged thus : —
Hyp. If two numbers are each an odd number.
Concl. Then their product is an odd number.
5°. The statement " If A is B then C is D " may be im
mediately put into the form —
If C is not D then A is not B,
which is called the contrapositive of the former.
The truth of a theorem establishes the truth of its contra
GENERAL CONSIDERATIONS. 3
positive, and vice versa, and hence if either is proved the
other is proved also.
6°. Two theorems are converse to one another when the
hypothesis and conclusion of the one are respectively the
conclusion and hypothesis of the other.
Ex. If an animal is a horse it has four legs.
Converse. If an animal has four legs it is a horse.
As is readily seen from the foregoing example, the truth of
a theorem does not necessarily establish the truth of its con
verse, and hence a theorem and its converse have in general
to be proved separately. But on account of the peculiar
relation existing between the two, a relation exists also
between the modes of proof for the two. These are known
as the direct and indirect modes of proof. And if any
theorem which admits of a converse can be proved directly
its converse can usually be proved indirectly. Examples
will occur hereafter.
7°. Many geometric theorems are so connected with their
converses that the truth of the theorems establishes that of
the converses, and vice versa.
The necessary connection is expressed in the Rule of
Identity i its statement being : —
If there is but 07ie X and one V, and if it is proved
that X is V, then it follows that V is X.
Where X and Y stand for phrases such as may form the
hypotheses or conclusions of theorems, and the " is " between
them is to be variously interpreted as " equal to," " corre
sponds to," etc.
Ex. Of two sides of a triangle only one can be the greater,
and of the two angles opposite these sides only one can be
the greater. Then, if it is proved that the greater side is
opposite the greater angle it follows that the greater angle is
opposite the greater side.
4 SYNTHETIC GEOMETRY.
In this example there is but one X (the greater side) and
one Y (the greater angle), and as X is (corresponds to or is
opposite) Y, therefore Y is (corresponds to or is opposite) X.
8°. A Corollary is a theorem deduced from some other
theorem, usually by some qualification or restriction, and
occasionally by some amplification of the hypothesis. Or a
corollary may be derived directly from an axiom or from a
definition.
As a matter of course no sharp distinction can be drawn
between theorems and corollaries.
Ex. From the theorem, " The product of two odd numbers
is an odd number," by making the two numbers equal we
obtain as a corollary, " The square of an odd number is an
odd number."
Exercises.
State the contrapositives and the converses of the follow
ing theorems : —
1. The sum of two odd numbers is an even number.
2. A diameter is the longest chord in a circle.
3. Parallel lines never meet.
4. Every point equidistant from the endpoints of a line
segment is on the right bisector of that segment.
SECTION I.
THE LINE AND POINT.
9°. Space may be defined to be that which admits of
length or distance in every direction ; so that length and
direction are fundamental ideas in studying the geometric
properties of space.
THE LINE AND POINT. 5
Every material object exists in, and is surrounded by
space. The limit which separates a material object from
the space which surrounds it, or which separates the space
occupied by the object from the space not occupied by it, is
a surface.
The surface of a blackboard is the limit which separates
the blackboard from the space lying without it. This surface
can have no thickness, as in such a case it would include a
part of the board or of the space without or of both, and
would not be the dividing Hmit.
io°. A flat surface, as that of a blackboard, is a plane
surface, or a Plaiie,
Pictures of geometric relations drawn on a plane surface
as that of a blackboard are usually called Plane Geometric
Figures^ because these figures lie in or on a plane.
Some such figures are known to every person under such
names as " triangle," " square," " circle," etc.
11°. That part of mathematics which treats of the properties
and relations of plane geometric figures is Plane Geometry.
Such is the subject of this work.
The plane upon which the figures are supposed to lie will
be referred to as the plane, and unless otherwise stated all
figures will be supposed to lie in or on the same plane.
12°. The Line. When the crayon is drawn along the
blackboard it leaves a visible mark. This mark has breadth
and occupies some of the surface upon which it is drawn,
and by way of distinction is called a physical line. By
continually diminishing the breadth of the physical line we
make it approximate to the geometric line. Hence we may
consider the geometric line as being the limit towards which
a physical line approaches as its breadth is continually
diminished. We may consequently consider a geometric
line as length abstracted from every other consideration.
6 SYNTHETIC GEOMETRY.
This theoretic relation of a geometric line to a physical
one is of some importance, as whatever is true for the
physical line, independently of its breadth, is true for the
geometric line. And hence arguments in regard to geometric
lines may be replaced by arguments in regard to physical
lines, if from such arguments we exclude everything that
would involve the idea of breadth.
The diagrams employed to direct and assist us in geo
metric investigations are formed of physical lines, but they
may equally well be supposed to be formed of threads, wires
or light rods, if we do not involve in our arguments any idea
of the breadth or thickness of the lines, threads, wires or
rods employed.
In the practical applications of Geometry the diagrams
frequently become material or represent material objects.
Thus in Mechanics we consider such things as levers, wedges,
wheels, cords, etc., and our diagrams become representations
of these things.
A pulley or wheel becomes a circle, its arms become radii
of the circle, and its centre the centre of the circle ; stretched
cords become straight lines, etc.
13°. The Point. A point marks position, but has no size.
The intersection of one line by another gives a point, called
the point of intersection.
If the lines are physical, the point is physical and has some
size, but when the lines are geometric the point is also
geometric.
14°. Straight Line. For want of a better definition we
may say that a straight line is one of which every part has
the same direction. For every part of a line must have some
direction, and when this direction is common to all the parts
of the line, the line is straight.
The word " direction " is not in itself definable, and when
applied to a line in the absolute it is not intelligible. But
THE LINE AND POINT. 7
every person knows what is meant by such expressions as
"the same direction," "opposite direction," etc., for these
express relations between directions, and such relations are
as readily comprehended as relations between lengths or
other magnitudes.
The most prominent property, and in fact the distinctive
property of a straight line, is the absolute sameness which
characterizes all its parts, so that two portions of the same
straight line can differ from one another in no respect except
in length.
Def. — A plane figure made up of straight lines only is
called a rectilinear figure.
15°. A Curve is a line of which no part is straight ; or a
curve is a line of which no two adjacent parts have the same
direction.
The most common example of a curve is a circle or portion
of a circle.
Henceforward, the word "line," unless otherwise qualified,
will mean a straight line.
16°. The "rule" or "straightedge" is a strip of wood,
metal, or other solid with one ^d^g^^ made straight. Its
common use is to guide the pen or pencil in drawing lines
in Practical Geometry.
17°. A Plane is a surface such that the line joining any two
arbitrary points in it coincides wholly with the surface.
The planarity of a surface may be tested by applying the
rule to it. If the rule touches the surface at some points and
not at others the sujface is not a plane. But if the rule
touches the surface throughout its whole length, and in every
position and direction in which it can be applied, the surface
is a plane.
The most accurately plane artificial surface known is
probably that of a wellformed plane mirror. Examina
8 SYNTHETIC GEOMETRY.
tion of the images of objects as seen in such mirrors is
capable of detecting variations from the plane, so minute as
to escape all other tests.
1 8°. A surface which is not plane, and which is not com
posed of planes, is a curved surface. Such is the surface of
a sphere, or cylinder.
19°. The point, the line, the curve, the plane and the
curved surface are the elements which go to make up geo
metric figures.
Where a single plane is the only surface concerned, the
point and line lie in it and form a plane figure. But where
more than one plane is concerned, or where a curved surface
is concerned, the figure occupies space, as a cube or a sphere,
and is called a spatial figure or a solid.
The study of spatial figures constitutes Solid Geometry, or
the Geometry of Space, as distinguished from Plane Geometry.
20°. Given Point and Line. A point or line is said to be
given when we are made to know enough about it to enable
us to distinguish it from every other point or line ; and the
data which give a point or line are commonly said to
determine it.
A similar nomenclature applies to other geometric ele
ments.
The statement that a point or line lies in a plane does not
give it, but a point or line placed in the plane for future
reference is considered as being given. Such a point is
usually called an origin^ and such a line a datum line^ an
initial lijie, a pri^ne vector ^ etc.
21°. Def. I. — A line considered merely as a geometric
element, and without any limitations, is an indefinite line.
2. — A limited portion of a line, especially when any refer
ence is had to its length, is 2i finite line, or a linesegment, or
simply a segment.
THE LINE AND POINT. 9
That absolute sameness (14°) which characterizes every
part of a line leads directly to the following conclusions : —
(i) No distinction can be made between any two segments
of the same line equal in length, except that of position
in the line.
(2) A line cannot return into, or cross itself.
(3) A line is not necessarily limited in length, and hence,
in imagination, we may follow a line as far as we
please without coming to any necessary termination.
This property is conveniently expressed by saying
that a line extends to infinity.
3. — The hypothetical endpoints of any indefinite line are
said to be points at infinity. All other points are finite
points.
22°. Notation. A point is denoted by a single letter where
ever practicable, as " the point A."
An indefinite line is also denoted by a single letter as "the
line L," but in this case the letter l
has no reference to any point. a b
A segment is denoted by naming its end points, as the
" segment AB," where A and B are the e7id points. This is
a biliteral., or twoletter notation.
A segment is also denoted by a single letter, when the
limits of its length are supposed to be known, as the " seg
ment rt." This is a uniliteral, or oneletter notation.
The term "segment" involves the notion of some finite
length. When length is not under consideration, the term
" line " is preferred.
Thus the " Hne AB " is the indefinite line having A and B
as two points upon it. But the "segment AB" is that portion
of the line which lies between A and B.
23°. In dealing with a hnesegment, we frequently have to
consider other portions of the indefinite line of which the
segment is a part.
lO SYNTHETIC GEOMETRY.
As an example, let it be required to divide the segment AB
, __, , into two parts whereof one shall
^ ^ ^ c be twice as long as the other.
To do this we put C in such a position that it may be twice as
far from one of the endpoints of the segment, A say, as it is
from the other, B. But on the indefinite line through A and
B we may place C so as to be twice as far from A as from B.
So that we have two points, C and C, both satisfying the
condition of being twice as far from A as from B.
Evidently, the point C does not divide the segment AB in
the sense commonly attached to the word divide. But on
account of the similar relations held by C and C to the end
points of the segment, it is convenient and advantageous to
consider both points as dividing the segment AB.
When thus considered, C is said to divide the segment
internally and C to divide it externally in the same manner.
24°. Axiom. — Through a given point only one line can
pass in a given direction.
Let A be the given point, and let the segment AP mark
— V. the given direction. Then, of all the lines
that can pass through the point A, only one
can have the direction AP, and this one must lie along and
coincide with AP so as to form with it virtually but one line.
Cor. I. A finite point and a direction determine one
line.
Cor. 2. Two given finite points determine one line. For, if
A and P be the points, the direction AP is given, and hence
the line through A and having the direction AP is given.
Cor. 3. Two lines by their intersection determine one finite
point. For, if they determined two, they would each pass
through the same two points, which, from Cor. 2, is impossible.
Cor. 4. Another statement of Cor. 2 is — Two lines which
have two points in common coincide and form virtually but
one line.
THE LINE AND POINT. II
25°. Axiom. — A straight line is the shortest distance be
tween two given points.
Although it is possible to give a reasonable proof of this
axiom, no amount of proof could make its truth more
apparent.
The following will illustrate the axiom. Assume any two
points on a thread taken as a physical line. By separating
these as far as possible, the thread takes the form which we
call straight, or tends to take that form. Therefore a straight
finite line has its endpoints further apart than a curved line
of equal length. Or, a less length of line will reach from one
given point to another when the line is straight than when it
is curved.
Def. — The distance between two points is the length of the
segment which connects them or has them as endpoints.
26°. Superpositiojt. — Comparison of Figjcres.—V\[e assume
that space is homogeneous, or that all its parts are alike, so
that the properties of a geometric figure are independent of its
position in space. And hence we assume that a figure may
be supposed to be moved from place to place, and to be turned
around or over in any way without undergoing any change
whatever in its form or properties, or in the relations existing
between its several parts.
The imaginary placing of one figure upon another so
as to compare the two is called siiperpositioti. By superposi
tion we are enabled to compare figures as to their equality or
inequality. If one figure can be superimposed upon another
so as to coincide with the latter in every part, the two figures
are necessarily and identically equal, and become virtually
one figure by the superposition.
27°. Two linesegments can be compared with respect to
length only. Hence a line is called a magnitude of one
dimension.
Two segments are equal when the endpoints of one can be
12 SYNTHETIC GEOMETRY.
made to coincide with the endpoints of the other by super
position.
28°. Def. — The stun of two segments is that segment which
is equal to the two when placed in line with one endpoint in
each coincident.
Let AB and DE be two segments, and on the line of which
D E AB is a segment let BC be equal to
^ DE. Then AC is the stem of AB
"a b c and DE.
This is expressed symbolically by writing
AC = AB + DE,
where = denotes equality in length, and + denotes the
placing of the segments AB and DE in line so as to have one
common point as an endpoint for each. The interpretation
of the whole is, that AC is equal in length to AB and DE
together.
29°. Def, — The difference between two segments is the
segment which remains when, from the longer of the segments,
a part is taken away equal in length to the shorter.
Thus, if AC and DE be two segments of which AC is the
longer, and if BC is equal to DE, then AB is the difference
between AC and DE.
This is expressed symbolically by writing
AB=ACDE,
which is interpreted as meaning that the segment AB is
shorter than AC by the segment DE.
Now this is equivalent to saying that AC is longer than
AB by the segment DE, or that AC is equal to the sum of
AB and DE.
Hence when we have AB = AC DE
we can write AC = ABlDE.
We thus see that in using these algebraic symbols, =, +,
and , a term, as DE, may be transferred from one side of
THE LINE AND POINT. 1 3
the equation to another by changing its sign from + to  or
vice versa.
Owing to the readiness with which these symbolic expres
sions can be manipulated, they seem to represent simple alge
braic relations, hence beginners are apt to think that the work
ing rules of algebra must apply to them as a matter of
necessity. It must be remembered, however, that the formal
rules of algebra are founded upon the properties of numbers,
and that we should not assume, without examination, that
these rules apply without modification to that which is not
number.
This subject will be discussed in Part II.
30°. Def. — That point, in a linesegment, which is equi
distant from the endpoints is the 7niddle point of the segment.
It is also called the internal point of bisection of the seg
ment, or, when spoken of alone, simply \h^ point of bisection.
Exercises.
1. If two segments be in line and have one common end
point, by what name will you call the distance between
their other endpoints .'*
2. Obtain any relation between " the sum and the differ
ence" of two segments and "the relative directions "of
the two segments, they being in line.
3. A given linesegment has but one middle point
4. In Art. 23°, if C becomes the middle point of AB, what
becomes of C 1
5. In Art. 30° the internal point of bisection is spoken of.
What meaning can you give to the " external point of
bisection " 1
14 SYNTHETIC GEOMETRY.
SECTION 11.
RELATIONS OF TWO LINES.— ANGLES.
31°. When two lines have not the same direction they are
said to make an angle with one another, and an angle is a
c differejice in direction.
'^ Illustratio7i.'—'L^\. A and B
represent two stars, and E the
A' *A position of an observer's eye.
Since the lines EA and EB, which join the eye and the
stars, have not the same direction they make an angle with
one another at E.
1. If the stars appear to recede from one another, the angle
at E becomes greater. Thus, if B moves into the position of
C, the angle between EA and EC is greater than the angle
between EA and EB,
Similarly, if the stars appear to approach one another, the
angle at E becomes smaller ; and if the stars become coinci
dent, or situated in the same line through E, the angle at E
vanishes.
Hence an angle is capable of continuous increase or
diminution, and is therefore a magnitude. And, being
magnitudes, angles are capable of being compared with one
another as to greatness, and hence, of being measured.
2. If B is moved to B', any point on EB, and A to A', any
point on EA, the angle at E is not changed. Hence increas
ing or diminishing one or both of the segments which form
an angle does not affect the magnitude of the angle.
Hence, also, there is no community in kind between an
angle and a linesegment or a line.
Hence, also, an angle cannot be measured by means of
linesegments or lines.
RELATIONS OF TWO LINES. — ANGLES. 1 5
32°. Def. — A line which changes its direction in a plane
while passing through a fixed point in the plane is said to
rotate about the point.
The point about which the rotation takes place is the pole^
and any segment of the rotating line, having the pole as an
endpoint, is a radius vector.
Let an inextensible thread fixed at O
be kept stretched by a pencil at P.
Then, when P moves, keeping the
thread straight, OP becomes a radius
vector rotating about the pole O. ^ "^
When the vector rotates from direction OP to direction
OP' it describes the angle between OP and OP'. Hence we
have the following : —
Def. I. — The angle between two lines is the I'otation neces
sary to bring one of the lines into the direction of the other.
The word " rotation," as employed in this definition, means
the amount of turning effected, and not the process of turning.
Def. 2. — For convenience the lines OP and OP', which, by
their difference in direction form the angle, are called the
arms of the angle, and the point O where the arms meet is
ih.^ vertex.
Cor. From 31°, 2, an angle does not in any way depend
upon the lengths of its arms, but only upon their relative
directions.
33°. Notation of Angles. — i. The symbol z. is used for the
word " angle."
2. When two segments meet at a vertex the angle between
them may be denoted by a single letter ^^^^^^B
placed at the vertex, as the ^O , or by
a letter with or without an arch of dots,
as /^S ; or by three letters of which the
extreme ones denote points upon the arms of the angle and
the middle one denotes the vertex, as lAOB.
l6 SYNTHETIC GEOMETRY.
3. The angle between two lines, when the vertex is not
pictured, or not referred to, is expressed by l{L. M), or LM,
where L and M denote the lines in the oneletter notation
(22°); or z.(AB, CD), where AB and CD denote the Hnes in
the twoletter notation.
Two angles are equal when the arms of the one
may be made to coincide in direction respec
tively with the arms of the other ; or when
the angles are described by the same rotation.
Thus, if, when C is placed upon O, and
O'A' is made to lie along OA, O'B' can also
be made to lie along OB, the ^lA'O'B' is equal
to lAOB. This equality is symbolized thus :
ZA'0'B'=^A0B.
Where the sign = is to be interpreted as
indicating the possibility of coincidence by superposition.
35". Sum and Difference of Angles. — The sutn of two
angles is the angle described by a radius vector which
describes the two angles, or their equals, in succession.
sp' Thus if a radius vector starts from co
incidence with OA and rotates into
direction OP it describes the zJVOP.
If it next rotates into direction OP' it
d A describes the /.POP'. But in its whole
rotation it has described the zAOP'. Therefore,
^AOP' = ^AOPH^POP'.
Similarly, ^AOP = ^AOP'^POP'.
Z>i?/:— When two angles, as AOP and POP', have one arm
in common lying between the remaining arms, the angles are
adjacent angles.
36°. Def. — A radius vector which starts from any given
direction and makes a complete rotation so as to return to its
original direction describes a circumangley ox perigon.
RELATIONS OF TWO LINES.— ANGLES. 1 7
Onehalf of a circumangle is a straight angle, and one
fourth of a circumangle is a right angle.
37°. Theorem. — If any number of lines meet in a point,
the sum of all the adjacent angles formed is a circumangle.
OA, OB, OC, ..., OF are lines meeting
in O. Then
^AOB + ^BOCf^COD + ...f^FOA
= a circumangle. D.
Proof. — A radius vector which starts
from coincidence with OA and rotates into
the successive directions, OB, OC, ..., 1^ ^
OF, OA describes in succession the angles AOB, BOC, ...,
EOF, FOA.
But in its complete rotation it describes a circumangle (36°).
/AOB + ^BOC + ... + Z.FOA = a circumangle. q.e.d.
Cor. The result may be thus stated : —
The sum of all the adjacent angles about a point in
the plane is a circumangle.
38°. Theorem. — The sum of all the adjacent angles on one
side of a line, and about a point in the
line is a straight angle.
O is a point in the line AB ; then
^AOC + iLCOB = a straight angle.
Proof. — Let A and B be any two points
in the line, and let the figure formed by X*'
AB and OC be revolved about AB without displacing the
points A and B, so that OC may come into a position OC.
Then (24°, Cor. 2) O is not displaced by the revolution,
^AOC = ^AOC', and ^BOC = ^BOC';
.lAOC + :LBOC = £.AOC' + ^BOC',
and since the sum of the four angles is a circumangle {zT^^
therefore the sum of each pair is a straight angle (36°). q.e.d.
Cor. I. The angle between the opposite directions of a line
is a straight angle.
B
1 8 SYNTHETIC GEOMETRY.
Cor. 2. If a radius vector be rotated until its direction is
reversed it describes a straight angle. And conversely, if a
radius vector describes a straight angle its original direction
is reversed.
Thus, if OA rotates through a straight angle it comes into
the direction OB. And conversely, if it rotates from direction
OA to direction OB it describes a straight angle.
39°. When two lines L and M cut one
another four angles are formed about the
point of intersection, any one of which
may be taken to be the angle between
the lines.
These four angles consist of two pairs of opposite or
vertical angles, viz., A, A', and B, B', A being opposite A',
and B being opposite B'.
40°. Theorem. — The opposite angles of a pair formed by
two intersecting lines are equal to one another.
Proof.— ^A + ^B = a straight angle (38°)
and Z.A' + A.B = a straight angle. (38°)
lA=^A',
and ^B = ^B'. q.e.d.
Def. I. — Two angles which together make up a straight
angle are supplementary to one another, and one is called the
supplement of the other. Thus, A is the supplement of B',
and B of A'.
Cor. If aA = aB, then Z.A' = ^B=z.B', and all four angles
are equal, and each is a right angle (36°).
Therefore, if two adjacent angles formed by two intersect
ing lines are equal to each other, all four of the angles so
formed are equal to one another, and each is a right angle.
Def. 2. — When two intersecting lines form a right angle at
their point of intersection, they are said to be perpendicular
to one another, and each is perpendicular to the other.
RELATIONS OF TWO LINES. — ANGLES. IQ
Perpendicularity is denoted by the symbol X, to be read
"perpendicular to" or "is perpendicular to."
A right angle is denoted by the symbol ~\.
The symbol ± also denotes two right angles or a straight
angle.
De/. 3. — When two angles together make up a right angle
they are complementary to one another, and each is the
C07nplement of the other.
The right angle is the simplest of all angles, for when two
lines form an angle they form four angles equal in opposite
pairs. But if any one of these is a right angle, all four are
right angles.
Perpendicularity is the most important directional relation
in the applications of Geometry.
Def. 4. — An acute angle is less than a right angle, and an
obtuse angle is greater than a right angle, and less than two
right angles.
41°. From (36°) we have
I circumangle = 2 straight angles
=4 right angles.
In estimating an angle numerically it may be expressed in
any one of the given units.
If a right angle be taken as the unit, a circumangle is
expressed by 4, i.e. four right angles, and a straight angle
by 2.
Angles less than a right angle may be expressed, approxi
mately at least, by fractions, or as fractional parts of the
right angle.
For practical purposes the right angle is divided into 90
equal parts called degrees ; each degree is divided into 60
equal parts called minutes ; and each minute into 60 equal
parts called seconds.
Thus an angle which is oneseventh of a circumangle
contains fiftyone degrees, twentyfive minutes, and fortytwo
20 SYNTHETIC GEOMETRY.
seconds and sixsevenths of a second. This is denoted as
follows : — 51° 25' 42f".
42°. Theorem. — Through a given point in a line only one
perpendicular can be drawn to the line.
The hne OC is _L AB, and OD is any
other line through O.
Then OD is not _L AB.
Proof.— 'Y\i^ angles BOC and COA
are each right angles (40°, Def. 2).
Therefore BOD is not a right angle, and OD is not ± AB.
But OD is any line other than OC.
Therefore OC is the only perpendicular. q.e.d.
Def. — The perpendicular to a linesegment through its
middle point is the right bisector of the segment.
Since a segment has but one middle point (30°, Ex. 3), and
since but one perpendicular can be drawn to the segment
through that point,
.*. a Hitesegmeiit has but one right bisector,
43°. Def. — The lines which pass through the vertex of an
angle and make equal angles with the arms, are the bisectors
of the angle. The one which lies within the angle is the
internal bisector, and the one lying without is the external
bisector.
Let AOC be a given angle ; and
■£ let EOF be so drawn that ^AOE
= Z.EOC.
EF is the internal bisector of the
angle AOC.
^ Also, let GOH be so drawn that
D ^COG = ^HOA.
HG is the external bisector of the angle AOC.
z.COG = ^HOA (hyp.)
and z.HOA = z_GOB, (40°)
^COG=^GOB;
RELATIONS OF TWO LINES. — ANGLES. 21
and the external bisector of AOC is the internal bisector of
its supplementary angle, COB, and vice versa.
The reason for calling GH a bisector of the angle AOC is
given in the definition, viz., GH makes equal angles with the
arms. Also, OA and OC are only parts of indefinite lines,
whose angle of intersection may be taken as the Z.AOC or
as the lCOB.
44°. Just as in 23° we found two points which are said
to divide the segment in the same manner, so we may find
two lines dividing a given angle in the same manner, one
dividing it internally, and the other externally.
Thus, if OE is so drawn that the Zj\OE is double the
/_EOC, some line OG may also be drawn so that the ^AOG
is double the Z.GOC.
This double relation in the division of a segment or an
angle is of the highest importance in Geometry.
45°. Theorem. — The bisectors of an angle are perpendicular
to one another.
EF and GH are bisectors of the ZAOC ;
then EF is ± GH.
Proof.
and
adding,
But
Exercises.
1. Three lines pass through a common point and divide the
plane into 6 equal angles. Express the value of each
angle in right angles, and in degrees.
2. OA and OB make an angle of 30°, how many degrees are
there in the angle made by OA and the external bisector
of the angle AOB.^
LS.OC
= i^OC, V OE
is a
bisector ;
LCOG:
=^COB, V OG
is a
bisector ;
Z.EOG:
= ^AOB.
Z.AOB
is a straight angle,
(38°,
Cor. i)
^EOG
is a right angle.
(36°)
22 SYNTHETIC GEOMETRY.
3. What is the supplement of 13° 27' 42"? What is its
complement ?
4. Two lines make an angle a with one another, and the
bisectors of the angle are drawn, and again the bisectors
of the angle between these bisectors. What are the
angles between these latter lines and the original ones ?
5. The lines L, M intersect at O, and through O, V and M'
are drawn J_ respectively to L and M. The angle be
tween L' and M' is equal to that between L and M.
SECTION III.
THREE OR MORE POINTS AND LINES.
THE TRIANGLE.
46°. Theorem. — Three points determine at most three lines ;
and three lines determine at most three points.
Proof I. — Since (24°, Cor. 2) two points determine one
line, three points detemiine as many lines
as we can form groups from three points
taken two and two.
Let A, B, C be the points ; the groups
are AB, BC, and CA.
Therefore three points determine at most three lines.
2. — Since (24°, Cor. 3) two lines determine one point, three
lines determine as many points as we can form groups from
three lines taken two and two.
But if L, M, N be the Hnes the groups are LM, MN, and
NL.
Therefore three lines determine by their intersections at
most three points.
THREE OR MORE POINTS AND LINES.
23
47°. Theorem. — Four points determine at most six lines ; and
four lines determine at most six points.
Proof.—!. Let A, B, C, D be the four
points. The groups of two are AB, AC,
AD, BC, BD, and CD ; or six in all.
Therefore six lines at most are deter
mined.
2. Let L, M, N, K be the lines. The
groups of two that can be made are KL,
KM, KN, LM, LN, and MN ; or six in all.
Therefore six points of intersection at
most are determined.
Cor. In the first case the six lines determined pass by
threes through the four points. And in the second case the
six points determined lie by threes upon the four lines.
This reciprocality of property is very suggestive, and in the
higher Geometry is of special importance.
Ex. Show that 5 points determine at most 10 lines, and 5
lines determine at most 10 points. And that in the first case
the lines pass by fours through each point ; and in the latter,
the points lie by fours on each line.
48°. Def. — A triangle is the figure formed by three lines
and the determined points, or by three points and the deter
mined lines.
The points are the vertices of the triangle, and the line
segments which have the points as endpoints are the sides.
The remaining portions of the determined lines are usually
spoken of as the "sides produced." But in many cases
generality requires us to extend the term \
" side " to the whole line. ^
Thus, the points A, B, C are the
vertices of the triangle ABC.
The segments AB, BC, CA are the
sides. The portions AE, BF, CD, etc.,
extending outwards as far as required, arc the sides produced.
24 SYNTHETIC GEOMETRY.
The triangle is distinctive in being the rectilinear figure for
which a given number of lines determines the same number
of points, or vice versa.
Hence when the three points, forming the vertices, are
given, or when the three lines or linesegments forming the
sides are given, the triangle is com
pletely given.
This is not the case with a rectilinear
figure having any number of vertices
other than three.
If the vertices be four in number,
with the restriction that each vertex is
determined by the intersection of two
sides, any one of the figures in the
^ margin will satisfy the conditions.
Hence the giving of the four vertices of such a figure is
not sufficient to completely determine the figure.
49°. Def.—i. The angles ABC, BCA, CAB are the in
ternal angles of the triangle, or simply the angles of the
triangle.
2. The angle DCB, and others of like kind, are external
angles of the triangle.
3. In relation to the external angle DCB, the angle BCA
is the adjacent interjial angle, while the angles CAB and
ABC are opposite iJiternal angles.
4. Any side of a triangle may be taken as its base, and then
the angles at the extremities of the base are its basal angles,
and the angle opposite the base is the vertical angle. The
vertex of the vertical angle is the vertex of the triangle
when spoken of in relation to the base.
50°. Notation. — The symbol A is commonly used for the
word triangle. In certain cases, which are always readily
apprehended, it denotes the area of the triangle.
THREE OR MORE POINTS AND LINES. 25
The angles of the triangle are denoted usually by the
capital letters A, B, C, and the sides opposite by the cor
responding small letters a, b^ c.
51°. Def. — When two figures compared by superposition
coincide in all their parts and become virtually but one figure
they are said to be congruent.
Congruent figures are distinguishable from one another
only by their position in space and are said to be identically
equal.
Congruence is denoted by the algebraic symbol of identity,
= ; and this symbol placed between two figures capable of
congruence denotes that the figures are congruent.
Closed figures, like triangles, admit of comparison in two
ways. The first is as to their capability of perfect coinci
dence ; when this is satisfied the figures are congruent. The
second is as to the magnitude or extent of the portions of the
plane enclosed by the figures. Equality in this respect is
expressed by saying that the figures are equal.
When only one kind of comparison is possible, as is the
case with linesegments and angles, the word equal is used.
CONGRUENCE AMONGST TRIANGLES.
52°. Theorem. — Two triangles are congruent when two
sides and the included angle in the one are respectively equal
to two sides and the included angle in the other.
If AB=A'B" the triangles y^B /^B'
BC = B'C \ are congru
and z.B = z.B' J ent.
Proof.— V\2.CQ AABC on A ~c a^ C
AA'B'C so that B coincides with B', and BA lies along B'A'.
z.B = /lB', BC lies along B'C, (34°)
and V AB = A'B' and BC = B'C';
A coincides with A' and C with C, (27")
26 SYNTHETIC GEOMETRY.
and .*. AC lies along A'C ; (24°, Cor. 2)
and the As coinciding in all their parts are congruent. (51°)
q.e.d.
Cor. Since two congruent triangles can be made to coin
cide in all their parts, therefore —
When two triangles have two sides and the included angle
in the one respectively equal to two sides and the included
angle in the other, all the parts in the one are respectively
equal to the corresponding parts in the other.
53°. Theorem. — Every point upon the right bisector of a
segment is equidistant from the endpoints of the segment
AB is a linesegment, and P is any point
on its right bisector PC. Then PA = PB.
Proof.— In the As APC and BPC,
AC = CB, (42°, Def.)
z_ACP = z_BCP, (42°, Def.)
and PC is common to both As ;
AAPC = ABPC, (52°)
and.. PA = PB. (52°, Cor.) ^.^.^/.
Def. I. — A triangle which has two sides equal to one an
other is an isosceles triangle.
Thus the triangle APB is isosceles.
The side AB, which is not one of the equal sides, is called
the base.
Cor. I. Since the AAPC=ABPC,
^A=^B.
Hence the basal angles of an isosceles triangle are equal to
one another.
Cor. 2. From (52°, Cor.), ^APC = ZBPC ;
Therefore the right bisector of the base of an isosceles tri
angle is the internal bisector of the vertical angle. And since
these two bisectors are one and the same line the converse is
true.
THREE OR MORE POINTS AND LINES.
27
Def. 2. — A triangle in which all the sides are equal to one
another is an equilateral triangle.
Cor. 3. Since an equilateral triangle is isosceles with re
spect to each side as base, all the angles of an equilateral
triangle are equal to one another ; or, an equilateral triangle
is equiangular.
54°. Theorem. — Every point equidistant from the end
points of a linesegment is on the right bisector of that
segment. (Converse of 53°.)
PA = PB. Then P is on the right bisector
ofAB. Q^
Proof. — If P is not on the right bisector of
AB, let the right bisector cut AP in Q.
Then QA=QB, (53°)
but PA=PB, (hyp.)
QP = PBQB,
or PB = QP + QB,
which is not true. (25°, Ax.)
Therefore the right bisector of AB does not cut AP ; and
similarly it does not cut BP ; therefore it passes through P,
or P is on the right bisector. q.e.d.
This form of proof should be compared with that of Art.
53°, they being the kinds indicated in 6°.
This latter or indirect form is known as proof by reductio
ad absurdum (leading to an absurdity). In it we prove the
conclusion of the theorem to be true by showing that the
acceptance of any other conclusion leads us to some relation
which is absurd or untrue.
55°. Def. — The linesegment from a vertex of a triangle to
the middle of the opposite side is a median of the triangle.
Cor. I. Every triangle has three medians.
Cor. 2. The median to the base of an isosceles triangle is
28 SYNTHETIC GEOMETRY.
the right bisector of the base, and the internal bisector of the
vertical angle. (53°j Cor. 2.)
Cor. 3. The three medians of an equilateral triangle are
the three right bisectors of the sides, and the three internal
bisectors of the angles.
56°. Theorem.— \i two angles of a triangle are equal to one
another, the triangle is isosceles, and the equal sides are
opposite the equal angles. (Converse of 53°,
Cor. I.)
z.PAB=^PBA, then PA = PB.
Proof. — If P is on the right bisector of
AB, PA=PB. (53°)
If P is not on the right bisector, let AP
A fc B cut the right bisector in Q.
Then QA=QB, and ^QAB=z.QBA. (53° and Cor. i)
But ^PBA = ^QAB; (hyp.)
z.PBA = ^QBA,
which is not true unless P and Q coincide.
Therefore if P is not on the right bisector of AB, the
Z.PAB cannot be equal to the ZPBA.
But they are equal by hypothesis ;
P is on the right bisector,
and PA=PB. q.e.d.
Cor. If all the angles of a triangle are equal to one another,
all the sides are equal to one another.
Or, an equiangular triangle is equilateral.
57°. From 53° and 56° it follows that equality amongst the
sides of a triangle is accompanied by equality amongst the
angles opposite these sides, and conversely.
Also, that if no two sides of a triangle are equal to one
another, then no two angles are equal to one another, and
conversely.
THREE OR MORE POINTS AND LINES.
29
Def. — A triangle which has no two sides equal to one
another is a scalene triangle.
Hence a scalene triangle has no two angles equal to one
another.
58°. Theorem. — If two triangles have the three sides in the
one respectively equal to the three sides in the other the
triangles are congruent. b b'
If A'B' = AB]the As ABC
B'C' = BC hand A'B'C are
C'A' = CA J congruent. A<
Proof.— iMXVi the AA'B'C
over and place A' on A, and
A'C along AC, and let B' fall at some point D.
A'C' = AC,C falls ate, (27°)
and AADC is the AA'B'C in its reversed position.
Since AB = AD and CB  CD,
A and C are on the right bisector of BD, and AC is the right
bisector of BD. (54°)
Z.BAC =z.DAC ; (53°, Cor. 2)
and the As BAC and DAC are congruent. (52°)
AABC = AA'B'C. q.e.d.
59°. Theorem. — If two triangles have two angles and the
included side in the one equal respectively to two angles and
the included side in the other, the triangles are congruent.
If lM^lK'X
and
A'C'=ACJ
the As ABC and A'B'C are congruent.
A c A' c'
Proof. — Place A' on A, and A'C along AC.
30
SYNTHETIC GEOMETRY.
Because A'C' = AC, C coincides with C ;
and V lA.'=lI^, A'B' lies along AB ;
and V lC=lC, CV>' lies along CB ;
B' coincides with B,
and the triangles are congruent.
(27°)
(24°, Cor. 3)
q.e.d.
60°. Theorem. — An external angle of a triangle is greater
than an internal opposite angle.
The external angle BCD is greater
than the internal opposite angle ABC or
BAG.
Proof. — Let BF be a median produced
until FG = BF.
Then the As ABF and GGF have
BF = FG, (construction)
AF = FC, • (55°)
and z.BFA = ^GFC. (40°)
AABF=ACGF, (52°)
and z.FCG = ^BAC. (52°, Cor.)
But /LACE is greater than Z.FCG.
z.ACEis>^BAC.
Similarly, aBCD is>^ABC,
and ^BGD=^ACE. (40°)
Therefore the ^s BCD and ACE are each greater than each
of the Z.S ABC and BAG. q.e.d.
61°. Theorem. — Only one perpendicular can be drawn to a
line from a point not on the line.
Proof.— L^t B be the point and AD the line ;
and let BG be _L to AD, and BA be any line
other than BG.
Then aBCD is>z.BAG, (60°)
^BAG is not a ~,
and BA is not ± to AD.
But BA is any line other than BG ;
BC is the only perpendicular from B to AD. q.e.d
A
THREE OR MORE POINTS AND LINES. 3 1
Cor. Combining this result with that of 42° we have —
Through a given point only one perpendicular can be drawn
to a given line.
62°. Theorem.— Oi any two unequal sides of a triangle
and the opposite angles —
1. The greater angle is opposite the longer side.
2. The longer side is opposite the greater angle.
1. BAis>BC; b
then Z.C is > ^A.
Proof.— L^t BD = BC.
Then the A^DC is isosceles, ** "^^^
and z_BDC = z.BCD. (53°, Cor. i)
But z.BDCis>z.A, (60°)
and ^BCAis>^BCD;
/LBCAis>z.BAC;
or, Z.C is>ZA. q.e.d.
2. lQ is > ^A ; then AB is > BC.
Proof. — From the Rule of Identity (7°), since there is but
one longer side and one greater angle, and since it is shown
(i) that the greater angle is opposite the longer side, therefore
the longer side is opposite the greater angle. q.e.d.
Cor. I. In any scalene triangle the sides being unequal to
one another, the greatest angle is opposite the longest side,
and the longest side is opposite the greatest angle.
Also, the shortest side is opposite the smallest angle, and
conversely.
Hence if A, B, C denote the angles, and a, b, c the sides
respectively opposite, the order of magnitude of A, B, C is
the same as that of a., b, c.
63°. Theorem. — Of all the segments between a given point
and a line not passing through the point —
1. The perpendicular to the line is the shortest.
2. Of any two segments the one which meets the line
I SYNTHETIC GEOMETRY.
further from the perpendicular is the longer ; and con
versely, the longer meets the line further from the
perpendicular than the shorter does.
3. Two, and only two segments can be equal, and they lie
upon opposite sides of the perpendicular.
P P is any point and BC a line not pass
ing through it, and PA is _L to BC.
I. PA which is J_ to BC is shorter
than any segment PB which is not ±
"b to BC.
ZPAC = ^PAB=~1 (hyp.)
z.PACis>^PBC; (60°)
^PABis>^PBC,
and .. PB is > PA. (62°, 2) q.e.d.
2. AC is > AB, then also PC is > PB.
P;^^— Since AC is > AB, let D be the point in AC
sothatAD = AB.
Then A is the middle point of BD, and PA is the right
bisector of BD. (42°, Def.)
PD = PB (53°)
and aPDB=^PBD. (53°, Cor i)
But ^PDBis>^PCB;
A.PBD is>z.PCB,
and PCis>PB. (62°, 2)
The converse follows from the Rule of Identity. q.e.d.
3. Proof. — In 2 it is proved that PD = PB. Therefore two
equal segments can be drawn from any point P to the line
BC ; and these lie upon opposite sides of PA.
No other segment can be drawn equal to PD or PB. For
it must lie upon the same side of the perpendicular, PA, as
one of them. If it lies further from the perpendicular than
this one it is longer, (2), and if it lies nearer the perpendicular
it is shorter. Therefore it must coincide with one of them
and is not a third line. q.e.d
THREE OR MORE POINTS AND LINES. 33
■l^^/'—The length of the perpendicular segment between
any point and a line is the distance of the point from the
line.
64°. Theorevt.~\i two triangles have two angles in the one
respectively equal to two angles in the other, and a side
opposite an equal angle in each equal, the triangles are
congruent.
If /A'=^Althen the As
^C' = aC \h:^'C and ABC
and A'B' = AbJ are congruent.
Proof.— V\2iQ^ A' on A, and
A'B' along AB.
A'B' = AB, B' coincides with B.
Also, *.• z.A' = ^A, A'C lies along AC.
Now if O does not coincide with C, let it fall at some other
point, D, on AC.
Then, •.' AB = A'B', AD = A'C, and lA==lM,
.'. AA'B'C = AABD, (52°)
and ^ADB = z.C'. (52°, Cor.)
But lC=lC, (hyp.)
.. ^ADB = ^C,
which is not true unless D coincides with C.
Therefore C must fall at C, and the As ABC and A'B'C
are congruent.
The case in which D may be supposed to be a point on
AC produced is not necessary. For we may then super
impose the AABC on the AA'B'C.
65°. Theorem. — If two triangles have two sides in the one
respectively equal to two sides in the other, and an angle
opposite an equal side in each equal, then —
1. If the equal angles be opposite the longer of the two
sides in each, the triangles are congruent.
2. If the equal angles be opposite the shorter of the two
C
34
SYNTHETIC GEOMETRY.
sides in each, the triangles are not necessarily con
gruent.
A'B' = AB,
B'C = BC,
^A'=^A.
I. If BCis>AB,
AA'B'C = AABC.
Proof.— Since BC is > AB, therefore B'C is > A'B'.
Place A' on A and A'C along AC.
^A' = ^A, and A'B' = AB,
B' coincides with B. (34°, 27°)
Let BP be _L AC ;
then B'C cannot lie between BA and BP (63°, 2), but must
lie on the same side as BC ; and being equal to BC, the lines
B'C and BC coincide (63°, 3), and hence
AA'B'C' = AABC. q.eJ.
2. If BC is < AB, the As
A'B'C and ABC may or may
not be congruent.
Since AB is > BC,
(63°, 2)
Proof.
PA is > PC.
Let PD = PC,
then BD = BC.
Now, let AA'B'C be superimposed on AABC so that A'
coincides with A, B' with B, and A'C lies along AC. Then,
since we are not given the length of A'C, B'C may coincide
with BC, and the As A'B'C and ABC be congruent;
or B'C may coincide with BD, and the triangles A'B'C and
ABC be not congruent. q.e.d.
Hence when two triangles have two sides in the one
respectively equal to two sides in the other, and an angle
opposite one of the equal sides in each equal, the triangles
are not necessarily congruent unless some other relation
exists between them.
THREE OR. MORE POINTS AND LINES. 35
The first part of the theorem gives one of the suffici
ent relations. Others are given in the following cor
ollaries.
Cor. I. If lC is a~J> BC and BD (2nd Fig.) coincide along
BP, and the As ABD and ABC become one and the same.
Hence C must fall at C, and the As A'B'C and ABC are
congruent.
Cor. 2. The Z.BDA is supplementary to BDC and therefore
to BCA. And •.• lBDA is > Z.BPA, .'. ^BDA is greater
than a right angle, and the /.BCA is less than a right
angle.
Hence if, in addition to the equalities of the theorem, the
angles C and C are both equal to, or both greater or both
less than a right angle, the triangles are congruent.
De/.— Angles which are both greater than, or both equal to,
or both less than a right angle are said to be 0/ the same
affection.
66°. A triangle consists of six parts, three sides and three
angles. When two triangles are congruent all the parts in
the one are respectively equal to the corresponding parts in
the other. But in order to establish the congruence of two
triangles it is not necessary to establish independently the
respective equality of all the parts ; for, as has now been
shown, if certain of the corresponding parts be equal the
equality of the remaining parts and hence the congruence of
the triangles follow as a consequence. Thus it is sufficient
that two sides and the included angle in one triangle shall be
respectively equal to two sides and the included angle in
another. For, if we are given these parts, we are given con
sequentially all the parts of a triangle, since every triangle
having two sides and the included angle equal respectively to
those given is congruent with the given triangle.
Hence a triangle is given when two of its sides and the
angle between them are given.
36 SYNTHETIC GEOMETRY.
A triangle is given or detertnined by its elements being
given according to the following table : —
1. Three sides, (58°)
2. Two sides and the included angle, (52°)
3. Two angles and the included side, (59°)
4. Two angles and an opposite side, (64°)
5. Two sides and the angle opposite the longer side, (65°)
When the three parts given are two sides and the angle
opposite the shorter side, two triangles satisfy the conditions,
whereof one has the angle opposite the longer side supple
mentary to the corresponding angle in the other.
This is known as the ambiguous case in the solution of
triangles.
A study of the preceding table shows that a triangle is
completely given when any three of its six parts are given,
with two exceptions : —
(i) The three angles ;
(2) Two sides and the angle opposite the shorter of the
two sides.
67°. Theorem. — If two triangles have two sides in the one
respectively equal to two sides in the other, but the included
angles and the third sides unequal, then
1. The one having the greater included angle has the
greater third side.
2. Conversely, the one having the greater third side has the
greater included angle.
^8' A'B' = AB and B'C' = BC, and
I. ^ABC is > /lA'B'C,
then AC is > A'C.
Proof. — Let A' be placed upon
A and A'B' along AB.
E c "^^, Since A'B' = AB, B' falls on B.
"^ Let C fall at some point D.
Then ABD is A'B'C in its new position.
THREE OR MORE POINTS AND LINES. 37
Let BE bisect the ^DBC and meet AC in E.
Join DE.
Then, in the As DBE and CBE,
DB = BC, (hyp.)
^DBE = z.CBE, (constr.)
and BE is common.
ADBE = ACBE,
and DE = CE.
But AC = AE + EC = AE + ED,
which is greater than AD.
ACis>A'C'. q.eJ.
2. AC is > A'C, then ^ABC is greater than ^A'B'C
Proof. — The proof of this follows from the Rule of Identity.
68°. Theorem. — i. Every point upon a bisector of an angle
is equidistant from the arms of the angle.
2. Conversely, every point equidistant from the arms of
an angle is on one of the bisectors of the angle.
1. OP and OQ are bisectors of the angle AOB, and PA,
PB are perpendiculars from P upon
the arms. Then
PA=PB. ''X X^^P
Proof.— T\i^ As POA and FOB a'
are congruent, since they have two
angles and an opposite side equal in
each (64°); .. PA=PB.
If Q be a point on the bisector OQ it is shown in a similar
manner that the perpendiculars from Q upon the arms of the
angle AOB are equal. q.e.d.
2. If PA is i. to OA and PB is ± to OB, and PA = PB,
then PO is a bisector of the angle AOB.
Proof— 1\i^ As POA and POB are congruent, since they
have two sides and an angle opposite the longer equal in
each (65°, i) ; .'. z.POA = ^POB,
and PO bisects the ^AOB.
38 SYNTHETIC GEOMETRY.
Similarly, if the perpendiculars from Q upon OA ana OB
are equal, QO bisects the i^BOA', or is the external bisector
of the ^AOB. q.e.d.
LOCUS.
69°. A locus is the figure traced by a variable point, which
takes all possible positions subject to some constraining
condition.
If the point is confined to the plane the locus is one or
more lines, or some form of curve.
Illustration. — In the practical process of drawing a line or
curve by a pencil, the point of the pencil becomes a variable
(physical) point, and the line or curve traced is its locus.
In geometric applications the point, known as iht generat
ing point, moves according to some law.
The expression of this law in the Symbols of Algebra is
known as the equation to the locus.
Cor. I. The locus of a point in the plane, equidistant from
the endpoints of a given linesegment, is the right bisector
of that segment.
This appears from 54°.
Cor. 2. The locus of a point in the plane, equidistant from
two given lines, is the two bisectors of the angle formed by
the lines.
This appears from 68°, converse.
Exercises.
1. How many lines at most are determined by 5 points }
by 6 points ? by 12 points ?
2. How many points at most are determined by 6 lines "i by
12 lines .?
3. How many points are determined by 6 lines, three of which
pass through a common point 1
THREE OR MORE POINTS AND LINES. 39
4. How many angles altogether are about a triangle? How
many at most of these angles are different in magni
tude ? What is the least number of angles of different
magnitudes about a triangle ?
5. In Fig. of 53°, if Q be any point on PC, APAQ=APBQ.
6. In Fig. of 53°, if the APCB be revolved about PC as an
axis, it will become coincident with APCA.
7. The medians to the sides of an isosceles triangle are
equal to one another.
8. Prove 58° from the axiom " a straight line is the shortest
distance between two given points."
9. Show from 60° that a triangle cannot have two of its
angles right angles.
10. If a triangle has a right angle, the side opposite that angle
is greater than either of the other sides.
11. What is the locus of a point equidistant from two sides
of a triangle ?
12. Find the locus of a point which is twice as far from one
of two given lines as from the other.
13. Find the locus of a point equidistant from a given line
and a given point.
SECTION IV.
PARALLELS, ETC
70°. Def. — Two lines, in the same plane, which do not
intersect at any finite point are parallel.
Next to perpendicularity, parallelism is the most important
directional relation. It is denoted by the symbol , which is
to be read " parallel to " or " is parallel to " as occasion may
require.
The idea of parallelism is identical with that of sameness
40 SYNTHETIC GEOMETRY.
of direction. Two linesegments may differ in length or in
direction or in both.
If, irrespective of direction, they have the same length,
they are equal ; if, irrespective of length, they have the same
direction, they are parallel ; and if both length and direction
are the same they are equal and parallel. Now when two
segments are equal one may be made to coincide with the
other by superposition without change of length, whether
change of direction is required or not. So when they
are parallel one may be made to coincide with the other
without change of direction, whether change of length is
required or not.
Axiom. — Through a given point only one line can be
drawn parallel to a given line.
This axiom may be derived directly from 24°.
71°. Theorem. — Two lines which are perpendicular to the
same line are parallel.
L and M are both ± to N,
then L is  to M.
Proof. — If L and M meet at any point, two
. perpendiculars are drawn from that point to
the line N.
But this is impossible (61°).
Therefore L and M do not meet, or they are parallel.
Cor. All lines perpendicular to the same line are parallel
to one another.
72°. Theorem. — Two lines which are parallel are perpen
dicular to the same line, or they have a common perpendicular.
(Converse of 71°.)
L is II to M, and L is ± to N ;
then M is JL to N.
Proof.— \i M is not _L to N, through any point P in M, let
K be ± to N.
PARALLELS, ETC. 4 1
Then K is  to L. (71°)
But M is II to L. (hyp.)
Therefore K and M are both jj to L,
which is impossible unless K and M coincide. (70°, Ax.)
Therefore L and M are both J_ to N,
or N is a common perpendicular.
'j'^. Def. — A line which crosses two or more lines of
any system of lines is a transversal. E/
Thus EF is a transversal to the lines ^ ^/ g
AB and CD.
In general, the angles formed by ^
a transversal to any two lines are ^/h
distinguished as follows — /f
a and e^ c and^, b and/, d and h are pairs of corresponding
angles.
c and f^ e and d are pairs of alternate angles.
c and <?, ^and/are pairs of interadjacent angles.
74°. When a transversal crosses parallel lines —
1. The alternate angles are equal in pairs.
2. The corresponding angles are equal in pairs.
3. The sum of a pair of interadjacent angles is a straight
angle. g^
AB is II to CD and EF is a transversal, a p / b_
1. ^AEF = ^EFD.
Proof. — Through O, the middle point
of EF, draw PQ a common _L to AB
and CD. (72°)
Then AOPE = AOQF ; (64°)
zAEF = Z.EFD. q.e.d.
Similarly the remaining alternate angles are equal.
2. ^AEG = ^CFE, etc.
Proof.— Z.AE G = supplement of zJVEF, (40°, Def. i)
and z_CFE = supplement of lEFD.
42 SYNTHETIC GEOMETRY.
But zJVEF=^EFD ; (74°, i)
^AEG = ^CFE. q.e.d.
Similarly the other corresponding angles are equal in pairs.
3. z_AEF + ^CFE=±.
Proof.— ^AEF=^EFD, (74°,
and £.CFE + ^EFD=_L; (38°)
zAEF+^CFE=±. q.e.d.
Cor. It is seen from the theorem that the equality of a pair
of alternate angles determines the equality in pairs of corre
sponding angles, and also determines that the sum of a pair
of interadjacent angles shall be a straight angle. So that the
truth of any one of the statements i, 2, 3 determines the truth
of the other two, and hence if any one of the statements be
proved the others are indirectly proved also.
75°. Theorem. — If a transversal to two lines makes a pair
of alternate angles equal, the two lines are parallel. (Con
verse of 74° in part.)
If ^AEF = ZEFD, AB and CD are parallel.
/'/w/— Draw PQ as in 74°, ± to AB,
AOPE = AOQF; (59°)
.. :lOPE=^OQF=~l
Q and.. AB is 1 to CD. {7i°)q.e.d.
Cor. It follows from 74° Cor. that if a pair of corresponding
angles are equal to one another, or if the sum of a pair of
interadjacent angles is a straight angle, the two lines are
parallel.
76°. Theoreju. — The sum of the internal angles of a tri
angle is a straight angle.
ABC is a A ;
the^A + ^B + ^C=±.
A c D Proof.— l.^X CE be  to AB, and
D be any point on AC produced.
A
P /
B
/^
c
f/
PARALLELS, ETC. 43
Then BC is a transversal to the parallels AB and CE ;
zlABC=a.BCE. (74°, i)
Also, AC is a transversal to the same parallels ;
z.BAC=^ECD. (74°, 2)
^BC + ^BAC=^BCD
= supplement of Z.BCA.
lA + lB + lC=±. q.e.d.
Cor. An external angle of any triangle is equal to the sum
of the opposite internal angles. (49°, 3)
For ^BCD=^A+ZB.
T]". From the property that the sum of the three angles of
any triangle is a straight angle, and therefore constant, we
deduce the following —
1. When two angles of a triangle are given the third is
given also ; so that the giving of the third furnishes
no new information.
2. As two parts of a triangle are not sufficient to deter
mine it, a triangle is not determined by its three angles,
and hence one side, at least, must be given (66°, i).
3. The magnitude of any particular angle of a triangle does
not depend upon the size of the triangle, but upon
the form only, i.e., upon the relations amongst the
sides.
4. Two triangles may have their angles respectively equal
and not be congruent. But such triangles have the
same form and are said to be similar.
5. A triangle can have but one obtuse angle ; it is then
called an obtuseangled triangle.
A triangle can have but one right angle, when it is called
a rightangled triangle.
All other triangles are called acuteangled triangles, and
have three acute angles.
6. The acute angles in a rightangled triangle are comple
mentary to one another.
44
SYNTHETIC GEOMETRY.
78°. Theorem. — If a line cuts a given line it cuts every
parallel to the given line.
Let L cut M, and let N be any parallel
to M. Then L cuts N.
Proof. — If L does not cut N it is  to N.
But M is II to N. Therefore through the same point P two
lines L and M pass which are both  to N.
But this is impossible ; (70°, Ax.)
L cuts N.
And N is any line  to M.
L cuts every line  to M. q.e.d.
79°. Theorem. — If a transversal to two lines makes the
sum of a pair of interadjacent angles less than a straight
angle, the two lines meet upon that side
of the transversal upon which these inter
adjacent angles lie.
GH is a transversal to AB and CD,
and ^BEF^EFD<±.
Then AB and CD meet towards B and D.
Proof. — Let LK pass through E making Z.KEF = Z.EFC.
Then LKistoCD.
But AB cuts LK in E,
it cuts CD. (78°)
Again, •.* EB lies between the parallels, and AE does not,
the point where AB meets CD must be on the side BD of
the transversal. q.e.d.
Cor. Two lines, which are respectively perpendicular to
two intersecting lines, intersect at some finite point.
80°. Def — I. A closed figure having four
lines as sides is in general called a quadrangle
'c or quadrilateral.
in Thus ABCD is a quadrangle.
2. The linesegments AC and BD which join opposite
vertices are the diagonals of the quadrangle.
PARALLELS, ETC. 45
3. The quadrangle formed when two parallel lines intersect
two other parallel lines is a para lie logra7?t, and is usually
denoted by the symbol 1 7 .
81°. Theorem. — In any parallelogram —
1. The opposite sides are equal to one another.
2. The opposite internal angles are equal to one another.
3. The diagonals bisect one another. A_
AB is II to CD, and AC is  to BD, and
AD and BC are diagonals. Z'''^^^
1. Then AB = CDandAC = BD. c d
Proof. — •.• AD is a transversal to the parallels AB and CD,
^CDA=^DAB. (74°, i)
and •.' AD is a transversal to the parallels AC and BD,
^CAD=^ADB. (74°, i)
Hence, ACAD = ABDA. (59°)
AB = CD and AC = BD. q.e.d.
2. iLCAB = z.BDCand^ACD = ^DBA.
Proof.— It is shown in i that z.CAD=^DB and Z.BAD
= ^DC;
.*. by adding equals to equals,
^CAB = ^CDB.
Similarly, ^ACD = ^ABD. q.e.d.
3. AO = OD and BO = OC.
Proof The AAO C = AD O B ; (59°)
AO = OD andBO = OC. q.e.d.
82°. Def I.— A parallelogram which has two adjacent sides
equal is a rhombus. B^
Cor. I. Since AB = BC (hyp.)
= DC(8i°, i)=AD. A<
Therefore a rhombus has all its sides
equal to one another.
Cor. 2. Since AC is the right bisector of BD,
and BD the right bisector of AC,
46 SYNTHETIC GEOMETRY.
Therefore the diagonals of a rhombus bisect one another
at right angles.
Def. 2. — A parallelogram which has one right angle is a
rectangle^ and is denoted by the symbol en.
Cor. 3. Since the opposite angle is a ~~1, (81°, 2)
and the adjacent angle is a ""], (74°) 3)
Therefore a rectangle has all its angles right angles.
Cor. 4. The diagonals of a rectangle are equal to one
another.
Def. 3. — A rectangle with two adjacent sides equal is a
square^ denoted by the symbol □.
Cor. 5. Since the square is a particular form of the rhombus
and a particular form of the rectangle,
Therefore all the sides of a square are equal to one another;
all the angles of a square are right angles ; and the diagonals
of a square are equal, and bisect each other at right angles.
84°. Theorem. — If three parallel lines intercept equal seg
ments upon any one transversal they do so upon every
transversal.
AE is a transversal to the three parallels
AB, CD, and EF, so that AC = CE, and
BF is any other transversal. Then BI)
= DF.
Proof. — Let GDH passing through D
be II to AE.
Then AGDC and CDHE are Z=7s. (80°, 3)
GD=AC = CE = DH. (81°, i)
Also, ^GBD =^DFH, •.' AG is  to EF, (74°, i)
and ^BDG = ^FDH ; (40°)
ABDG = AFDH, (64°)
and BD = DF. q.e.d.
Def— The figure ABFE is a trapezoid.
Therefore a trapezoid is a quadrangle having only two
PARALLELS, ETC.
47
sides parallel. The parallel sides are the major and minor
bases of the figure.
Cor. I. Since
and
2CD=AG + EH,
=AB+BG+EFHF
BG=HF;
CD = (AB + EF).
Or, the linesegment joining the middle points of the non
parallel sides of a trapezoid is equal to onehalf the sum of
the parallel sides.
Cor. 2. When the transversals meet upon
one of the extreme parallels, the figure
AEF' becomes a A and CD' becomes a
line passing through the middle points of
the sides AE and AF', and parallel to the
base EF'.
Therefore, i, the line through the middle point of one side
of a triangle, parallel to a second side, bisects the third side.
And, 2, the line through the middle points of two sides of
a triangle is parallel to the third side.
85°. Theorem. — The three medians of a triangle pass
through a common point.
CF and AD are medians intersecting in O.
Then BO is the median to AC.
Proof.— L^\. BO cut AC in E, and let
AG II to FC meet BO in G. Join CG.
Then, BAG is a A and FO passes through
the middle of AB and is  to AG,
.. O is the middle of BG. (84^ Cor 2)
Again, DO passes through the middle
points of two sides of the ACBG,
CG is II to AO or OD ;
AOCG is a E=7,
and AE = EC;
BO is the median to AC.
(84°, Cor. 2)
(81°, 3)
q.e.d.
48 SYNTHETIC GEOMETRY.
Def. — When three or more lines meet in a point they are
said to be concurrent.
Therefore the three medians of a triangle are concurrent.
Def. 2. — The point of concurrence, O, of the medians of a
triangle is the centi'oid oi the triangle.
Cor. Since O is the middle point of BG, and E is the middle
point of OG, (8i°, 3)
OE = OB,
= 1EB.
Therefore the centroid of a triangle divides each median
at twothirds of its length from its vertex.
86°. Theorem. — The three right bisectors of the sides of a
B triangle are concurrent.
/ Proof. — Let L and N be the right
bisectors of BC and AB respectively.
Then L and N meet in some point O.
A ^ (79°, Cor.)
Since L is the right bisector of BC, and N of AB, O is
equidistant from B and C, and is also equidistant from A
and B. (53°)
Therefore O is equidistant from A and C, and is on the
right bisector of AC. (54°)
Therefore the three right bisectors meet at O. q.e.d.
Cor. Since two lines L and N can meet in only one point
(24°, Cor. 3), O is the only point in the plane equidistant
from A, B, and C.
Therefore only one finite point exists in the plane equi
distant from three given points in the plane.
Def. — The point O, for reasons given hereafter, is called
the ctrcumce7itre of the triangle ABC.
^j". Def. — The line through a vertex of a triangle per
pendicular to the opposite side is the perpendicular to that
side, and the part of that line intercepted within the triangle
is the altitude to that side.
PARALLELS, ETC. 49
Where no reference to length is made the word altitude is
often employed to denote the indefinite line forming the
perpendicular.
Hence a triangle has three altitudes, one to each side.
88°. Theorem. — The three altitudes of a triangle are con
current.
Proof.— Let ABC be a triangle.
Complete the zz::7s, ACBF,
ABDC, and ABCE.
Then •.• FB is 1 to AC,
and BD is  to AC,
FBD is one line, (70°, Ax.)
and FB = BD.
Similarly, DCE is one line and DC = CE,
and EAF is one line and EA = AF.
Now, ••• AC is II to FD, the altitude to AC is J_ to FD and
passes through B the middle point of FD. (72°)
Therefore the altitude to AC is the right bisector of FD,
and similarly the altitudes to AB and BC are the right bisec
tors of DE and EF respectively.
But the right bisectors of the sides of the ADEF are
concurrent (86°), therefore the altitudes of the AABC are
concurrent. q.e.d.
Def, — The point of concurrence of the altitudes of a tri
angle is the orthocentre of the triangle.
Cor. I. If a triangle is acuteangled {j']°^ 5), the circum
centre and orthocentre both lie within the triangle.
2. If a triangle is obtuseangled, the circumcentre and
orthocentre both lie without the triangle.
3. If a triangle is rightangled, the circumcentre is at the
middle point of the side opposite the right angle, and the
orthocentre is the rightangled vertex.
Def. — The side of a rightangled triangle opposite the right
angle is called the hypothenuse.
D
50
SYNTHETIC GEOMETRY.
89°. The definition of 80° admits of three different figures,
viz. : —
I. The normal quadrangle (i) in which each of the in
ternal angles is less than a straight angle. When not
(i) (2) (3)
otherwise qualified the term quadrangle will mean this
figure.
2. The quadrangle (2) in which one of the internal angles,
as at D, is greater than a straight angle. Such an angle in a
closed figure is called a reentrant angle. We will call this
an inverted quadrangle.
3. The quadrangle (3) in which two of the sides cross one
another. This will be called a crossed quadrangle.
In each figure AC and BD are the diagonals ^ so that both
diagonals are within in the normal quadrangle, one is within
and one without in the inverted quadrangle, and both are
without in the crossed quadrangle.
The general properties of the quadrangle are common to
all three forms, these forms being only variations of a more
general figure to be described hereafter.
90°. Theorem. — The sum of the internal angles of a quad
rangle is four right angles, or a circumangle.
Proof. — The angles of the two As ABD and CBD make
up the internal angles of the quadrangle.
But these are ± + _L ; (76°)
therefore the internal angles of the quadrangle are together
equal to four right angles. q.e.d.
Cor. This theorem applies to the inverted quadrangle as is
readily seen.
PARALLELS, ETC. 5 1
91°. Theorem. — If two lines be respectively perpendicular
to two other lines, the angle between the first two is equal
or supplementary to the angle between the last two.
BC is _L to AB
and CD is _L to AD.
Then ^(BC . CD) is equal or supplemen
tary to 4AB . AD).
Proof. — ABCD is a quadrangle, and the
Ls> at B and D are right angles ;
^BAD+^BCD = J_,
or Z.BCD is supplementary to Z.BAD.
But /.BCD is supplementary to ^ECD ;
and the ^(BC . CD) is either the angle BCD or DCE. (39°)
Z.(BC . CD) is = or supplementary to ^BAD. q.e.d.
Exercises.
1. ABC is a A, and A', B', C are the vertices of equilateral
As described outwards upon the sides BC, CA, and
AB respectively. Then AA'=BB' = CC'. (Use 52°.)
2. Is Ex. I true when the equilateral As are described
" inwardly " or upon the other sides of their bases ?
3. Two lines which are parallel to the same line are parallel
to one another.
4. L' and M' are two lines respectively parallel to L and M.
Thez.(L'.M')=^(L.M).
5. On a given line only two points can be equidistant from
a given point. How are they situated with respect to
the perpendicular from the given point 1
6. Any side of a A is greater than the difference between
the other two sides.
7. The sum of the segments from any point within a A to
the three vertices is less than the perimeter of the A
8. ABC is a A and P is a point within on the bisector of
z_A. Then the difference between PB and PC is less
than that between AB and AC, unless the A is isosceles.
52 SYNTHETIC GEOMETRY.
9. Is Ex. 8 true when the point P is without the A? but on
the same bisector ?
10. Examine Ex. 8 when P is on the external bisector of A,
and modify the wording of the exercise accordingly.
11. CE and CF are bisectors of the angle between AB and
CD, and EF is parallel to AB. Show that EF is
bisected by CD.
12. If the middle points of the sides of a A be joined two
and two, the A is divided into four congruent As.
13. From any point in a side of an equilateral A lines are
drawn parallel to the other sides. The perimeter of
the I 7 so formed is equal to twice a side of the A
14. Examine Ex. 13 when the point is on a side pro
duced.
15. The internal bisector of one angle of a A and the ex
ternal bisector of another angle meet at an angle which
is equal to onehalf the third angle of the A
16. O is the orthocentre of the AABC. Express the angles
AOB, BOC, and COA in terms of the angles A, B,
and C.
17. P is the circumcentre of the AABC. Express the angles
APB, BPC, and CPA in terms of the angles A, B,
and C.
1 8. The joins of the middle points of the opposite sides of
any quadrangle bisect one another.
T9. The median to the hypothenuse of a rightangled triangle
is equal to onehalf the hypothenuse.
20. If one diagonal of a 1 7 be equal to a side of the figure,
the other diagonal is greater than any side.
21. If any point other than the point of intersection of the
diagonals be taken in a quadrangle, the sum of the
linesegments joining it with the vertices is greater
than the sum of the diagonals.
22. If two rightangled As have the hypothenuse and an
acute angle in the one respectively equal to the like
parts in the other, the As are congruent.
PARALLELS, ETC. 53
23. The bisectors of two adjacent angles of a 1 7 are JL to
one another.
24. ABC is a A The angle between the external bisector
of B and the side AC is (C  A).
25. The external bisectors of B and C meet in D. Then
^BDC = i(B + C).
26. A line L which coincides with the side AB of the AABC
rotates about B until it coincides with BC, without at
any time crossing the triangle. Through what angle
does it rotate }
27. The angle required in Ex. 26 is an external angle of the
triangle. Show in this way that the sum of the three
external angles of a triangle is a circumangle, and that
the sum of the three internal angles is a straight angle.
28. What property of space is assumed in the proof of Ex. 27?
29. Prove 76° by assuming that AC rotates to AB by crossing
the triangle in its rotation, and that AB rotates to CB,
and finally CB rotates to CA in like manner.
SECTION V.
THE CIRCLE.
92°. Def. I. — A Circle is the locus of a point which, movin<
in the plane, keeps at a constant ____^==— b
distance from a fixed point in the
plane.
The compasses, whatever be their
form, furnish us with two points, A "^^a
and B, which, from the rigidity of _
the instrument, are supposed to
preserve an unvarying distance ^ ^
from one another. Then, if one of the points A is fixed,
while the other B moves over the paper or other plane
54 SYNTHETIC GEOMETRY.
surface, the moving point describes a physical circle. The
limit of this physical circle, when the curved line has its
thiclcness diminished endlessly, is the geometric circle.
Def. 2. — The fixed point is the centre of the circle, and the
distance between the fixed and moveable points is the radius
of the circle.
The curve itself, and especially where its length is under
consideration, is commonly called the circumference of the
circle.
The symbol employed for the circle is 0.
93° From the definitions of 92° we deduce the following
corollaries : —
1. All the radii of a are equal to one another.
2. The is a closed figure ; so that to pass from a point
within the figure to a point without it, or vice versa^ it is
necessary to cross the curve.
3. A point is within the 0, on the 0, or without the 0,
according as its distance from the centre is less, equal to, or
greater than the radius.
4. Two 0s which have equal radii are congruent ; for, if
the centres coincide, the figures coincide throughout and form
virtually but one figure.
Def. — Circles which have their centres coincident are
called concentric circles.
94°. Theore7n.—h.X\VLQ can cut a circle in two points, and in
two points only.
Proof Since the is a closed curve (93°, 2), a line which
cuts it must lie partly within the and partly without. And
the generating point (69°) of the line must cross the in
passing from without to within, and again in passing from
within to without.
.*. a line cuts a at least twice if it cuts the at all.
THE CIRCLE.
55
Again, since all radii of the same are equal, if a line
could cut a three times, three equal segments could be
drawn from a given point, the centre of the 0, to a given line.
And this is impossible (63°, 3).
Therefore a line can cut a only twice. q.e.d.
Cor. I. Three points on the same circle cannot be in line ;
or, a circle cannot pass through three points which are in line.
95°. Def. I. — A line which cuts a is a secant or secantline.
Def. 2. — The segment of a secant
included within the is a chord.
Thus the line L, or AB, is a
secant, and the segment AB is a
chord. (21°)
The term chord whenever involv
ing the idea of length means the segment having its end
points on the circle. But sometimes, when length is not
involved, it is used to denote the whole secant of which it
properly forms a part.
Def. 3. — A secant which passes through the centre is a
centreline, and its chord is a diameter.
Where length is not implied, the term diameter is some
times used to denote the centreHne of which it properly
forms a part.
Thus M is a centreline and CD is a diameter.
96°. Theorejn. — Through any three points not in line —
1. One circle can be made to pass. *0
2. Only one circle can be made to pass.
Proof. — Let A, B, C be three points
not in line.
Join AB and BC,and let L and M be the a"
right bisectors of AB and BC respectively.
I. Then, because AB and BC intersect at B,
L and M intersect at some point O, (79°
Cor.)
56 SYNTHETIC GEOMETRY.
and O is equidistant from A, B, and C. (86°)
.•. the with centre at O, and radius equal to OA, passes
through B and C. q.e.d.
2. Any O through A, B, and C must have its centre equally
distant from these three points.
But O is the only point in the plane equidistant from A, B,
and C. (86°, Cor.)
And we cannot have two separate 0s having the same
centre and the same radius. (93°, 4)
.•. only one circle can pass through A, B, and C. q.e.d.
Cor. I. Circles which coincide in three points coincide
altogether and form one circle.
Cor. 2. A point from which more than two equal segments
can be drawn to a circle is the centre of that circle.
Cor. 3. Since L is a centreline and is also the right
bisector of AB,
.•. the right bisector of a chord is a centre line.
Cor. 4. The AAOB is isosceles, since OA=OB. Then, if
D be the middle of AB, OD is a median to the base AB and
is the right bisector of AB. (55°, Cor. 2)
.'. a centreline which bisects a chord is perpendicular to
the chord.
Cor. 5. From Cor. 4 by the Rule of Identity,
A centre line which is perpendicular to a chord bisects the
chord.
.*. the right bisector of a chord, the centreline bisecting
the chord, and the centreline perpendicular to the chord are
one and the same.
97°. From 92°, Def., a circle is given when the position of
its centre and the length of its radius are given. And, from
96°, a circle is given when any three points on it are given.
It will be seen hereafter that a circle is determined by three
points even when two of them become coincident, and in
higher geometry it is shown that three points determine a
THE CIRCLE. 57
circle, under certain circumstances, when all three of the
points become coincident.
Def. — Any number of points so situated that a circle can
pass through them are said to be concyclic^ and a rectilinear
figure (14°, Def.) having its vertices concyclic is said to be
inscribed in the circle which passes through its vertices, and
the circle is said to circumscribe the figure.
Hence the circle which passes through three given points
is the circumcircle of the triangle having these points as ver
tices, and the centre of that circle is the circumcentre of the
triangle, and its radius is the circumradius of the triangle.
(86°, Def.)
A hke nomenclature applies to any rectilinear figure having
its vertices concyclic.
98°. Theorem. — If two chords bisect one another they are
both diameters.
If AP = PD and CP = PB, then P is the '''
centre.
Proof. — Since P is the middle point of
both AD and CB (hyp.), therefore the right
bisectors of AD and CB both pass through P.
But these right bisectors also pass through the centre ;
(96°, Cor. 3) .'. P is the centre. (24°, Cor. 3) q.e.d.
99°. Theorem. — Equal chords are equally distant from the
centre ; and, conversely, chords equally dis ^
tant from the centre are equal.
. If AB = CD and OE and OF are the per
pendiculars from the centre upon these
chords, then OE = OF ; and conversely, if
OE = OF, then AB = CD. ^
Proof. — Since OE and OF are centre lines J. to AB and CD,
AB and CD are bisected in E and F. (96°, Cor. 5)
.'. in the As OBE and ODF
OB = OD, EB = FD,
58
SYNTHETIC GEOMETRY.
and they are rightangled opposite equal sides,
AOBE = AODF,
and OE = OF.
Conversely, by the Rule of Identity, if OE = OF, then
AB = CD. q.e.d.
ioo°. Theorem. — Two secants which make equal chords
A ^•'" — ^B ^ p make equal angles with the centreline
through their point of intersection.
AB = CD, and PO is a centreHne
through the point of intersection of
AB and CD. Then
^APd=ACPO.
Proof.— L^t OE and OF be J_ to
AB and CD from the centre O.
Then OE = OF, (99")
AOPE = AOPF, (65°)
and Z.APO=^CPO. q.e.d.
Cor. I. •.* E and F are the middle
points of AB and CD, (96°, Cor. 5)
.. PE = PF, PA = PC, and PB = PD.
Hence, secants which make equal chords make two pairs of
equal linesegments between their point of intersection and
the circle.
Cor. 2. From any point two equal linesegments can be
drawn to a circle, and these make equal angles with the
centreline through the point.
101°. As all circles have the same form, two circles which
have equal radii are equal and congruent (93°, 4), (51°).
Hence equal and cojigruent are equivalent terms when
applied to the circle.
Def. I, — Any part of a circle is an arc.
The word equal when applied to arcs means congruence
or capability of superposition. Equal arcs come from the
same circle or from equal circles.
THE CIRCLE.
59
Def. 2.— A line which divides a figure into two parts such
that when one part is revolved about the line it may be made
to fall on and coincide with the other part is an axis of
symmetry of the figure.
102°. Theorem.
the circle.
A centreline is an axis of symmetry of
A^ — ^^B
Proof. — Let AB and CD be equal
chords meeting at P, and let PHOG
be a centre line.
Let the part of the figure which lies q>
upon the F side of PG be revolved
about PG until it comes to the plane
on the E side of PG.
Then/ ^GPA = ^GPC, (100°)
.*. PC coincides with PA.
And •.• PB = PD
and PA = PC, (100°, Cor. i)
.*. D coincides with B,
and C coincides with A.
And the arc HCG, coinciding in three points with the arc
HAG, is equal to it, and the two arcs become virtually but
one arc. (96°, Cor. i)
Therefore PG is an axis of symmetry of the O, and divides
it into two equal arcs. q.e.d.
Def — Each of the arcs into which a centreline divides the
circle is a semicircle.
Any chord, not a centreline, divides the circle into unequal
arcs, the greater of which is called the major arc, and the
other the minor arc.
Cor. I. By the superposition of the theorem we see that
arc AB = arc CD, arc HB=arc HD, arc GA=arc GC,
arc BDCA = arc DBAC. (ist Fig.)
But the arcs BDCA and AB are the major and minor arcs
60 SYNTHETIC GEOMETRY.
to the chord AB, and the arcs DBAC and CD are major and
minor arcs to the chord CD.
Therefore equal chords determine equal arcs, major being
equal to major and minor to minor.
Cor. 2. Equal arcs subtend equal angles at the centre.
103°. Theorem. — Parallel secants intercept equal arcs on a
E circle.
If AB is II to CD,
then arc A C = arc D B.
/d Proof. — Let EF be the centreline _L to AB.
Then EF is J_ to CD also. (72^)
When EBDF is revolved about EF,
B comes to coincidence with A, and D with C, and the arc
BD with the arc AC,
arc AC = arc DB. q.e.d.
Cor. Since the chord AC = chord BD,
Therefore parallel chords have the chords joining their
endpoints equal.
Exercises.
1. Any plane closed figure is cut an even number of times by
an indefinite line.
2. In the figure of Art. 96°, if A, B, and C shift their relative
positions so as to tend to come into line, what becomes
of the point O ?
3. In the same figure, if ABC is a right angle where is the
point O }
4. Given a circle or a part of a circle, show how to find its
centre.
5. Three equal segments cannot be drawn to a circle from
a point without it.
6. The vertices of a rectangle are concyclic.
7. If equal chords intersect, the segments of one between the
THE CIRCLE. 6 1
point of intersection and the circle are respectively
equal to the corresponding segments of the other.
8. Two equal chords which have one endpoint in common
lie upon opposite sides of the centre.
9. If AB and CD be parallel chords, AD and BC, as also
AC and BD, meet upon the right bisector of AB or CD.
10. Two secants which make equal angles with a centreline
make equal chords in the circle if they cut the circle.
(Converse of 100°)
11. What is the axis of symmetry of (a) a square, (d) a
rectangle, (c) an isosceles triangle, (d) an equilateral
triangle ? Give all the axes where there are more than
one.
12. When a rectilinear figure has more than one axis of
symmetry, what relation in direction do they hold to
one another ?
13. The vertices of an equilateral triangle trisect its circum
circle.
14. A centreline perpendicular to a chord bisects the arcs
determined by the chord.
1 5. Show how to divide a circle (a) into 6 equal parts, {d) into
8 equal parts.
16. If equal chords be in a circle, one pair of the connecters
of their endpoints are parallel chords.
(Converse of 103°, Cor.)
THE PRINCIPLE OF CONTINUITY.
104°. The principle of continuity is one of the most prolific
in the whole range of Mathematics.
Illustrations of its meaning and application in Geometr}^
will occur frequently in the sequel, but the following are
given by way of introduction.
I. A magnitude is cojitinuous throughout its extent.
Thus a line extends from any one point to another without
62 SYNTHETIC GEOMETRY.
breaks or interruptions ; or, a generating point in passing
from one position to another must pass through every inter
mediate position.
2. In Art. 53° we have the theorem — Every point on the
right bisector of a segment is equidistant from the endpoints
of the segment.
In this theorem the Imiiting condition in the hypothesis is
that the point must be on the right bisector of the segment.
Now, if P be any point on the right bisector, and we move
P along the right bisector, the limiting condition is not at
any time violated during this motion, so that P remains con
tijtuously equidistant from the endpoints of the segment
during its motion.
We say then that the property expressed in the theorem is
continuous while P moves along the right bisector.
3. In Art. 97° we have the theorem — The sum of the in
ternal angles of a quadrangle is four right angles.
The limiting condition is that the figure shall be a quad
A rangle, and that it shall have in
ternal angles.
Now, let ABCD be a quadrangle.
Then the condition is not violated
if D moves to D^ or D2. But in
^ the latter case the normal quad
rangle ABCD becomes the inverted quadrangle ABCDg,
and the theorem remains true. Or, the theorem is continu
ously true while the vertex D moves anywhere in the plane,
so long as the figure remains a quadrangle and retains four
internal angles.
Future considerations in which a wider meaning is given
to the word " angle " will show that the theorem is still true
even when D, in its motion, crosses one of the sides AB or
BC, and thus produces the crossed quadrangle.
The Principle of Continuity avoids the necessity of proving
theorems for different cases brought about by variations in
the disposition of the parts of a diagram, and it thus gener
THE CIRCLE. 63
alizes theorems or relieves them from dependence upon the
particularities of a diagram. Thus the two figures of Art.
100° differ in that in the first figure the secants intersect
without the circle, and in the second figure they intersect
within, while the theorem applies with equal generality to both.
The Principle of Continuity may be stated as follows : —
When a figure, which involves or illustrates some geometric
property, can undergo change, however small, in any of its
parts or in their relations without violating the conditions
upon which the property depends, then the property is con
tinuous while the figure undergoes any amount of change of
the same kind within the range of possibility.
105°. Let AB be a chord dividing the into unequal arcs,
and let P and Q be any points upon
the major and minor arcs respectively.
(102°, Def )
Let O be the centre.
1. The radii OA and OB form two
angles at the centre, a major angle
denoted by a and a minor angle de
noted by ^. These together make up
a circumangle.
2. The chords PA, PB, and QA, QB form two angles at
the circle, of which APB is the tninor angle and AQB is the
major angle.
3. The minor angle at the circle, APB, and the minor angle
at the centre, /3, stand upon the minor arc, AQB, as a base.
Similarly the major angles stand upon the major arc as base.
4. Moreover the ZAPB is said to be z;z the arc APB, so
that the minor angle at the circle is in the major arc, and the
major angle at the circle is in the minor arc.
5. When B moves towards B' all the minor elements
increase and all the major elements decrease, and when B
comes to B' the minor elements become respectively equal to
the major, and there is neither major nor minor.
64 SYNTHETIC GEOMETRY.
When B, moving in the same direction, passes B', the
elements change name, those which were formerly the minor
becoming the major and vice versa.
io6°. Theorem. — An angle at the circle is onehalf the
corresponding angle at the centre, major corresponding to
major and minor to minor.
p Z.AOB minor is 2Z.APB.
/^ /\ \ N^ Proof. — Since AAPO is isosceles,
/ / a'l \'Y .'. /.OAP = z.OPA,(53°,Cor.i)
/j^/sC\ and^0AP + ^0PA=2^0PA.
}^:^^T^y ^"^ zj\OC=^OAP + ^OPA,
\>1 y/^ (76°, Cor.)
^^~p^ .*. ^A0C = 2^0PA.
Similarly ^B0C = 2^0PB ;
..adding, ^AOB minor = 2^APB. g.e.d.
The theorem is thus proved for the minor angles. But
since the limiting conditions require only an angle at the
circle and an angle at the centre, the theorem remains true
while B moves along the circle. And when B passes B'
the angle APB becomes the major angle at the circle, and
the angle AOB minor becomes the major angle at the
centre.
the theorem is true for the major angles.
Cor. I. The angle in a given arc is constant. (105°, 4)
Cor. 2. Since ^APB = z.AOB minor,
and ^AOB = i^OB major,
and '.• ^AOB minor + Z.AOB major = 4 right angles (37°)
Z.APB + ^QB = a straight angle.
And APBQ is a coney clic quadrangle.
Hence a concyclic quadrangle has its opposite internal
angles supplementary. (40°, Def. i)
Cor. 3. D being on AQ produced,
lBQD is supplementary to ^AOB.
THE CIRCLE.
65
But Z.APB is supplementary to ^A(2B,
^APB=ZBOD.
Hence, if one side of a concyclic quadrangle
be produced, the external angle is equal to a
the opposite internal angle.
Cor. 4. Let B come to B'. (Fig. of 106°)
Then ^AOB' is a straight angle,
Z.APB' is a right angle.
But the arc APB' is a semicircle,
(102°, Def )
Therefore the angle in a semicircle is a right angle.
opposite
107°. Theorem. — A quadrangle which has its
angles supplementary has its vertices concyclic.
(Converse of 106°
ABCD is a quadrangle whereof the ^lADC
is supplementary to ^ABC ; then a circle
can pass through A, B, C, and D.
Proof. — If possible let the through A,
B, and C cut AD in some point P.
Join P and C.
Then Z_APC is supplementary to Z.ABC, (106°, Cor. 2)
and zADC is supplementary to Z.ABC,
z.APC=^DC,
which is not true.
.*. the cannot cut AD in any point other than D,
Hence A, B, C, and D are concyclic.
Cor. I. The hypothenuse of a rightangled triangle is the
diameter of its circumcircle. (88°, 3, Def. ; 97°, Def.)
Cor. 2. When P moves along the the AAPC (last figure)
has its base AC constant and its vertical angle APC constant.
Therefore the locus of the vertex of a triangle which has a
constant base and a constant vertical angle is an arc of a
circle passing through the endpoints of the base.
This property is employed in the trammel which is used to
describe an arc of a given circle.
E
(hyp.)
(60^)
g.e.d.
66
SYNTHETIC GEOMETRY.
It consists of two rules (i6°) L and M joined at a
L ^y^2\ determined angle. When it is made
to slide over two pins A and B, a
pencil at P traces an arc passing
through A and B.
io8°. Theorem. — The angle between two intersecting se
cants is the sum of those angles in the circle which stand
on the arcs intercepted between the
secants, when the secants intersect
within the circle, and is the difference
of these angles when the secants
intersect without the circle.
lAPC=^ABC + z_BCD, {i%)
^APC = aABC^BCD. CfIJ)
Proof.i. z.APC=^PBC+^PCB, (6o°)
.. aAPC = z.ABC + £.BCD.
2. /.ABC=^APC + ^BCP,
.. ^APC = ^ABC^BCD.
q.e.d.
Exercises.
1. If a sixsided rectilinear figure has its vertices concyclic,
the three alternate internal angles are together equal
to a circumangle.
2. In Fig. 105°, when B comes to O, BQ vanishes ; what is
the direction of BQ just as it vanishes 1
3. Two chords at right angles determine four arcs of which a
pair of opposite ones are together equal to a semi
circle.
4. A, B, C, D are the vertices of a square, and A, E, F of an
equilateral triangle inscribed in the same circle.
What is thQ angle between the lines BE and DF?
between BF and ED ?
THE CIRCLE.
67
SPECIAL SECANTS— TANGENT.
Let P be a fixed point on the S and O a variable
109
one.
The position of the secant L, cut
ting the circle in P and Q, depends
upon the position of O.
As Q moves along the the secant
rotates about P as pole. While Q
makes one complete revolution along
the the secant L passes through two
special positions. The first of these is when O is farthest
distant from P, as at Q', and the secant L becomes a centre
line. The second is when Q comes into coincidence with P,
and the secant takes the position TT' and becomes a tangent.
Def. I.— A tangent to a circle is a secant in its limiting
position when its points of intersection with the circle become
coincident.
That the tangent cannot cut or cross the is evident.
For if it cuts the at P it must cut it again at some other
point. And since P represents two points we would have the
absurdity of a line cutting a circle in three points. (94°)
Def. 2.— The point where P and Q meet is called Xh^ point
of contact. Being formed by the union of two points it repre
sents both, and is therefore a double point.
From Defs. i and 2 we conclude —
1. A point of contact is a double point.
2. As a line can cut a only twice it can touch a only once.
3. A line which touches a cannot cut it.
4. A is determined by two points if one of them is a
given point of contact on a given line ; or, only one
circle can pass through a given point and touch a
given line at a given point. (Compare 97°.)
6S
SYNTHETIC GEOMETRY
iio°. Theorem. — A centreline and a tangent to the same
point on a circle are perpendicular to one another.
L' is a centreline and T a tan
iQ gent, both to the point P. Then L'
is _L to T.
Proof. — '.* T has only the one point
P in common with the 0, every point
of T except P lies without the 0. .'.
T' if O is the centre on the line L', OP
is the shortest segment from O to T.
OP, or L', is ± to T. (63°, i) q.c.cL
Cor. I. Tangents at the endpoints of a diameter are
parallel.
Cor. 2. The perpendicular to a tangent at the point of con
tact is a centreline. (Converse of the theorem.)
Cor. 3. The perpendicular to a diameter at its endpoint is
a tangent.
111°. Theorem. — The angles between a tangent and a
D chord from the point of contact are
respectively equal to the angles in the
opposite arcs into which the chord
divides the circle.
TP is a tangent and PQ a chord to
the same point P, and A is any point on
the 0. Then
^QPT=^QAP.
Let PD be a diameter.
^QAP = ^QDP,
/.DQP is a "X
^DPQ is comp. of Z.QPT,
^DPQ is comp. of ^ODP,
^QDP=i.QPT=z.QAP.
Similarly, the ^QPT'=^QBP.
Proof.
Then
and
Also
and
(106°, Cor. i)
(106°, Cor. 4)
(40°, Def. 3)
{n% 6)
q.e.d.
THE CIRCLE.
69
112°. Theorem. — Two circles can intersect in only two
points.
Proof. — If they can intersect in three points, two circles
can be made to pass through the same three points. But
this is not true. (96°)
.'. two circles can intersect in only two points.
Cor. Two circles can touch in only one point. For a
point of contact is equivalent to two points of intersection.
113°. Theorem. — The common centreline of two intersect
ing circles is the right bisector
of their common chord.
O and O' are the centres of S
and S', and AB is their common
chord. Then 00' is the right
bisector of AB.
(50
Proof. Since AO = BO,
and AO' = BO',
.*. O is on the right bisector of AB.
Similarly O' is on the right bisector of AB,
.*. 00' is the right bisector of AB.
Cor. I. By the principle of continuity, OO' always bisects
AB. Let the circles separate until A and B coincide. Then
the circles touch and 00' passes through the point of contact.
Def. — Two circles which touch one another have external
contact when each circle lies without the other, and internal
contact when one circle lies within the other.
Cor. 2. Since 00' (Cor. i) passes through the point of
contact when the circles touch one another —
{(.i) When the distance between the centres of two circles
is the sum of their radii, the circles have external
contact.
{b) When the distance between the centres is the difference
of the radii, the circles have internal contact.
70
SYNTHETIC GEOMETRY.
(c) When the distance between the centres is greater than
the sum of the radii, the circles exclude each other
without contact.
(d) When the distance between the centres is less than
the difference of the radii, the greater circle includes
the smaller without contact.
(e) When the distance between the centres is less than the
sum of the radii and greater than their difference, the
circles intersect.
14°. Theo7em. — From any point without a circle two
tangents can be drawn to the
circle.
Proof.— 1.^1 S be the and P
the point. Upon the segment
PO as diameter let the ©S' be
described, cutting 0S in A and
B. Then PA and PB are both
tangents to S.
For Z.OAP is in a semicircle and is a "~' (106°, Cor. 4)
.. AP is tangent to S. (110°, Cor. 3)
Similarly BP is tangent to S.
Cor. I. Since PO is the right bisector of AB, (113°)
PA = PB. (53°)
Hence calling the segment PA the tangent from P to the
circle, when length is under consideration, we have — The two
tangents from any point to a circle are equal to one another.
Def — The line AB, which passes through the points of
contact of tangents from P, is called the chord of contact for
the point P.
115°. Def I. — The angle at which two circles intersect is
the angle between their tangents at the point of intersection.
Def 2. When two circles intersect at right angles they are
said to cut each other orthogonally.
THE CIRCLE.
71
The same term is conveniently applied to the intersection
of any two figures at right angles.
Cor. I. If, in the Fig. to 114°, PA be made the radius of a
circle and P its centre, the circle will cut the circle S ortho
gonally. For the tangents at A are respectively perpendicular
to the radii.
Hence a circle S is cut orthogonally by any circle having
its centre at a point without S and its radius the tangent
from the point to the circle S .
116°. The following examples furnish theorems of some
importance.
Ex. I. Three tangents touch the circle
S at the points A, B, and C, and inter
sect to form the AA'B'C. O being the
centre of the circle,
A.A0C = 2^A'0C'.
Proof.— AC'=BC',
and BA' = CA', (114°, Cor. i)
AAOC' = ABOC, and ABOA' = ACOA'
^BOC' = ^AOC, and ^COA'=^BOA',
zAOC = 2^A'OC'. q.e.d.
Similarly ^AOB = 2/.A'OB', and ^B0C = 2^B'0C'.
If the tangents at A and C are fixed, and the tangent at B
is variable, we have the following theorem : —
The segment of a variable tangent intercepted by two fixed
tangents, all to the same circle, subtends a fixed angle at the
centre.
Ex. 2. If four circles touch two and two externally, the
points of contact are concyclic.
Let A, B, C, D be the centres of the circles, and P, Q, R,
S be the points of contact.
Then AB passes through P, BC through Q, etc. (113°, Cor. i )
J 2 SYNTHETIC GEOMETRY.
Now, ABCD being a quadrangle,
^A + i.B + ^C + ^D=4~s. (90°)
But the sum of all the internal angles of the four As APS,
BQP, CRQ, and DSR is 8~ls, and
z.APS=^ASP, ^BPQ=/.BQP, etc.,
z.APS + ^BPQ + ^CRQ + /.DRS = 2~ls.
Now * /.SPQ = 2~]s(^APS + £.BPQ),
^QRS = 2 "Is  (^CRQ + Z.DRS),
^SP0 + /.QRS = 2"ls.
and P, Q, R, S are concyclic (107°). q.e.d.
Ex. 3. If the common chord of two intersecting circles
subtends equal angles at the two circles, the circles are equal.
AB is the common chord, C, C points upon the circles, and
^ACB = .lACB.
Let O, O' be the centres. Then ^AOB = ^AO'B. (106°)
And the triangles OAB and O'AB being isosceles are con
gruent, .'. OA = O^A, and the circles are equal. (93°, 4)
Ex. 4. If O be the orthocentre of a AABC, the circum
circles to the As ABC, AOB, BOC, COA
are all equal.
*.' AX and CZ are ± respectively to
BC and AB,
£.CBA = sup. of ^XOZ
= sup. of ^COA.
But D being any point on the arc
AS2C, ^CDA is the sup. of ^COA.
^CBA = ^CDA,
and the 0s S and 83 are equal by Ex. 3.
In like manner it may be proved that the 0S is equal to the
03 S3 and Si.
Ex. 5. If any point O be joined to the vertices of a AABC,
the circles having OA, OB, and OC as diameters intersect
upon the triangle.
Proof.— VixTs.^ OX _L to BC and OY ± to AC.
THE CIRCLE.
73
•/ z.OXB=~~, the on OB as diameter passes through X.
(107°, Cor. I)
Similarly the on OC as diameter passes through X.
Therefore the 0s on OB and OC intersect in X ; and in like
manner it is seen that the 0s on OC and OA intersect in Y,
and those on OA and OB intersect in Z, the foot of the X
from O to AB.
Ex. 6. The feet of the medians and the feet of the altitudes
in any triangle are six concyclic points, and the circle bisects
that part of each altitude lying
between the orthocentre and the
vertex.
D, E, F are the feet of the
medians, i.e., the middle points
of the sides of the AABC. Let
the circle through D, E, F cut
the sides in G, H, K.
Now FD is II to AC and ED is  to AB,
^FDE = ^FAE.
But ^FDE = ^FHE,
.. AAFH is isosceles, and AF = FH = FB
^AHB=n,
and H is the foot of the altitude from B.
Similarly, K and G are feet of the altitudes from C and A.
Again, ^KPH = ^KFH = 2/.KAH. And A, K, O, H are
concyclic (107°), and AO is a diameter of the circumcircle,
therefore P is the middle point of AO.
Similarly, O is the middle point of BO, and R of CO.
Def. — The circle S passing through the nine points D, E,
F, G, H, K, and P, Q, R, is called the 7iinepoiiits circle of
the AABC.
Cor. Since the ninepoints circle of ABC is the circum
circle of ADEF, whereof the sides are respectively equal to
half the sides of the AABC, therefore the radius of the nine
points circle of any triangle is onehalf that of its circumcircle.
(84°, Cor. 2)
(106°, Cor. I)
(106°, Cor. 4)
74 SYNTHETIC GEOMETRY.
Exercises.
1. In 105° when P passes B where is the Z.APB ?
2. A, B, C, D are four points on a circle whereof CD is a
diameter and E is a point on this diameter. If
z_AEB = 2^ACB, E is the centre.
3. The sum of the alternate angles of any octagon in a circle
is six right angles.
4. The sum of the alternate angles of any concyclic polygon
of 2;/ sides is 2{n — i) right angles.
5. If the angle of a trammel is 60° what arc of a circle will
it describe ? what if its angle is 7t° ?
6. Trisect a right angle and thence show how to draw a
regular 12sided polygon in a circle.
7. If r, r' be the radii of two circles, and d the distance
between them, the circles touch when d=r±r'.
8. Give the conditions under which two circles have 4, 3, 2,
or I common tangent.
9. Prove Ex. 2, 116°, by drawing common tangents to the
circles at P, Q, R, and S.
I o. A variable chord passes through a fixed point on a circle,
to find the locus of the middle point of the chord.
11. A variable secant passes through a fixed point, to find
the locus of the middle point of the chord determined
by a fixed circle.
12. In Ex. II, what is the locus of the middle point of the
secant between the fixed point and the circle ?
13. In a quadrangle circumscribed to a circle the sums of the
opposite sides are equal in pairs ; and if the vertices
be joined to the centre the sums of the opposite angles
at the centre are equal in pairs.
14. If a hexagon circumscribe a circle the sum of three
alternate sides is equal to that of the remaining
three.
15. If two circles are concentric, any chord of the outer
which is tangent to the inner is bisected by the point
THE CIRCLE. 75
of contact ; and the parts intercepted on any secant
between the two circles are equal to one another.
1 6. If two circles touch one another, any line through the
point of contact determines arcs which subtend equal
angles in the two circles.
17. If any two lines be drawn through the point of contact of
two touching circles, the lines determine arcs whose
chords are parallel.
18. If two diameters of two touching circles are parallel, the
transverse connectors of their endpoints pass through
the point of contact.
19. The shortest chord that can be drawn through a given
point within a circle is perpendicular to the centreline
through that point.
20. Three circles touch each other externally at A, B, and C.
The chords AB and AC of two of the circles meet the
third circle in D and E. Prove that DE is a diameter
of the third circle and parallel to the common centre
line of the other two.
21. A line which makes equal angles with one pair of oppo
site sides of a concyclic quadrangle makes equal angles
with the other pair, and also with the diagonals.
22. Two circles touch one another in A and have a common
tangent BC. Then lBAC is a right angle.
23. OA and OB are perpendicular to one another, and AB is
variable in position but of constant length. Find the
locus of the middle point of AB.
24. Two equal circles touch one another and each touches
one of a pair of perpendicular lines. What is the locus
of the point of contact of the circles ?
25. Two lines through the common points of two intersecting
circles determine on the circles arcs whose chords are
parallel.
26. Two circles intersect in A and B, and through B a secant
cuts the circles in C and D. Show that Z.CAD is
constant, the direction of the secant being variable.
jG SYNTHETIC GEOMETRY.
27. At any point in the circumcircle of a square one of the
sides subtends an angle three times as great as that
subtended by the opposite side.
28. The three medians of any triangle taken in both length
and direction can form a triangle.
SECTION VI.
CONSTRUCTIVE GEOMETRY,
INVOLVING THE PRINCIPLES OF THE FIRST FIVE
SECTIONS, ETC.
117°. Constructive Geometry applies to the determination
of geometric elements which shall have specified relations to
given elements.
Constructive Geometry is Practical when the determined
elements are physical, and it is Theoretic when the elements
are supposed to be taken at their limits, and to be geometric
in character. (12°)
Practical Constructive Geometry, or simply " Practical
(Geometry," is largely used by mechanics, draughtsmen, sur
veyors, engineers, etc., and to assist them in their work
numerous aids known as " Mathematical Instruments " have
been devised.
A number of these will be referred to in the sequel.
In "Practical Geometry " the " Rule " (16°) furnishes the
means of constructing a line, and the " Compasses " (92°) of
constructing a circle.
In Theoretic Constructive Geometry we assume the ability
to construct these two elements, and by means of these we
are to determine the required elements.
118'. To test the " Rule."
Place the rule on a plane, as at R, and draw a line AB
CONSTRUCTIVE GEOxMETRY. 7/
along its edge. Turn the rule into the position R'. If the
edge now coincides with the line ^ ; v^_^
the rule is true. ^^(7 r ^^ —
This test depends upon the pro
perty that two finite points A and B determine one line.
(24°, Cor. 2)
Def. — A construction proposed is in general called a
proposition (2°) and in particular 7ip7'oblem.
A complete problem consists of (i) the statement of what
is to be done, (2) the construction, and (3) the proof that the
construction furnishes the elements sought.
119°. Problem. — To construct the right bisector of a given
line segment.
Let AB be the given segment.
Construction. — With A and B as centres
and with a radius AD greater than half of a"
AB describe circles.
Since AB is < the sum of the radii and
> their difference, the circles will meet in
two points P and Q. (ii3°> Cor. 2, e)
The line PQ is the right bisector required.
Proof. — P and O are each equidistant from A and B and
.'. they are on the right bisector of AB ; (54°)
.*. PQ is the right bisector of AB.
Cor. I. The same construction determines C, the middle
point of AB.
Cor. 2. If C be a given point on a line, and we take A and
B on the line so that CA=CB, then the right bisector of the
segment AB passes through C and is J_ to the given line.
.•. the construction gives the perpendicular to a given line
at a given point in the line.
120°. Problem. — To draw a perpendicular to a given line
from a point not on the line.
I \
I \
I >
^ — Id
7S
SYNTHETIC GEOMETRY.
Let L be the given line and P be the point.
Constr. — Draw any line through
P meeting L at some point A.
Bisect AP in C (119°, Cor. i), and
with C as centre and CP as radius
describe a circle.
If PA is not i. to L, the will
cut L in two points A and D.
Then PD is the J. required.
Proof. — PDA is the angle in a semicircle,
^PDA is a "~I (106°, Cor. 4)
Cor. Let D be a given point in L. With any centre C
and CD as radius describe a circle cutting L again in some
point A. Draw the radius ACP, and join D and P. Then
DP is ± to L.
.*. the construction draws a X to L at a given point in L.
(Compare 119°, Cor. 2)
Cor. 2. Let L be a given line and C a given point.
To draw through C a line parallel to L.
With C as the centre of a circle, construct a figure as
given. Bisect PD in E (119°, Cor. i). Then CE is  to L.
For C and E are the middle points of two sides of a triangle
of which L is the base. (84°, Cor. 2)
121°. The Square. — The square consists of two rules with
their edges fixed permanently at
right angles, or of a triangular
plate of wood or metal having
Dtwo of its edges at right angles.
To test a square.
Draw a line AB and place the
square as at S, so that one edge
coincides with the line, and along
the other edge draw the line
CD.
Next place the square in the position S'. If the edges can
CONSTRUCTIVE GEOMETRY.
79
now be made to coincide with the two lines the square
is true.
This test depends upon the fact that a right angle is one
half a straight angle.
The square is employed practically for drawing a line J_ to
another line. —
Cor. I. The square is employed to
draw a series of parallel lines, as in
the figure.
Cor. 2. To draw the bisectors of an angle by means of the
square.
Let AOB be the given angle. Take OA = OB, and at A
and B draw perpendiculars to OA and OB.
Since AOB is not a straight angle, these perpendiculars
meet at some point C. (79°, Cor.)
Then OC is the internal bisector of ^lAOB. For the tri
angles AOC and BOC are evidently congruent.
^AOC = ^BOC.
The line drawn through O _L to OC is the external bisector.
122°. Problem. — Through a given point in a line to draw a
line which shall make a given angle
with that line.
Let P be the given point in the
line L, and let X be the given
angle.
Constr.— Yxom any point B
the arm OB draw a ± to the arm OA.
Make PA' = OA, and at A' draw the J. A'B' making
A'B' = AB, PB' is the line required.
Proof. — The triangles OBA and PB'A' are evidently con
gruent, and .*. z.BOA = X = ^B'PA'.
Cor. Since PA' might have been taken to the left of P, the
problem admits of two solutions. When the angle X is a
right angle the two solutions become one.
(120°)
8o
SYNTHETIC GEOMETRY.
123°. The Protractor. — This instrument has different forms
depending upon the accuracy re
quired of it. It usually consists of
a semicircle of metal or ivory divided
into degrees, etc. (41°). The point C
is the centre. By placing the straight
edge of the instrument in coincidence with a given line AB
so that the centre falls at a given point C, we can set off any
angle given in degrees, etc., along the arc as at D. Then
the line CD passes through C and makes a given angle
with AB.
124°. Problem.
Given the sides of a triangle to construct it.
Constr. — Place the three sides of
the triangle in line, as AB, 15 C,
CD.
With centre C and radius CD
D describe a circle, and with centre
B and radius BA describe a circle.
Let E be one point of intersection of these circles.
Then A^EC is the triangle required.
Proof.— V,^ = V>K and CE = CD.
Since the circles intersect in another point E', a second
triangle is formed. But the two triangles being congruent
are virtually the same triangle.
Cor. I. When AB = BC=? C A the triangle is equilateral.
(53°, Def. 2)
In this case the circle AE passes through C and the circle
DE through B, so that B and C become the centres and BC
a common radius.
Cor. 2. When BC is equal to the sum or difference of AB
and CD the circles touch (113°, Def.) and the triangle takes
the limiting form and becomes a line.
When BC is greater than the sum or less than the differ
CONSTRUCTIVE GEOMETRY. 8 1
ence of AB and CD the circles do not meet (113°, Def.) and
no triangle is possible.
Therefore that three linesegments may form a triangle,
each one must be less than the sum and greater than the
difference of the other two.
125°. The solution of a problem is sometimes best effected
by supposing the construction made, and then by reasoning
backwards from the completed figure to some relation
amongst the given parts by means of which we can make
the construction.
This is analogous to the process employed for the solution
of equations in Algebra, and a more detailed reference will
be made to it at a future stage.
The next three problems furnish examples
126°. Problein. — To construct a triangle when two sides
and the median to the third side are given.
Let a and b be two sides and n
the median to the third side.
Suppose ACB is the required
triangle having CD as the given
median.
By completing the ^^ACBC
and joining DC, we have DC c'
equal to CD and in the same line, and BC=AC (81°) ; and
the triangle CCB has CC = 2;/, CB = rt:, and BC=AC = ^,
and is constructed by 124°.
Thence the triangle ACB is readily constructed.
Cor. Since CC is twice the given median, and since the
possibility of the triangle ACB depends upon that of CCB,
therefore a median of a triangle is less than onehalf the
sum, and greater than onehalf the difference of the conter
minous sides. (124°, Cor. 2)
F
82
SYNTHETIC GEOMETRY.
127°. Problem. — To trisect a given linesegment, i.e.^ to
divide it into three equal parts.
Cojistr. — Let AB be the segment.
Through A draw any line CD and make
Af •^^ 3^B AC=AD. Bisect DB in E, and join CE,
E cutting AB in F,
D Then AF is AB.
Proof. — CBD is a A and CE and BA are two medians.
AF=AB. (85°, Cor.)
Bisecting FB gives the other point of division.
128°. Problem. — To construct a A when the three medians
/,
are given.
Let /, m, n be the
medians, and suppose
to be the required
Then
BE = ?«, and CF = ;2,
given
ABC
triangle,
c
AD =
AO = f/, OB = ;;z, and OY = \n,
.'. in the AAOB we have two sides and the median to the
third side given. Thence AAOB is constructed by 126°
and 127°.
Then producing FO until 0C = 2F0, C is the third vertex
of the triangle required.
Ex. To describe a square whose sides shall pass through
four given points.
Let P, Q, R, S be the given points, and
suppose ABCD to be the square required.
Join P and Q upon opposite sides of the
square, and draw QG  to BC. Draw SX
J_ to PQ to meet BC in E, and draw EF
II to CD. Then AQPG = AFSE,
and SE = PO.
Hence the construction : —
Join any two points PQ, and through a third point S draw
CONSTRUCTIVE GEOMETRY. S^
SX ± to PQ. On SX take SE = PQ and join E with the
fourth point R. ER is a side of the square in position and
direction, and the points first joined, P and Q, are on oppo
site sides of the required square.
Thence the square is readily constructed.
Since SE may be measured in two directions along the
line SX, two squares can have their sides passing through
the same four points P, Q, R, S, and having P and Q on
opposite sides.
Also, since P may be first connected with R or S, two
squares can be constructed fulfilling the conditions and
having P and Q on adjacent sides.
Therefore, four squares can be constructed to have their
sides passing through the same four given points.
CIRCLES FULFILLING GIVEN CONDITIONS.
The problems occurring here are necessarily of an elemen
tary character. The more complex problems require relations
not yet developed.
129°. Problem. — To describe a circle to touch a given line
at a given point. m
P is a given point in the line L.
Co7istr. — Through P draw M J_ to L.
A circle having any point C, on M, as
centre and CP as radius touches L at P.
Proof. — L is ± to the diameter at its
endpoint, therefore L is tangent to the circle. (110°, Cor. 3)
Def. — As C is a7iy point on M, any number of circles may
be drawn to touch L at the point P, and all their centres lie
on M.
Such a problem is indefinite because the conditions are
not sufficient to determine a particular circle. If the circle
84 SYNTHETIC GEOMETRY.
varies its radius while fulfilling the conditions of the problem,
the centre moves along M ; and M is called the centrelocus
of the variable circle.
Hence the centrelocus of a circle which touches a fixed
line at a fixed point is the perpendicular to the line at that
point.
Cor. If the circle is to pass through a second given point
Q the problem is definite and the circle is a particular one,
since it then passes through three fixed points, viz., the double
point P and the point Q. (109°) 4)
In this case Z.COP=Z_CPQ.
But ^CPQ is given, since P, Q, and the line L are given.
.*. z_CQP is given and C is a fixed point.
130°. Problem. — To describe a circle to touch two given
nonparallel lines.
Let L and M be the lines inter
secting at O.
Draw N, N, the bisectors of the
angle between L and M.(i2i°, Cor. 2)
From C, any point on either bi
sector, draw CA X to L.
The circle with centre C and radius CA touches L, and if
CB be drawn X to M, CB = CA. (68°)
Therefore the circle also touches M.
As C is any point on the bisectors the problem is indefinite,
and the centrelocus of a circle which touches two intersecting
lines is the two bisectors of the angle between the lines .
131°. Prohle77i. — To describe a circle to touch three given
lines which form a triangle.
L, M, N are the lines forming the triangle .
Constr. — Draw Ij, E^, the internal and external bisectors
of the angle A ; and I2, Eg, those of the angle B.
^A + /.Bis<J_, .. ^BAOfiLABO is<~l
CONSTRUCTIVE GEOMETRY.
8S
.*. Ii and I2 meet at some point O (79°) and are not X to one
another and therefore Ej and Eg meet at some point O3.
m\ /m (79°, Cor.)
Also Ii and Eg meet at some point Oj, and similarly Ig and
El meet at O2.
The four points O, O^, Og, O3 are the centres of four circles
each of which touches the three lines L, M, and N.
Proof. — Circles which touch M and N have \ and E^ as
their centrelocus (130°), and circles which touch N and L
have Ig and Eg as their centrelocus.
.'. Circles which touch L, M, and N must have their
centres at the intersections of these loci.
But these intersections are O, O^, Og, and O3,
.*. O, Oi, Og, and O3 are the centres of the circles required.
The radii are the perpendiculars from the centres upon any
one of the lines L, M, or N.
Cor. I. Let I3 and E3 be the bisectors of the Z.C. Then,
since O is equidistant from L and M, I3 passes through O. (68°)
.*. the three internal bisectors of the angles of a triangle
are concurrent.
Cor. 2. Since O3 is equidistant from L and M, I3 passes
through O3. (68°)
86 SYNTHETIC GEOMETRY.
.*. the external bisectors of two angles of a triangle and the
internal bisector of the third angle are concurrent.
Def. I. — When three or more points are in line they are
said to be collinear.
Cor. 3. The line through any two centres passes through a
vertex of the AABC.
.*. any two centres are collinear with a vertex of the A
The lines of collinearity are the six bisectors of the three
angles A, B, and C.
Def. 2 — With respect to the AABC, the circle touching
the sides and having its centre at O is called the i7iscribed
circle or simply the incircle of the triangle.
The circles touching the lines and having centres at Oj,
O2, and O3 are the escribed or excircles of the triangle.
REGULAR POLYGONS.
132°. Def. I. — A closed rectilinear figure without reentrant
angles (89°, 2) is in general called z. polygon.
They are named according to the number of their sides as
follows : —
3, triangle or trigon ;
4, quadrangle, or tetragon, or quadrilateral ;
5, pentagon ; 6, hexagon ; 7, heptagon ;
8, octagon ; 10, decagon ; 12, dodecagon ; etc.
The most important polygons higher than the quadrangle
are regular polygons.
Def 2. — A regular polygon has its vertices concyclic, and
all its sides equal to one another.
The centre of the circumcircle is the centre of the polygon.
133". Theorem. — If ;/ denotes the number of sides of a
CONSTRUCTIVE GEOMETRY. 8/
regular polygon, the magnitude of an internal angle is
( 2  4 J right angles.
\ nf
Proof. — Let AB, BC be two consecu
tive sides of the polygon and O its centre.
Then the triangles AOB, BOG are
isosceles and congruent.
^OAB=^OBA/.OBC = etc.,
^OAB + ^OBA = /.ABC.
But ^OAB + ^OBAJ_zAOB,
and (132°, Def. 2) ^ob = 4 right angles^
z.ABC = (24) right angles,
i^)}
90"
q.e.d.
Cor. The internal angles of the regular polygons expressed
in right angles and in degrees are found, by putting proper
values for «, to be as follows : —
Equilateral triangle, f 60° Octagon, . f 135°
Square, i 90° Decagon, . f 144°
Pentagon, . . . . f 108° Dodecagon, f 150°
Hexagon, . . . . ^ 120°
134°. Problem. — On a given linesegment as side to con
struct a regular hexagon. A^
Let AB be the given segment.
Constr. — On AB construct the equi pj
lateral triangle AOB (124°, Cor. i), and
with O as centre describe a circle through
A, cutting AO and BO produced in D e."
and E. Draw FC, the internal bisector of Z.AOE.
ABCDEF is the hexagon.
Proof.— ^AOB = ^EOD = O
Z.A0E=O and A0F = O
^A0B = /.B0C = ^C0D = etc. = n.
88
SYNTHETIC GEOMETRY.
And the chords AB, BC, CD, etc., being sides of congruent
equilateral triangles are all equal.
Therefore ABCDEF is a regular hexagon.
Cor. Since AOB is an equilateral triangle, AB=AO ;
.*. the side of a regular hexagon is equal to the radius of
its circumcircle.
135°. Pi'oblem. — To determine which species of regular
polygons, each taken alone, can fill the plane.
That a regular polygon of any species may be capable of
filling the plane, the number of right angles in its internal
angle must be a divisor of 4. But as no internal angle can
be so great as two right angles, the only divisors, in 133°,
Cor., are f, i, and , which give the quotients 6, 4, and 3.
Therefore the plane can be filled by 6 equilateral triangles,
or 4 squares, or 3 hexagons.
It is worthy of note that, of the three regular polygons
which can fill the plane, the hexagon includes the greatest
area for a given perimeter. As a consequence, the hexagon
is frequently found in
Nature, as in the cells of
bees, in certain tissues of
plants, etc.
Ex. I. Let D, E, F be
points of contact of the in
circle, and P, P', P", R, R',
R", etc., of the ex circles.
(131")
ThenAP = AP',CP' = CP",
andBP = BP",(ii4°,Cor.i)
.. AP' + AP
=AB+BC+AC
and, denoting the perimeter of the triangle by is^ we have
AP = AF=J,
CONSTRUCTIVE GEOMETRY. 89
CP' = J'^ = CP", BP=jr=BP".
Similarly, AR = ^^ = AR", BR' = sa = BR", etc.
Again, CD = CE = ^AE = ^AF = ^(^BF)
CE = CD=j^=BP".
Similarly, AE = AF = j  ^ = BR", etc.
These relations are frequently useful.
If we put Az to denote the distance of the vertex A from
the adjacent points of contact of the incircle, and A^, Ac to
denote its distances from the points of contact of the excircles
upon the sides d and c respectively, we have
A/= B^ = C <^ = i  ^'r,
Bi = Ca = Ac=sd,
Cz=Ad=Ba = s — c.
Exercises.
1. In testing the straightness of a " rule " three rules are
virtually tested. How ?
2. To construct a rectangle, and also a square.
3. To place a given linesegment between two given lines
so as to be parallel to a given line.
4. On a given line to find a point such that the lines joining
it to two given points may make equal angles with the
given line.
5. To find a point equidistant from three given points.
6. To find a line equidistant from three given points. How
many lines ?
7. A is a point on line L and B is not on L. To find a point
P such that PA±PB may be equal to a given segment.
8. On a given line to find a point equidistant from two
given points.
9. Through a given point to draw a line which shall form an
isosceles triangle with two given lines. How many
solutions ?
90 SYNTHETIC GEOMETRY.
10. Through two given points on two parallel lines to draw
two lines so as to form a rhombus.
11. To construct a square having one of its vertices at a
given point, and two other vertices lying on two given
parallel lines.
12. Through a given point to draw a line so that the intercept
between two given parallels may be of a given length.
13. To construct a triangle when the basal angles and the
altitude are given.
14. To construct a rightangled triangle when the hypothen
use and the sum of the sides are given.
1 5. To divide a linesegment into any number of equal parts.
16. To construct a triangle when the middle points of its
sides are given.
17. To construct a parallelogram when the diagonals and
one side are given.
18. Through a given point to draw a secant so that the chord
intercepted by a given circle may have a given length.
19. Draw a line to touch a given circle and be parallel to a
given line. To be perpendicular to a given line.
20. Describe a circle of given radius to touch two given lines.
21. Describe a circle of given radius to touch a given circle
and a given line.
22. Describe a circle of given radius to pass through a given
point and touch a given circle.
23. Describe a circle of given radius to touch two given circles.
24. To inscribe a regular octagon in a circle.
25. To inscribe a regular dodecagon in a circle.
26. A, B, C, D, ..., are consecutive vertices of a regular
octagon, and A, B', C, D', ..., of a regular dodecagon
in the same circle. Find the angles between AC and
B'C; between BE' and B'E. (Use io8°.)
27. Show that the plane can be filled by
{a) Equilateral triangles and regular dodecagons.
{b) Equilateral triangles and squares.
{c) Squares and regular octagons.
PART II
PRELIMINARY.
136°. Def. I. — The area of a plane closed figure is the por
tion of the plane contained within the figure, this portion
being considered with respect to its extent only, and without
respect to form.
A closed figure of any form may contain an area of any
given extent, and closed figures of different forms may con
tain areas of the same extent, or equal areas.
Def. 2. — Closed figures are equal to one another when they
include equal areas. This is the definition of the term
" equal " when comparing closed figures.
Congruent figures are necessarily equal, but equal figures
are not necessarily congruent. Thus, a A and a [=3 may have
equal areas and therefore be equals although necessarily
having different forms.
137°. Areas are compared by superposition. If one area
can be superimposed upon another so as exactly to cover it,
the areas are equal and the figures containing the areas are
equal. If such superposition can be shown to be impossible
the figures are not equal.
In comparing areas we may suppose one of them to
be divided into any requisite number of parts, and these
parts to be afterwards disposed in any convenient order,
since the whole area is equal to the sum of all its parts.
91
92 SYNTHETIC GEOMETRY.
Illustration. — ABCD is a square.
Then the AABC^AADC, and they are therefore equal.
6 ^ Now, if AD and DE be equal and in
line, the As ADC and EDC are con
gruent and equal.
Therefore the AABC may be taken
"e from its present position and be put into
the position of CDE. And the square ABCD is thus trans
formed into the A ACE without any change of area ;
QABCD^AACE.
It is evident that a plane closed figure may be considered
from two points of view.
1. With respect to the character and disposition of the
lines which form it. When thus considered, figures group
themselves into triangles, squares, circles, etc., where the
members of each group, if not of the same form, have at
least some community of form and character.
2. With respect to the areas enclosed.
When compared from the first point of view, the capability
of superposition is expressed by saying that the figures are
congruent. When compared from the second point of view,
it is expressed by saying that the figures are equal.
Therefore congruence is a kind of higher or double
equality, that is, an equality in both form and extent of area.
This is properly indicated by the triple lines ( = ) for con
gruence, and the double lines ( = ) for equality.
138°. Def. — The altitude of a figure is the linesegment
which measures the distance of the farthest point of a figure
from a side taken as base.
The terms base and altitude are thus correlative. A tri
angle may have three different bases and as many corre
sponding altitudes. ^1°)
In the rectangle (82°, Def. 2) two adjacent sides being
perpendicular to one another, either one may be taken as the
COMPARISON OF AREAS. 93
base and the adjacent one as the altitude. The rectangle
having two given segments as its base and altitude is called
the rectangle on these segments.
Notation. — The symbol im stands for the word rectangle
and I 7 iox parallelogram.
Rectangles and parallelograms are commonly indicated by
naming a pair of their opposite vertices.
SECTION I.
COMPARISON OF AREAS—RECTANGLES,
PARALLELOGRAMS, TRIANGLES.
139°. Theorem. — i. Rectangles with equal bases and equal
altitudes are equal.
2. Equal rectangles with equal bases have equal altitudes.
3. Equal rectangles with equal altitudes have equal bases.^
c
Q
D
G
I. In the OS BD and FH, if \
AD = EH,
and AB = EF, ^
then [=]BD=c=iFH. ^
Proof. — Place E at A and EH along
AD. Then, as ^FEH =z.BAD ="], EF will lie along AB.
And because EH = AD and EF = AB, therefore H falls at
D and F at B, and the two cds are congruent and therefore
equal. q.e.d.
2. If[=iBD = i=iFH andAD = EH, then AB = EF.
Proof.— \{ EF is not equal to AB, let AB be > EF.
Make AP = EF and complete the cuPD.
Then oP D = aF H , by the first part,
but i=iBD=aFH, (hyp.)
94 SYNTHETIC GEOMETRY.
i=iPD = [=iBD, which is not true,
.*. AB and EF cannot be unequal, or
AB = EF. q,e.d.
3. lf[=iBD=[=iFH and AB = EF, then AD = EH.
Proof.— l^et AB and EF be taken as bases and AD and
EH as altitudes (138°), and the theorem follows from the
second part. q.e.d.
Cor. In any rectangle we have the three parts, base, alti
tude, and area. If any two of these are ^iven the third is
given also.
140°. Theorem. — A parallelogram is equal to the rectangle
E .B F c on its base and altitude.
/ AC is a I 7 whereof AD is the base
and DF is the altitude.
Then nZ7AC=i=i on AD and DF.
/•r^^/— Complete the cuADFE by drawing AE ± to CB
produced.
Then AAEB = ADFC, '.• AE = DF, AB = DC,
and ^EAB = ^FDC;
.*. ADFC may be transferred to the position AEB, and
^ZIjABCD becomes the cuAEFD,
ZZZ7AC = [=i on AD and DF. q.e.d.
Cor. I . Parallelograms with equal bases and equal altitudes
are equal. For they are equal to the same rectangle.
Cor. 2. Equal parallelograms with equal bases have equal
altitudes, and equal parallelograms with equal altitudes have
equal bases.
Cor. 3. If equal parallelograms be upon the same side of
the same base, their sides opposite the common base are
in line.
141°. Theorem. — A triangle is equal to onehalf the rect.
angle on its base and altitude.
COMPARISON OF AREAS.
95
ABC is a triangle of which AC is the base and BE the
altitude. b P
Then AABC = (=i on AC and BE.
Pr^^/— Complete the £I^ABDC, of
which AB and AC are adjacent sides, a
Then AABC = ADCB,
AABC=£ZI7AD=i=] on AC and BE. (140°) q.e.d.
Cor. I. A triangle is equal to onehalf the parallelogram
having the same base and altitude.
Cor. 2. Triangles with equal bases and equal altitudes are
equal. For they are equal to onehalf of the same rectangle.
Cor. 3. A median of a triangle bisects the area. For the
median bisects the base.
Cor. 4. Equal triangles with equal bases have equal alti
tudes, and equal triangles with equal altitudes have equal bases.
Cor. 5. If equal triangles be upon the same side of the
same base, the line through their vertices is parallel to their
common base.
142°. Theorem. — If two triangles are upon opposite sides
of the same base —
1. When the triangles are equal, the base bisects the seg
ment joining their vertices ;
2. When the base bisects the segment joining their vertices,
the triangles are equal. (Converse of i.)
ABC and ADC are two triangles upon
opposite sides of the common base AC.
A'
I. If
then
AABC = AADC,
BH = HD.
Proof.— Lq\. be and DF be altitudes,
Then •.* AABC = AADC, .'. BE = DF,
AEBH=AFDH, andBH = HD.
2. IfBH = HD, then AABC=AADC.
q.e.d.
X Y
96 SYNTHETIC GEOMETRY.
Proof.— Since BH = HD, .'. AABH=AADH,
and ACBH=ACDH. (141°, Cor. 3)
/.adding, AABC=AADC. q.e.d.
143. Def. — By the stem or difference of two closed figures is
meant the sum or difference of the areas of the figures.
If a rectangle be equal to the sum of two other rectangles
its area may be so superimposed upon the others as to cover
both.
144°. Theorem. — If two rectangles have equal altitudes,
their sum is equal to the rectangle on their common altitude
and the sum of their bases.
'^ D F Proof.— Lei the as X and Y, having
equal altitudes, be so placed as to have
their altitudes in common at CD, and so
that one en may not overlap the other.
Then ^BDC=z.CDF="]»
BDF is a Hne. (38°, Cor. 2)
Similarly ACE is a line.
But BD is II to AC, and BA is  to DC  to FE ; therefore
AF is the m on the altitude AB and the sum of the bases
AC and CE ; and the [=]AF=cdAD +0CF. q.e.d.
Cor. I. If two triangles have equal altitudes, their sum is
equal to the triangle having the same altitude and having a
base equal to the sum of the bases of the two triangles.
Cor. 2. If two triangles have equal altitudes, their sum is
equal to onehalf the rectangle on their common altitude
and the sum of their bases.
Cor. 3. If any number of triangles have equal altitudes,
their sum is equal to onehalf the rectangle on their common
altitude and the sum of their bases.
In any of the above, "base" and "altitude" are inter
changeable.
COMPARISON OF AREAS. . 97
145°. Theorem. — Two lines parallel to the sides of a
parallelogram and intersecting upon a diagonal divide the
parallelogram into four parallelograms such that the two
through which the diagonal does not pass are equal to one
another.
In the£IZ7ABCD,EFis  to AD and
GH is II to BA, and these intersect at
O on the diagonal AC.
Then /=Z7BO = Z=Z70D.
/'r^^/— AABC = AADC, and AAEO=AAHO,
and AOGC = AOFC ; (141°, Cor. i)
but £IZ7BO=AABCAAEOAOGC,
and C=70D=AADCAAH0A0FC.
'B0 = /=Z70D. q.e.d.
Cor. I. £Z7BF = ^=7GD.
Cor. 2. If £Z:7B0 = £Z70D, O is on the diagonal AC.
(Converse of the theorem.)
For if O is not on the diagonal, let the diagonal cut EF in
O'. Then Z=II7BO' = iCZ70'D. (145°)
But £ZI7B0' is < ^IZ7BO, and .CZJO'D is > ZZZ70D ;
.'. I 7 BO is > I 7 OD, which is contrary to the hypothesis;
.'. the diagonal cuts EF in O.
Ex. Let ABCD be a trapezoid.
(84°, Def ) In line with AD make
DE = BC, and in line with BC make
CFAD.
Then BF = AE and BFEA is a
EX
But the trapezoid CE can be superimposed on the trape
zoid DB, since the sides are respectively equal, and
^F=A, and :lE = B, etc.
trapezoid BD=/^I7BE,
or, a trapezoid is equal to onehalf the rectangle on its alti
tude and the sum of its bases.
G
98 SYNTHETIC GEOMETRY.
Exercises.
1. To construct a triangle equal to a given quadrangle.
2. To construct a triangle equal to a given polygon.
3. To bisect a triangle by a line drawn through a given
point in one of the sides.
4. To construct a rhombus equal to a given parallelogram,
and with one of the sides of the parallelogram as
its side.
5. The three connectors of the middle points of the sides
of a triangle divide the triangle into four equal
triangles.
6 Any line concurrent with the diagonals of a parallelogram
bisects the parallelogram.
7. The triangle having one of the nonparallel sides of a
trapezoid as base and the middle point of the opposite
side as vertex is onehalf the trapezoid.
8. The connector of the middle points of the diagonals of a
quadrangle is concurrent with the connectors of the
middle points of opposite sides.
9. ABCD is a parallelogram and O is a point within. Then
AA0B+AC0D=^ZIZ7.
What does this become when O is without ?
TO. ABCD is a parallelogram and O is a point within. Then
AAOC = AAODAAOB.
What does this become when O is without? (This
theorem is important in the theory of Statics.)
11. Bisect a trapezoid by a line through the middle point of
one of the parallel sides. By a line through the
middle point of one of the nonparallel sides.
12. The triangle having the three medians of another tri
angle as its sides has threefourths the area of
the other.
COMPARISON OF AREAS.
99
POLYGON AND CIRCLE.
146°. Def. — The sum of all the sides of a polygon is called
its perimeter^ and when the polygon is regular every side is
at the same distance from the centre. This distance is the
apothem of the polygon.
Thus if ABC D... LA be a regu
lar polygon and O the centre
(132°, Def. 2), the triangles OAB,
OBC, ... are all congruent, and Lj
OP = OQ = etc. o
AB + BC + CD + ... + LA is the perimeter and OP, per
pendicular upon AB, is the apothem.
147°. Theorem. — A regular polygon is equal to onehalf the
rectangle on its apothem and perimeter.
Proof.— T\i^ triangles AOB, BOC, ... LOA have equal
altitudes, the apothem OP, .'. their sum is onehalf the cu on
OP and the sum of their bases AB + BCf... LA. (144°, Cor. 3)
But the sum of the triangles is the polygon, and the sum of
their bases is the perimeter.
.*. a regular polygon = [Z] on its apothem and perimeter.
148°. Of a limit. — A limit or limiting value of a variable
is the value to which the variable by its variation can be
made to approach indefinitely near,
but which it can never be made to
pass.
Let ABCD be a square in its cir
cumcircle. If we bisect the arcs AB,
BC, CD, and DA in E, F, G, and H,
we have the vertices of a regular
octagon AEBFCGDHA. Now, the
area of the octagon approaches nearer
to that of the circle than the area of the square does ; and the
lOO SYNTHETIC GEOxMKTRY.
perimeter of the octagon approaches nearer to the length of
the circle than the perimeter of the square does ; and the
apothem of the octagon approaches nearer to the radius of
the circle than the apothem of the square does.
Again, bisecting the arcs AE, EB, BF, etc., in I, J, K, etc.,
we obtain the regular polygon of i6 sides. And all the fore
going parts of the polygon of i6 sides approach nearer to
the corresponding parts of the circle than those of the
octagon do.
It is evident that by continually bisecting the arcs, we may
obtain a series of regular polygons, of which the last one may
be made to approach the circle as near as we please, but that
however far this process is carried the final polygon can never
become greater than the circle, nor can the final apothem
become greater than the radius.
Hence the circle is the limit of the perimeter of the regular
polygon when the number of its sides is endlessly increased,
and the area of the circle is the limit of the area of the poly
gon, and the radius of the circle is the limit of the apothem
of the polygon under the same circumstances.
149°. Theorem. — A circle is equal to onehalf the rectangle
on its radius and a linesegment equal in length to the circle.
Proof. — The is the limit of a regular polygon when the
number of its sides is endlessly increased, and the radius of
the is the limit of the apothem of the polygon.
But, whatever be the number of its sides, a regular polygon
is equal to onehalf the CD on its apothem and perimeter. (147°)
.'. a is equal to onehalf the [=] on its radius and a line
segment equal to its circumference.
Exercises.
I. Show that a regular polygon may be described about a
circle, and that the limit of its perimeter when the
number of its sides is increased indefinitely is the
circumference of the circle.
MEASUREMENT OF LENGTHS AN'B AREA^: lOI
2. The difference between the areas of tw.a re^uai: polygons^
one inscribed in a circle and the other circumscribed
about it, vanishes at the limit when the number of
sides of the polygons increases indefinitely.
3. What is the limit of the internal angle of a regular polygon
as the number of its sides is endlessly increased ?
SECTION 11.
MEASUREMENT OF LENGTHS AND AREAS.
150°. Def. — I. That part of Geometry which deals with
the measures and measuring of magnitudes is Metrical
Geometry.
2. To measure a magnitude is to determine how many unit
magnitudes of the same kind must be taken together to form
the given magnitude. And the number thus determined is
called the measure of the given magnitude with reference to
the unit employed. This number may be a whole or a frac
tional number, or a numerical quantity which is not arith
metically expressible. The word "number" will mean any
of these.
3. In measuring length, such as that of a linesegment, the
unit is a segment of arbitrary length called the unitlength.
In practical work we have several such units as an inch, a
foot, a mile, a metre, etc., but in the Science of Geometry the
unitlength is quite arbitrary, and results obtained through it
are so expressed as to be independent of the length of the
particular unit employed.
4. In measuring areas the unit magnitude is the area of the
square having the unitlength as its side. This area is the
unitarea. Hence the unitlength and unitarea are not both
102' ■ ' SYNTHETIC GEOMETRY.
kWLt^ry,, cforl if' tiiher is fixed the other is fixed also, and
determinable.
This relation between the unitlength and the unitarea is
conventional, for we might assume the unitarea to be the
area of any figure which is wholly determined by a single
segment taken as the unitlength : as, for example, an equi
lateral triangle with the unitlength as side, a circle with the
unitlength as diameter, etc. The square is chosen because
it offers decided advantages over every other figure.
For the sake of conciseness we shall symbolize the term
unitlength by ?/./. and unitarea by z^.a.
5. When two magnitudes are such that they are both
capable of being expressed arithmetically in terms of some
common unit they are coinviensurable^ and when this is not
the case they are mcomme7tsiirable.
Illus. — Let ABCD be a square, and let
EF and HG be drawn _L to BD, and EH
and FG _L to AC. Then EFGH is a
square (82°, Cor. 5), and the triangles
AEB, APB, BFC, BPC, etc., are all equal
H D G to one another.
If AB be taken as u.l.^ the area of the square AC is the
u.a.\ and if EF be taken as u.l., the area of the square
EG is the 21. a.
In the first case the measure of the square AC is i, and
that of EG is 2 ; and in the latter case the measure of the
square EG is i, and that of AC is \. So that in both cases
the measure of the square EG is double that of the square AC.
.'. the squares EG and AC are commensurable.
Now, if AB be taken as ?/./., EF is not expressible arith
metically, as will be shown hereafter.
.". AB and EF are incommensurable.
151°. Let AB be a segment trisected at E and F (127°),
and let AC be the square on AB. ThenAD = AB. And
p
1
2
3
4
'^S
6
7
8
9
G H
MEASUREMENT OF LENGTHS AND AREAS. IO3
if AD be trisected in the points K and M, and through
E and F s be drawn to AD, and through ^ e F
K and M s be drawn to AB, the figures
I, 2, 3, 4, 5, 6, 7, 8, 9 are all squares equal to
one another.
Now, if AB be taken as u.l.^ AC is the u.a. ;
and if AE be taken as w./., any one of the
small squares, as AP, is the u.a. And the segment AB con
tains AE 3 times, while the square AC contains the square
AP in three rows with three in each row, or 3'^ times.
.*. if any assumed u.l. be divided into 3 equal parts for a
new ?/./., the corresponding u.a. is divided into 3^ equal parts
for a new u.a. And the least consideration will show that this
is true for any whole number as well as 3.
.'. I. If an assumed u.l. be divided into n equal parts for a
new u.l.., the corresponding u.a. is divided into 11^ equal parts
for a new u.a.\ n denoting any whole number.
Again, if any segment be measured by the u.l. AB, and
also by the 21.I. AE, the measure of the segment in the latter
case is three times that in the former case. And if any area
be measured by the ti.a. AC, and also by the ti.a. AP, the
measure of the area in the latter case is 3^ times its measure
in the former case. And as the same relations are evidently
true for any whole number as well as 3,
.'. 2. If any segment be measured by an assumed u.l. and
also by th of the assumed u.l. as a new u.l.., the measure of
;/
the segment in the latter case is n times its measure in the
former. And if any area be measured by the corresponding
u.a.% the measure of the area in the latter case is n^ times its
measure in the former case ; n being any whole number.
This may be stated otherwise as follows : —
By reducing an assumed u.l. to ^th of its original length,
n
we increase the measure of any given segment 11 times, and
we increase the measure of any given area J1^ times ; ;/ being
a whole number.
104 SYNTHETIC GEOMETRY.
In all cases where a u.l. and a u.a. are considered together,
they are supposed to be connected by the relation of 1 50°,
3 and 4.
I52^ Theorem. — The number of unitareas in a rectangle
is the product of the numbers of unitlengths in two adjacent
sides.
The proof is divided mto three cases.
1. Let the measures of the adjacent sides with respect to
c the unit adopted be whole numbers.
Let AB contain the assumed 11. l. a
times, and let AD contain it b times.
Then, by dividing AB into a equal
parts and drawing, through each point
of division, lines  to AD, and by dividing AD into b equal
parts and drawing, through each point of division, lines  to
AB, we divide the whole rectangle into equal squares, of
which there are a rows with b squares in each row.
the whole number of squares is ab.
But each square has the u.l. as its side and is therefore the u.a.
u.a.s in AC=^?/./.s in AB x u.l.s in AD.
We express this relation more concisely by writing symbolic
ally [=]AC = AB.AD,
where [=:AC means" the number ofu.a.s in[=]AC,"and AB and
AD mean respectively "the numbers of w./.s in these sides."
And in language we say, the area of a rectangle is the pro
duct of its adjacent sides ; the proper interpretation of which
is easily given.
2. Let the measures of the adjacent sides with respect to
the unit adopted be fractional.
Then, ".• AB and AD are commensurable, some unit will
be an aliquot part of each (150°, 5). Let the new unit be
th of the adopted unit, and let AB contain p of the new
units, and AD contain ^ of them.
The measure of [z:AC in terms of the new 71. a. is ^a
MEASUREMENT OF LENGTHS AND AREAS. I05
(i 52°, i), and the measure of the ciiAC in terms of the adopted
unit is ^f (151°, 2)
But the measure of AB in terms of the adopted //./. is
'^, and of AD it is ^ . (151", 2)
n n
and ^=^.2,
or nAC^AB.AD.
Illus. — Suppose the measures of AB and AD to some
unitlength to be 3.472 and 4.631. By taking a //./. 1000
times smaller these measures become the whole numbers
3472 and 4631, and the number of corresponding u.a.s in the
rectangle is 3472 x 4631 or 16078832 ;
and dividing by looo^, the measure of the area with respect to
the original 2i.L is 16.078832 = 3.472 x 4.631.
3. Let the adjacent sides be incommensurable. There is
now no ti.l. that will measure both AB b c
and AD.
If [=]AC is not equal to AB . AD, let it
be equal to AB . AE, where AE has a a e h d
measure different from AD ; and suppose, first, that AE is <
AD, so that E lies between A and D.
With any u.l. which will measure AB, and which is less
than ED, divide AD into parts. One point of division at
least must fall between E and D ; let it fall at H. Complete
the rectangle BH.
Then AB and AH are commensurable, and
□BH^AB.AH,
but miBD^AB.AE; (hyp.)
and i=]BHis<[=!BD;
AB.AH is<AB.AE,
and AB being a common factor
AH is < AE ; which is not true.
.*. If [=iAC = AB . AE, AE cannot be < AD, and similarly
it may be shown that AE cannot be > AD ; .*. AE = AD, or
[=]AC=AB.AD. q.e.d.
i06 SYNTHETIC GEOMETRY.
153°. The results of the last article in conjunction with
Section I. of this Part give us the following theorems.
1. The area of a parallelogram is the product of its base
and altitude. (140°)
2. The area of a triangle is onehalf the product of its base
and altitude. (141°)
3. The area of a trapezoid is onehalf the product of its
altitude and the sum of its parallel sides. (M5°j Ex.)
4. The area of any regular polygon is onehalf the product
of its apothem and perimeter. (H7°)
5. The area of a circle is onehalf the product of its radius
and a linesegment equal to its circumference. (H9°)
Ex. I. Let O, O' be the centres of the incircle and of the
P excircle to the side BC (131°);
V y/^ \ and let OD, O'P" be perpen
^^" \ diculars on BC, OE, O'P' per
^(^;\d^^^^^^°' pendiculars on AC, and OF, O'P
yy^^^^^^\y^/ ^"^ AB Then
A ^ c*\~P' ^,^^1 0'P = 0'P' = 0'P''=r';
.. AABC = AAOBABOC + ACOA
= UB. OF + ^BC.ODtCA.OE (153°, 2)
= ^ r X perimeter = 7's,
where s is the half perimeter ;
/\ = rs.
Ex. 2. AABC = AAO'B + AAO'CABO'C
= \0'V . AB f ^O'P'. AC  0'P". BC
r=\r\b^ca) = r\sd),
where r' is the radius of the excircle to side a ;
l\=r\sa).
Similarly, A = ^"''■^  b)
MEASUREMENT OF LENGTHS AND AREAS. lO;
Exercises.
1. l=L+J +JL.
I ' If ' tit
r r r r
2. /\^ = rr'r"r"'.
3. What relation holds between the radius of the incircle
and that of an ex circle when the triangle is equiangular.'*
Note. — When the diameter of a circle is taken as the u.l.
the measure of the circumference is the inexpressible numeri
cal quantity symbolized by the letter tt, and which, expressed
approximately, is 3. 141 5926....
4. What is the area of a square when its diagonal is taken
as the u.l. ?
5. What is the measure of the diagonal of a square when
the side is taken as the ?/./.? (150°, 5)
6. Find the measure of the area of a circle when the di
ameter is the u.l. When the circumference is the u.l.
7. If one linesegment be twice as long as another, the
square on the first has four times the area of the
square on the second. (151°) 2)
8. If one linesegment be twice as long as another, the
equilateral triangle on the first is four times that on
the second. (141°)
9. The equilateral triangle on the altitude of another equilat
eral triangle has an area threefourths that of the other.
10. The three medians of any triangle divide its area into
six equal triangles.
1 1. From the centroid of a triangle draw three lines to the
sides so as to divide the triangle into three equal
quadrangles.
12. In the triangle ABC X is taken in BC, Y in CA, and Z
in AB, so that BX = iBC, CY = iCA, and AZ = ^AB.
Express the area of the triangle XYZ in terms of that
of ABC.
13. Generalize 12 by making BX = BC, etc.
14. Show that a = s(\ ^\ = {r" ■\r'
108 SYNTHETIC GEOMETRY.
SECTION III.
GEOMETRIC INTERPRETATION OF
ALGEBRAIC FORMS.
1 54''. We have a language of symbols by which to express
and develop mathematical relations, namely, Algebra. The
symbols of Algebra are quantitative and operative, and it is
very desirable, while giving a geometric meaning to the
symbol of quantity, to so modify the meanings of the sym
bols of operation as to apply algebraic forms in Geometry.
This application shortens and generalizes the statements of
geometric relations without interfering with their accuracy.
Elementary Algebra being generalized Arithmetic, its
quantitative symbols denote numbers and its operative sym
bols are so defined as to be consistent with the common
properties of numbers.
Thus, because 2 + 3 = 312 and 2.3 = 3.2, we say that
a^rb = b\a and ab=ba.
This is called the commutative law. The first example is
of the existence of the law in addition, and the second of its
existence in multiplication.
The commutative law in addition may be thus expressed : —
A sum is independent of the order of its addends ; and in
multiplication — A product is independent of the order of its
factors.
Again, because 2(3 + 4) = 2 . 3 f 2 . 4, we say that
a{b + c) — ab + ac.
This is called the distributive law and may be stated
thus : — The product of multiplying a factor by the sum of
several terms is equal to the sum of the products arising from
multiplying the factor by each of the terms.
These two are the only laws which need be here mentioned.
And any science which is to employ the forms of Algebra
INTERPRETATION OF ALGKBRAIC FORMS. IO9
must have that, whatever it may be, which is denoted by the
algebraic symbol of quantity, subject to these laws.
155°. As already explained in 22° we denote a single hne
segment, in the oneletter notation, by a single letter, as a,
which is equivalent to the algebraic symbol of quantity ; and
hence,
A single algebraic symbol of quantity is to be interpreted
geometrically as a line segment.
It must of course be understood, in all cases, that in em
ploying the twoletter notation for a segment (22°), as " AB,"
the two letters standing for a single linesegment are equiva
lent to but a single algebraic symbol of quantity.
The expression a^b denotes a segment equal in length to
those denoted by a and b together.
Similarly '2a = a\a^ and na means a segment as long as n
of the segments a placed together in line, n being any
numerical quantity whatever. (28°)
a — b^ when a is longer than b, is the segment which is left
when a segment equal to b is taken from a.
Now it is manifest that, if a and b denote two segments,
avb=b\a, and hence that the commutative law for addition
applies to these symbols when they denote magnitudes having
length only, as well as when they denote numbers.
1 56°. Line in Opposite Senses. — A quantitative symbol, a, is
in Algebra always affected with one of two signs, f or  ,
which, while leaving the absolute value of the symbol un
changed, impart to it certain properties exactly opposite in
character.
This oppositeness of character finds its complete interpreta
tion in Geometry in the opposite directions of every segment.
Thus the segment in the margin may be con a
sidered as extending/r<77/z A to B or from B to A. A B
With the twoletter notation the direction can be denoted
by the order of the letters, and this is one of the advantages
I TO SYNTHETIC GEOMETRV.
of this notation ; but with the oneletter notation, if we denote
the segment AB by + a^ we ;;/«j/ denote the segment B A
by a.
But as there is no absolute reason why one direction rather
than the other should be considered positive, we express the
matter by saying that AB and BA, or +a and a, denote
the same segment taken in opposite senses.
Hence the algebraic distinction of positive and negative as
applied to a single symbol of quantity is to be interpreted
geometrically by the oppositeness of direction of the segment
denoted by the symbol.
Usually the applications of this principle in Geometry are
confined to those cases in which the segments compared as
to sign are parts of one and the same line or are parallel.
Ex. I. Let ABC be any A and let BD be the altitude from
the vertex B.
Now, suppose that the sides AB and BC
undergo a gradual change, so that B may
move along the Hne BB' until it comes
into the position denoted by B'.
"a d c Then the segment AD gradually di
minishes as D approaches A ; disappears when D coincides
with A, in which case B comes to be vertically over A
and the A becomes rightangled at A ; reappears as D
passes to the left of A, until finally we may suppose that
one stage of the change is represented by the AAB'C with
its altitude B'D'.
Then, if we call AD positive, we must call AD' negative, or
we must consider AD and AD' as having opposite senses.
Again, from the principle of continuity (104°) the foot of
the altitude cannot pass from D on the right of A to D' on
the left of A without passing through every intermediate
point, and therefore passing through A. And thus the seg
ment AD must vanish before it changes sign.
This is conveniently expressed by saying that a line
INTERPRETATION OF ALGEBRAIC FORMS. 1 1 1
segjnent changes sign when it passes through zero; passing
through zero being interpreted as vanishing and reappearing
on the other side of the zeropoint.
Ex. 2. ABCD is a normal quadrangle. Consider the side
AD and suppose D to move along the line g
DA until it comes into the position D'.
The segments AD and AD' are opposite
in sense, and ABCD' is a crossed quad
rangle.
.•. the crossed quadrangle is derived from "^ ^ °
the normal one by changing the sense of one of the sides.
Similarly, if one of the sides of a crossed quadrangle be
changed in sense the figure ceases to be a crossed quadrangle.
Ex. 3. This is an example where segments which are par
allel but which are not in line have
opposite senses.
ABC is a A ^rid P is any point
within from which perpendiculars
PD, PE, PF are drawn to the sides.
Suppose that P moves to P'.
Then PF becomes P'F', and PF
and P'F' being in the same direc
tion have the same sense. Similarly PE becomes P'E',
and these segments have the same sense. But PD becomes
P'D' which is read in a direction opposite to that of PD.
Hence PD and P'D' are opposite in sense.
But PD and P'D' are perpendiculars to the same line from
points upon opposite sides of it, and it is readily seen that in
passing from P to P' the _LPD becomes zero and then changes
sense as P crosses the side BC.
Hence if by any continuous change in a figure a point
passes from one side of a line to the other side, the peipen
dicular from that point to the line changes sense.
Cor, If ABC be equilateral it is easily shown that
PD}PEhPF = a constant.
112 SYNTHETIC GEOMETRY.
And if we regard the sense of the segments this statement is
true for all positions of P in the plane.
157°. Product. — The algebraic form of a product of two
symbols of quantity is interpreted geometrically by the rect
angle having for adjacent sides the segments denoted by the
quantitative symbols.
This is manifest from Art. 152°, for in the form ab the
single letters may stand for the measures of the sides, and the
product ab will then be the measure of the area of the rect
angle.
If we consider ab as denoting a en having a as altitude and
b as base, then ba will denote the izu having b as altitude and
a as base. But in any en it is immaterial which side is taken
as base (138°) ; therefore ab^ba, and the form satisfies the
commutative law for multiplication.
Again, let AC be the segment b\Cy and AB be the segment
A D c a, so placed as to form the ni^a{b + c) or
" ^ I AF. Taking AD =^^, let DE be drawn
! II to AB. Then AE and DF are rect
F angles and DE = AB = rt:.
cziAE is nnaby and cziDF is nnac ;
n2a{b >rc) = nnab f \:nac^
and the distributive law is satisfied.
158°. We have then the two following interpretations to
which the laws of operation of numbers apply whenever such
operations are interpretable.
I. A si7tgle symbol of qua7itity denotes a line segment.
As the sum or difference of two linesegments is a segment,
the sum of any number of segments taken in either sense is a
segment.
Therefore any number of single symbols of quantity con
nected by + and  signs denotes a segment, as a\b^
abirc, ab + {c), etc.
INTERPRETATION OF ALGEBRAIC FORMS. II 3
For this reason such expressions or forms are often called
linear^ even in Algebra.
Other forms of linear expressions will appear hereafter.
2. The product form of two symbols of quantity denotes the
rectangle whose adjaceitt sides are the segments denoted by the
single symbols.
A rectangle encloses a portion of the plane and admits of
measures in two directions perpendicular to one another,
hence the area of a rectangle is said to be of two dimensions.
And as all areas can be expressed as rectangles, areas in
general are of two dimensions.
Hence algebraic terms which denote rectangles, such as ab^
{a\b)c, {airb){c+d\ etc., are often called rectangular terms,
and are said to be of two dimensions. a d c
Ex. Take the algebraic identity
a{b\c) = ab\ac.
The geometric interpretation gives — be f
If there be any three segments {a, b, c) the cu on the first
and the sum of the other two {b, c) is equal to the sum of the
czis on the first and each of the other two.
The truth of this geometric theorem is evident from an
inspection of a proper figure.
This is substantially Euclid, Book II., Prop. i.
159°. Square. — When the segment b is equal to the seg
ment a the rectangle becomes the square on a. When this
equality of symbols takes place in Algebra we write a^ for aa,
and we call the result the " square " of a, the term " square "
being derived from Geometry.
Hence the algebraic form of a square is interpreted geo
metrically by the square which has for its side the segment
denoted by the root symbol.
Ex. In the preceding example let b become equal to a^ and
a{a + ^) = ^ 4 iiCj
H
114 SYNTHETIC GEOMETRY.
which interpreted geometrically gives —
If a segment (a + c) be divided into two parts {a, c), the
rectangle on the segment and one of its parts {a) is equal to
the sum of the square on that part {aP) and the rectangle on
the two parts {ac).
This is Euclid, Book II., Prop. 3. The truth of the geo
metric theorem is manifest from a proper figure.
160°. Homogeneity. — Let a, b, c, d denote segments. In
the linear expressions a + b, ab, etc., and in the rectangular
expressions ab + cd, etc., the interpretations of the symbols +
and — are given in 28°, 29°, and 143°, and are readily in
telligible.
But in an expression such as ab + c we have no interpreta
tion for the symbol f if the quantitative symbols denote
linesegments. For ah denotes the area of a rectangle and c
denotes a segment, and the adding of these is not intelligible
in any sense in which we use the word " add."
Hence an expression such as ab + c is not interpretable
geometrically This is expressed by saying that — An alge
braic form has no geometric interpretation unless the form is
homogeneous, i.e., unless each of its terms denotes a geo
metric element of the same kind.
It will be observed that the terms "square," "dimensions,"
"homogeneous," and some others have been introduced into
Algebra from Geometry.
161°. Rectangles i7i Opposite Senses. — The algebraic term
ab changes sign if one of its factors changes sign. And to be
consistent we must hold that a rectangle changes sense
whenever one of its adjacent sides changes sense.
Thus the rectangles AB . CD and AB . DC are the same in
extent of area, but have opposite senses. And
AB.CD + AB.DC = o,
for the sum = AB(CD + DC),
and CD + DC = o. (156°)
INTERPRETATION OF ALGEBRAIC FORMS.
15
+
—
f
c
As the sense of a rectangle depends upon that of a line
segment there is no difficulty in determining when rectangles
are to be taken in different senses.
The following will illustrate this part of the subject : —
Let OA = OA'and OC = OC', and let bob,
the figures be rectangles.
ns OA. OC and OA'. OC have the
common altitude OC and bases equal
in length but opposite in sense. There
fore OA.'OC and OA'. OC are opposite in sense, and if we
call [zdOA. OC positive we must call cdOA'. OC negative.
Again, cds OC . OA' and OC . OA' have the common base
OA' and altitudes equal in length but opposite in sense.
Therefore cds OC. OA' and OC. OA' are opposite in sense,
and therefore ens OA. OC and OA'. OC are of the same sense.
Similarly ens OC . OA' and OC. OA are of the same sense.
These four czis are equivalent to the algebraic forms : —
+ a. + l>= +a^, a.\b= ab,
{■a.b= ab, a.b=+ab.
Ex. I. ABCD is a normal quadrangle whose opposite sides
meet in O, and OE, OF are altitudes
of the As DOC and AOB respectively.
The Od. ABCD
=ADOCAAOB,
= c=]DC.OE[=]AB.OF. (141°)
Now, let A move along AB to A'
(104°). Then O comes to O', F to F',
E to E', and O'E', O'F' become the altitudes of the As
DO'C and A'O'B respectively.
But O'E' and OE have the same sense, therefore DC . OE
and DC . OE' have the same sense.
Also, A'B is opposite in sense to AB, and O'F' is opposite
in sense to OF. (156°, Ex. 3)
AB . OF and A'B . O'V have the same sense ;
Od. A'BCD = ADO'CAA'0'B;
ii6
SYNTHETIC GEOMETRY.
/ 
yr
N
/\
\U
^
J
s
D
R
.^Q
or, the area of a crossed quadrangle must be taken to be the
difference between the two triangles which constitute it.
162°. Theorem. — A quadrangle is equal to onehalf the
parallelogram on its diagonals taken in both magnitude and
relative direction.
P B .Q ABCD is a quadrangle of which AC
and BD are diagonals. Through B and
D let PQ and RS be drawn  to AC, and
through A and C let PS and OR be
drawn  to BD. Then PORS is the
I 7 on the diagonals AC and BD in
both magnitude and direction.
Qd. ABCD = z=::7PQRS.
Proof.— Qi^. ABCD
=AABC + AADC(istFig.)
= AABC  AADC (2ndFig.)
A (161°, Ex.)
AABC = i£II7P0CA,
AADC = nZ7SRCA, (141°, Cor. i)
Qd. ABCD=^£ZI7P0RS in both figures.
This theorem illustrates the generality of geometric results
when the principle of continuity is observed, and segments
and rectangles are considered with regard to sense. Thus
the principle of continuity shows that the crossed quadrangle
is derived from the normal one (156°, Ex. 2) by changing the
sense of one of the sides.
This requires us to give a certain interpretation to the area
of a crossed quadrangle (161°, Ex. i), and thence the present
example shows us that all quadrangles admit of a common
expression for their areas.
163°. A rectangle is constructed upon two segments which
are independent of one another in both length and sense.
But a square is constructed upon a single segment, by using
INTERPRF.TATION OF ALGEBRAIC FORMS. 11/
it for each side. In other words, a rectangle depends upon
two segments while a square depends upon only one.
Hence a square can have only one sign, and this is the
one which we agree to call positive.
Hence a squtwe is always positive.
164°. The algebraic equation ab=cd tells us geometrically
that the rectangle on the segments a and b is equal to the
rectangle on the segments c and d.
But the same relation is expressed algebraically by the form
cd
therefore, since rz is a segment, the form ^ is linear and
o
denotes that segment which with a determines a rectangle
equal to cd.
Hence an expression such as  + — + — is linear.
cab
165°. The expression a^ = bc tells us geometrically that the
square whose side is a is equal to the rectangle on the seg
ments b and c.
But this may be changed to the form
a = J be.
Therefore since «: is a segment, the side of the square, the form
f^fbc is linear.
Hence the algebraic forin of the square root of the product
of two symbols of quantity is interpreted geometrically by the
side of the square which is equal to the rectangle on the
segments denoted by the quatititative symbols.
166°. The following theorems are but geometric interpreta
tions of wellknown algebraic identities. They may, however,
be all proved most readily by superposition of areas, and
thus the algebraic identity may be derived from the geo
metric theorem.
Ii8
SYNTHETIC GEOMETRY.
ab
b'
a«
ab
I. The square on the sum of two segments is equal to
the sum of the squares on the segments and twice the
a b rectangle on the segments.
2. The rectangle on the sum and differ
ence of two segments is equal to the
difference of the squares on these segments.
3. The sum of" the squares on the sum and on the differ
ence of two segments is equal to twice the sum of the squares
on the segments.
{a + bf + {a bf = 2(^2 + d'^)^ a>b.
4. The difference of the squares on the sum and on the
difference of two segments is equal to four times the rectangle
on the segments.
{a^bf{abf = ^ab, a>b.
Exercises.
I. To prove 4 of Art. 166°.
Let AH = ^ and HB = ^ be the segments, so that
A g H 6 B AB is their sum. Through H
draw HG 1 to EC, a side of the
square on AB. Make HG = rt,
and complete the square FGLE,
as in the figure, so that FG is
a — b.
Then AC is {a + bf and EG is
{a — by, and their difference is
the four rectangles AF, HK, CL, and DE ; but these
each have a and b as adjacent sides.
{a + bf{ab)'^=^ab.
State and prove geometrically {a  b)^= or + b^  2ab.
State and prove geometrically
{a\b){a{c) = a' + a{b + c) + bc.
State and prove geometrically by superposition of areas
ab b
ab
b
b
ab
E F
L G
6 ab
(«+b)'^
AREAL RELATIONS. 1 I9
{a\i?)\{abf\2{(i + b){ab)=^{2ay,v^hQVQ a and b
denote segments.
If a given segment be divided into any three parts the
square on the segment is equal to the sum of the
squares on the parts together with twice the sum of
the rectangles on the parts taken two and two.
Prove, by comparison of areas from the Fig. of Ex. i, that
{a\bf = 2b{a + b) + 2b{a'b)>r{abf, and state the
theorem in words.
SECTION IV.
AREAL RELATIONS.
167°. Def. — 1. The segment which joins two given points
is called the join of the points ; and where no reference is
made to length the Join of two points may be taken to mean
the hne determined by the points.
2. The foot of the perpendicular from a given point to a
given line is the orthogonal projection^ or simply the projec
tion^ of the point upon the line.
3. Length being considered, the join of the projection of
two points is the projection of the
join of the points.
Thus if L be a given line and P,
Q, two given points, and PP', QQ'
perpendiculars upon L ; PQ is the
join of P and Q, P' and Q' are the p^ Q^ l
projections of P and Q upon L, and the segment P'Q' is the
projection of PO upon L.
168°. Theorem.— T\iQ: sum of the projections of the sides of
20
SYxNTHETIC GEOMETRY.
any closed rectilinear figure, taken in cyclic order with
c respect to any line, is zero.
ABCD is a closed rectilinear
figure and L is any line. Then
Pr.AB + Pr.BC + Pr.CD + Pr.DA=o
A' B o D' L /'/T?^— Draw the perpendiculars
AA', BE', CC, DD', and the sum of the projections becomes
A'B' + B'C' + C'D' + D'A'.
But D'A' is equal in length to the sum of the three others and
is opposite in sense. .*. the sum is zero.
It is readily seen that since we return in every case to the
point from which we start the theorem is true whatever be
the number or disposition of the sides.
This theorem is of great importance in many investigations.
Cor. Any side of a closed rectilinear figure is equal to the
sum of the projections of the remaining sides, taken in cyclic
order, upon the line of that side.
Def. — In a rightangled triangle the side opposite the right
angle is called the hypothe?tuse, as distinguished from the
remaining two sides.
Theorem. — In any rightangled triangle the square
on one of the sides is equal to the rectangle
on the hypothenuse and the projection of
that side on the hypothenuse.
ABC is rightangled at B, and BD is ±
AC. Then AB2 = AC.AD.
Proof.— \.^\. AF be the □ on AC, and let
EH be II to AB, and AGHB be a en, since
Z_B is a ~
Then. ^GAB = z.EAC=~, (82^ Cor. 5)
z.CAB = ^EAG.
Also, AE = AC, (hyp.)
ACAB = AEAG, (64°)
and .. AG = AB, and AH is the n P" '^R
AREAL RELATIONS.
121
Now nAH=£IZ7ABLE = i=]ADKE, (140°)
i.e., AB'^ = AC.AD. q.e.d.
As this theorem is very important we give an alternative
proof of it.
Proof.— KY is the D on AC and AH
is the D on AB, and BD is _L AC.
^GAB=Z.CAE=~1. (82°, Cor. 5)G<
.*.
^GAC=z.BAE.
Also,
AG = AB,
and
AC = AE,
(82°, Cor^ 5)
.'.
AGAC=ABAE.
(5O
But
AGAC = inAH,
(141°)
and
ABAE=i=iAK,
DAH=dAK,
i.e..
AB2=AC.AD.
Cor. I. Since AB2 = AC . AD we have from symmetry
BC2=AC.DC,
.. adding, AB2 + BC2 = AC(AD + UC),
or AB2 + BC2 = AC2.
.'. The square on the hypothenuse of a rightangled triangle
is equal to the sum of the squai'es on the remaining sides.
This theorem, which is one of the most important in the
whole of Geometry, is said to have been discovered by
Pythagoras about 540 B.C.
Cor. 2. Denote the sides by a and c and the hypothenuse
by b, and let a^ and c^ denote the projections of the sides a
and c upon the hypothenuse.
Then a^ = af), c'^^c^b.,
and «2 + ^2_^2_
Cor. 3. Denote the altitude to the hypothenuse by p.
Then b=^c^\a^, and ADB and CDB are rightangled at D,
^2= ^^2 + ^^2 .,.2^^^^ ; (166°, i)
add 2/J2 to each side and
b^ + ip^ = c;^ 1/2 + a^ +/2 + 2r^.7„
122 SYNTHETIC GEOiMETRY.
or 6^ + rt2_.2^2_^2 + ^2 + 2^i^i. (Cor. l)
or BD2 = AD.DC,
i.e.^ the square on the altitude to the hypothenuse is equal to
the rectangle on the projections of the sides on the hypo
thenuse.
Def. — The side of the square equal in area to a given
rectangle is called the mean proportional or the geoinetric
mean between the sides of the rectangle.
Thus the altitude to the hypothenuse of a rightangled /\
is a geometric mean between the segments into which the
altitude divides the hypothenuse. (169°, Cor. 3)
And any side of the A is a geometric mean between the
hypothenuse and its projection on the hypothenuse. (169'')
170°. Theorem. — If the square on one side of a triangle is
equal to the sum of the squares on the remaining sides, the
triangle is rightangled at that vertex which is opposite the
side having the greatest square. (Converse of 169°, Cor.)
If AC2 = AB2 + BC^, the Z.B is a "1
Proof.— l.tt ADC be a i0 on AC.
AC=^ = AB2fBC2,
AB is < AC.
a chord AD can be found equal to AB.
// Then the AADC is rightangled at D.
IS (106", Cor. 4)
AC = AD2fDC2, (169°, Cor. i)
and AC2=AB2fBC2, and AD = AB. (hyp.)
DC = BC,
and AADC = AABC.
^D=^B = ~1. ^.^.^•
171°. Theorem 169° with its corollaries and theorem 170°
are extensively employed in the practical applications of
Geometry. If we take the three numbers 3, 4, and 5, we
AREAL RELATIONS.
123
have 52=32 + 42. Therefore if a triangle has its sides 3, 4,
and 5 feet, metres, miles, or any other ;/./., it is rightangled
opposite the side 5.
For the segments into which the altitude divides the
hypothenuse we have S^h^3'^ ^"^ 5'^i = 4^ whence ^?i =  and
=iJi
For the altitude itself, /2==u
whence p ■■
Problem. — To find sets of whole numbers which represent
the sides of rightangled triangles.
This problem is solved by any three numbers ,r, y^ and s^
which satisfy the condition ;ir2 =^2 4. ^2_
Let in and n denote any two numbers. Then, since
{7n^Vn^Y = {in'^n^f^{2inny, (166°, 4)
the problem will be satisfied by the numbers denoted by
)n^\n^^ m^ — fi^, and 2tfin.
The accompanying table, which may be extended at
pleasure, gives a number of sets of such numbers : —
2
3
4
5
6
7
8
9
10 ' ...
5 ' 10
15
26
37
50
65
82
lOI ...
I
3 6
8
10
12
14
16
18
20
,,
! 4 : 8
17
24
3S
48
63
80
99 1
,^3
20
29
40
53
68
85
104
2
i 12
16
20
24
28
32
36
40
..
5
12
21
32
45
60
77
96
..
25
34
45
S8
7?>
90
109
..
3 '
24
30
36
42
48
54
60
..
'
7
16
27
40
55
72
91
..
41
52
6^
80
97
116
..
4
9
48
S6
64
72
80
..
40
20
33
48
65
84
61
74
89
106
125 J
5
60
70
80
90
icoj
i 1
"
24
39
56
75 1
..
...
...
••
24 SYNTHETIC GEOMETRY.
172°. Let a, d, c be the sides of any triangle, and let b be
3 taken as base. Denote the projections
of a and r on /^ by a^ and q, and the
^a altitude to b by p. Then
(i) U^^ci^\a^^2.c^a^, (166°, i)
(2) ^2 = ^1^/2, (169°, Cor. I)
(3) ^^ = ^i2+/2.
1. By subtracting (2) from (3)
: . The difference between the squares upon two sides of a
triangle is equal to the difference of the squares on the projec
tions of these sides on the third side, taken in the same order.
Since all the terms are squares and cannot change sign
(163°), the theorem is true without any variation for all As.
2. By adding (i) and (2) and subtracting (3),
b'^\c'^a^ = ic^ + 2^1 ^1
= 2^^i, •.* b=c^\a^,
a'^ = b'' + c'^2bc^.
Now, since we have assumed that b=c^^]a^, where c^ and a^
are both positive, D falls between A and C, and the angle A
is acute.
.'. In any triangle the square on a side opposite an acute
angle is less than the sum of the sqicares upon the other two
sides by twice the rectangle ott one of these sides and the pro
jection of the other side upon it.
3. Let the angle A become obtuse. Then D, the foot of
the altitude to b, passes beyond A,
and c^ changes sign.
a .•. [=1(5^1 changes sign, (161°)
and a^ = b'^'Vc^^ibc^.
.'. The square on the side opposite
obttcse angle i7i an obtuse
angled triaftgle is greater than the sum of the squares on the
other two sides by twice the rectangle on one of these sides and
the projection of the other side upon it.
AREAL RELATIONS. 125
The results of 2 and 3 are fundamental in the theory of
triangles.
These results are but one ; for, assuming as we have done
that the nn^Ci is to be subtracted from d'^ + c^ when A is an
acute angle, the change in sign follows necessarily when A
becomes obtuse, since in that case the en changes sign because
one of its sides changes sign (161°); and in conformity to
algebraic forms  ( — 2bc^= + 2bc^.
Cor. If the sides a, b, c of a. triangle be given in numbers,
we have from 2 Ci~ ,
20
which gives the projection of c on k
If ^1 is + the lA is acute ;
if c^ is o the z_A is ~~\ ;
and if ^1 is  the lA is obtuse.
Ex. The sides of a triangle being 12, 13, and 4, to find the
character of the angle opposite side 13.
Let 13 = ^, and denote the other sides as you please, ^.^.,
^=12 and c=4. Then
. _ I2H4^I3 ^_ 3
1 :r, ~ Q'
24 6
and the angle opposite side 13 is obtuse.
173°. Theorem. — The sum of the squares on any two sides
of a triangle is equal to twice the sum of the squares on one
half the third side and on the median to b
that side.
BE is the median to AC. Then
AB2 + BC2=2(AE2 + EB2).
Proof.— l^^t D be the foot of the altitude ^ ^ ^
on AC. Consider the AABE obtuseangled at E, and
AB2 = AE2fEB2 + 2AE.ED. (172°, 3)
Next, consider the ACBE acuteangled at E, and
BC2 = EC2HEB22EC.ED. (172°, 2)
126 SYNTHETIC GEOMETRY.
Now, adding and remembering that AE = EC,
AB2 + BC2 = 2AE2 + 2EB2. q.e.d.
Cor. I. Denoting the median by in and the side upon which
it falls by b^ we have for the length of the median
4
Cor. 2. All the sides of an equilateral triangle are equal
and the median is the altitude to the base and the right
bisector of the base. (53", Cors. 2, 3)
.'. in an equilateral triangle,
ni^=p^ = \aP^ ox p = \asj 2i^ a being the side.
173°. Theorem. — The sum of the squares on the sides of a
quadrangle is equal to the sum of the
squares on the diagonals, and four times
the square on the join of the middle points
of the diagonals.
E, F are middle points of AC and BD.
Then 2(AB2) = AC2 + BD2+4EF2.
D Proof.— ]6m AF and CF.
Then AF is a median to AABD, and CF to ACRD.
AB2 + AD2=2BF2f2AF2, (172^^
and BC2 + CD2 = 2BF2 + 2CF2,
.. adding, 2(AB2) = 4BF2 + 2(AF2+CF2).
But EF is a median to AAFC.
AF2 + CF2 = 2CE2 + 2EF2, (172^)
2;( AB2) = 4BF2 + 4CE2 + 4EF2
= BD2 + AC2 + 4EF2. q.e.d.
Since squares only are involved this relation is true with
out any modification for all quadrangles.
Cor. I. When the quadrangle becomes a 1 7 the diagonals
bisect one another (81°, 3) and EF becomes zero.
.•. the sum of the squares on the sides of a parallelogram
is equal to the sum of the squares on its diagonals.
AREAL RELATIONS. I 2/
174°. Let ABC be an isosceles triangle and P be any point
in the base AC, and let D be the middle point of ^
the base, and therefore the foot of the altitude.
In the ABAP, acuteangled at A,
BP2=BA2 + AP22AP.AD, (172°, 2) /
BA2BP2=AP(2ADAP) /
= AP.PC. Q A P D C
If P moves to Q, AP becomes AQ and changes sign, BP
becomes BQ which is > BA, and thus both sides of the
equality change sign together as they pass through zero by
P passing A.
Now, of the two segments from B we always know which
is the greater by 63°, and if we write PA for AP the izdPA . PC
is positive when P is on the Q side of A. Hence, considering
the rectangle as being always positive, we may state the
theorem —
The differoice between the squares on a side of an isosceles
triangle and on the join of the vertex to any point in the base
is equal to the rectangle on the segments into which that poijit
divides the base.
175°. I. From 174° we have BA2BP2 = AP . PC. Now
BA is fixed, therefore the [=]AP . PC increases as BP de
creases. But BP is least when P is at D (63°, i), therefore
the cdAP . PC is greatest when P is at D.
Def I. — A variable magnitude, which by continuous
change may increase until a greatest value is reached and
then decrease, is said to be capable of a maximum, and the
greatest value reached is its maximum.
Thus as P moves from A to C the imAP . PC increases
from zero, when P is at A, to its maximum value, when P is
at D, and then decreases again to zero, when P comes to C.
And as AC may be considered to be any segment divided
at P,
.'. The maximum rectangle on the parts of a given segment
is formed by bisecting the segment;
128 SYNTHETIC GEOMETRY.
Or, of all rectangles with a given perimeter the square has
the greatest area.
2. AC2=(AP + PC)2 = AP2 + PC2 + 2AP.PC (166°, i)
= AP2 + PC2 + 2(AB2BP2). (174°)
But AC and AB are constant,
AP2 + PC2 decreases as BP2 decreases.
But BP is least when P is at D,
AP2 + PC2 is least when P is at D.
Def. 2. — A variable magnitude which by continuous change
decreases until it reaches a least value and then increases is
said to be capable of a miniinnm, and the least value attained
is called its miniimiin.
.'. The sum of the squares on the ttvo parts of a given seg
ment is a minimum when the segme7it is bisected.
I75j°. The following examples give theorems of importance.
B Ex. I. Let ABC be any triangle and BD the
altitude to side b. Then
y p\ \a . b' + C'^a
^1= zi ' (172°, Cor.)
A fa D c But P^^c^c^^ic^c^icc^,
and /\ = \bp. (153°, 2)
Now, .+., = (i±f)^^=(A±i±^(^±fZ).^
2b lb
and cc _ a''{bc f_ {a + bc){ab + c)
'^ 2b 2b '
)'^=:i6/\^ = {a + b + c){b^ca){c+ab){a + bc\
and by writing s for \{a + b + c), dindi accordingly sa for
\{b + c—a), etc., we obtain
A = slXsa){sb){s~c):
This important relation gives the area of the A in terms of
its three sides.
Ex. 2. Let ABC be an equilateral A Then the area may
be found from Ex. i by making a = b = c, when the reduced
2
expression becomes, A = '^ \^3
4
AREAL RELATIONS. 1 29
Ex. 3. To find the area of a regular octagon in terms of its
circumradius. A d
Let A, B, C be three vertices of the octagon
and O the centre. Complete the square OD,
and draw BE ± to OA.
Since lEOB^^~], and OB = r,
E0 = EBrV2,
and A0AB=^0A.EB=r.rV2=irV2.
But AOAB is oneeighth of the octagon,
Oct. = 2rV2.
Exercises.
1. ABC is rightangled at B, and E and F are middle points
of BA and BC respectively. Then 5AC2 = 4(CE2^ AF2).
2. ABC is rightangled at B and O is the middle of AC,
and D is the foot of the altitude from B. Then
2AC.OD = AB2BC2.
3. ABC is rightangled at B and, on AC, AD is taken equal
to AB, and on CA, CE is taken equal to CB. Then
ED2=2AE.DC.
4. The square on the sum of the sides of a rightangled tri
angle exceeds the square on the hypothenuse by twice
the area of the triangle.
5. To find the side of a square which is equal to the sum of
two given squares.
6. To find the side of a square which is equal to the differ
ence of two given squares.
7. The equilateral triangle described upon the hypothenuse
of a rightangled triangle is equal to the sum of the
equilateral triangles described on the sides.
8. ABC is a triangle having AB = CB, and AD is J_ upon
BC. Then AC2=2CB.CD.
9. Four times the sum of the squares on the three medians
of a triangle is equal to three times the sum of the
squares on the sides.
I
I30 SYNTHETIC GEOMETRY.
10. ABCD is a rectangle and P is any point. Then
PA2 + PC2=PB2 + PD2.
11. O is the centre of a circle, and AOB is a centreline.
OA=OB and C is any point on the circle. Then
AC2 + BC2 = a constant.
Define a circle as the locus of the point C.
12. AD is a perpendicular upon the line OB, and BE is a
perpendicular upon the line OA. Then OA . OE
= OB.OD.
13. Two equal circles pass each through the centre of the
other. If A, B be the centres and E, F be the points
of intersection, EF2 = 3AB2.
If EA produced meets one circle in P and AB pro
duced meets the other in Q, PQ2 = 7AB2.
14. ABC is a triangle having the angle A twothirds of a
right angle. Then AB2fAC2 = BC2 + AC. AB.
15. In the triangle ABC, D is the foot of the altitude to AC
and E is the middle point of the same side. Then
2ED.AC = AB2BC2.
16. AD is a line to the base of the triangle ABC, and O is
the middle point of AD. If AB2 + BD2=AC2 + CD2,
then OB = OC.
17. ABC is rightangled at B and BD is the altitude to AC.
Then AB. CD = BD . BC and AD . CB = BA. BD.
18. ABC is a triangle and OX, OY, OZ perpendiculars from
any point O on BC, CA, and AB respectively. Then
BX2 f C Y2 f AZ2 = CX2 + A Y2 1 BZl
A similar relation holds for any polygon.
19. AA^, BBi are the diagonals of a rectangle and P any point.
Then PA2+ PB2 + PAi2 + PBi2=AAj}4P02, where O
is the intersection of the diagonals.
20. ABC is a triangle, AD, BE, CF its medians, and P any
point. Then
PA2hPB2 + PC2 = PD2fPE2+PF2fi(AD + BE2fCF2),
or 2PA2 = 2PD2i 12;a22^ where in is a median.
21. If O be the centroid in 20,
AREAL RELATIONS. I31
PA2+PB2 + PC2 = 3P02 + MAD2 + BE2 + CF2),
or SPA2 = 3P02 + fS;/z2.
22. ABCD is a square and AA', BE', CC, DD' perpendicu
lars upon any line L. Then
(AA'2 + CC'2)~2BB'. DD' = area of the square.
23. The sum of the squares on the diagonals of any quad
rangle is equal to twice the sum of the squares on the
joins of the middle points of opposite sides.
24. ABCD is a trapezoid having AD parallel to BC. Then
AB2 + CD2 + 2AD.BC = AC2 + BD2.
25. If A, B, C be equidistant points in line, and D a fourth
point in same line, the difference between the squares
on AB and DB is equal to the rectangle on AD and
CD.
26. If A, B, C, D be any four points in line,
AD2 + BC2=:AC2fBD2 + 2AB.CD.
27. Any rectangle is equal to onehalf the rectangle on the
diagonals of the squares described on adjacent sides.
28. In the triangle ABC, D is any point in BC, E is the
middle point of AC and F of BC. Then
AB24AC2 = AD2 + 4EF2f2BD.DC.
29. The sides of a rectangle are a and b. If/ be the length
of the perpendicular from a vertex upon a diagonal and
q be the distance between the feet of the two parallel
perpendiculars so drawn,
ps!'cfiV^= ab and q^'c^T¥= b^ a^ {b > a),
what linesegment is denoted by s/ii^ + b'^ ?
30. ABCD is a square. P is a point in AB produced, and Q
is a point in AD. If the rectangle BP . QD is con
stant, the triangle PQC is constant.
31. If the lengths of the sides of a triangle be expressed by
x^+i, x^ I, and 2r, the triangle is rightangled.
32. If a and c be the sides of a rightangled triangle and p be
the altitude to the hypothenuse,
141=1.
^2 • ^2 ^2
132 SYNTHETIC GEOMETRY.
33. The triangle whose sides are 20, 15, and 12 has an
obtuse angle.
34. The area of an isosceles triangle is 8^15 and the side is
twice as long as the base. Find the length of the side
of the triangle.
35. What is the length of the side of an equilateral triangle
which is equal to the triangle whose sides are 13, 14,
and 15?
36. If AB is divided in C so that AC22BC2, then
AB2 + BC2=2AB.AC.
37. Applying the principle of continuity state the resulting
theorem when B comes to D in (i) the Fig. of 172°,
(2) the Fig. of 173°.
38. Applying the principle of continuity state the resulting
theorem when B comes to E in the Fig. of 173°.
39. The bisector of the right angle of a rightangled triangle
cuts the hypothenuse at a distance a from the middle
point, and the hypothenuse is lb. Find the lengths of
the sides of the triangle.
40. Construct an equilateral triangle having one vertex at a
given point and the remaining vertices upon two given
parallel lines.
41. A square of cardboard whose side is s stands upright
with one edge resting upon a table. If a lower corner
be raised vertically through a distance a^ through
what distance will the corner directly above it be
raised ?
42. What would be the expression for the area of a rectangle
if the area of the equilateral triangle having its side
the u.l. were taken as the u.a?.
43. The opposite walls of a house are 12 and 16 feet high and
20 feet apart. The roof is rightangled at the ridge
and has the same inclination on each side. Find the
lengths of the rafters.
44 Two circles intersect in P and Q. The longest chord
through P is perpendicular to PQ.
AREAL RELATIONS.
33
45. The largest triangle with a given perimeter is an equi
lateral triangle.
46. The largest triangle having its base and the sum of the
other two sides given is isosceles.
47. The largest polygon of given species and given perimeter
is regular.
48. The largest isosceles triangle with variable base has its
sides perpendicular to one another,
49. The largest rectangle inscribed in an acuteangled tri
angle and having one side lying on a side of the
triangle has its altitude onehalf that of the triangle.
50. L, M are two lines meeting in O, and P is any point.
APB is a variable hne cutting L in A and M in B.
The triangle AOB is least when P bisects AB.
EQUALITIES OF RECTANGLES ON SEGMENTS
RELATED TO THE CIRCLE.
176°. Theorem. — If two secants to the same circle inter
sect, the rectangle on the segments between the point of
intersection and the circle with respect to one of the secants
is equal to the corresponding rectangle with respect to the
other secant.
I. Let the point of intersection be
within the circle. Then
AP.PB = CP.PD.
Proof. — AOB is an isosceles triangle,
and P is a point on the base AB.
OA2OP2 = AP.PB. (174°)
Similarly, COD is an isosceles tri
angle, and P a point in the base CD,
OC2OP2=CP.PD.
But OC = OA,
AP.PB = CP.PD. q.e.d.
134
SYNTHETIC GEOMETRY.
Cor. I. (a) Let CD become a diameter and be _L to AB.
Then AP. PB becomes AP2, (96°, Cor. 5)
AP2 = CP.PD,
and denoting AP by c, CP by z/, and the
radius of the circle by r, this becomes
which is a relation between a chord of a 0,
the radius of the 0, and the distance CP,
commonly called the versed sine, of the arc AB.
{b) When the point of intersection P passes without the
we have still, by the principle of con
tinuity, AP . PB = CP . PD. But the as
being now both negative we make them
both positive by writing
PA.PB = PC.PD.
Cor. 2. When the secant PAB be
comes the tangent PT (109°), A and
B coincide at T, and PA.PB becomes
PT2, .. PT2 = PC.PD,
.e., if a tangent and a secant be drawn
from the same point to a circle^ the square on the tangent is
equal to the rectangle on the segments of the secant between
the point a7id the circle.
Cor. 3. Conversely, if T is on the circle and PT2=PC . PD,
PT is a tangent and T is the point of contact.
For, if the line PT is not a tangent it must cut the circle in
some second point T' (94°). Then
PT.PT' = PC.PD = PT2.
Therefore PT = PT', which is not true unless T and T' coin
cide. Hence PT is a tangent and T is the point of contact.
Cor. 4. Let one of the secants become a centreline as
PEF. Denote PT by /, PE by h, and the radius of the circle
byr. Then PT2 = PE.PF
becomes f^—hipr+h).
AREAL RELATIONS. 1 35
Exercises.
1. The shortest segment from a point to a circle is a portion
of the centreline through the point.
2. The longest segment from a point to a circle is a portion
of the centreline through the point.
3. If two chords of a circle are perpendicular to one another
the sum of the squares on the segments between the
point of intersection and the circle is equal to the
square on the diameter.
4. The span of a circular arch is 120 feet and it rises 15
feet in the middle. With what radius is it con
structed?
5. A conical glass is b inches deep and a inches across the
mouth. A sphere of radius r is dropped into it. How
far is the centre of the sphere from the bottom of the
glass ?
6. The earth's diameter being assumed at 7,960 miles, how
far over its surface can a person see from the top of a
mountain 3 miles high ?
7. How much does the surface of still water fall away from
the level in one mile ?
8. Two circles whose radii are 10 and 6 have their centres
12 feet apart. Find the length of their common chord,
and also that of their common tangent.
9. Two parallel chords of a circle are c and c^ and their
distance apart is d^ to find the radius of the
circle.
10. If V is the versed sine of an arc, k the chord of half the
arc, and r the radius, k^ = 2vr.
177°. Theorem. — If upon each of two intersecting lines a
pair of points be taken such that the rectangle en the seg
ments between the points of intersection and the assumed
points in one of the lines is equal to the corresponding rect
1.^6
SYNTHETIC GEOMETRY.
angle for the other Hne, the four assumed points are concyclic.
(Converse of 176°.)
L and M intersect in O, and
OA.OB = OC.OD.
Then A, B, C, and D are concydic.
Proof. — Since the ens are equal, if A and B lie upon the
same side of O, C and D must lie upon the same side of O ;
and if A and B He upon opposite sides of O, C and D must
lie upon opposite sides of O.
Let a pass through A, B, C, and let it cut M in a second
point E. Then OA.OB = OC.OE. (176°)
But OA.OB = OC.OD. (hyp.)
OD = OE,
and as D and E are upon the same side of O they must co
incide; .*. A, B, C, D are concyclic. q.e.d.
178°. Let two circles excluding each other without contact
have their centres at A and B, and let C be the point, on
their common centreline> which divides AB so that the
difference between the squares on the segments AC and CB
is equal to the difference between the squares on the con
terminous radii. Through C
draw the line PCD X to AB,
and from any point P on this
line draw tangents PT and PT'
to the circles.
Join AT and BT'.
Then, by construction,
AC2BC2 = AT2Br2.
But, since PC is an altitude in
BC2=AP2BP2, (172°, i)
AP2=AT2 + PT2,
BP2 = BT'2 + PT'2, (169°, Cor. i)
p'J'2_p'p'2
PT = PT'.
AREAL RELATIONS.
137
Therefore PCD is the locus of a point from which equal
tangents are drawn to the two circles.
Def. — This locus is called the radical axis of the circles,
and is a line of great importance in studying the relations of
two or more circles.
Cor. I. The radical axis of two circles bisects their com
mon tangents.
Cor. 2. When two circles intersect, their radical axis is
their common chord.
Cor. 3. When two circles touch externally, the common
tangent at the point of contact bisects the other common
tangents.
179°. The following examples give theorems of some im
portance.
Ex. I. P is any point without a circle and TT' is the chord
of contact (114°, Def.) for the point P.
TT' cuts the centreHne PO in O.
Then, PTO being a "]» (i 10°)
OQ.OP = OT2. (169°)
.*. the radius is a geometric mean be
tween the join of any poifit with the
centre and the perpendictdar from the
centre upon the chord of contact of the point.
Def. — P and Q are called inverse points with respect to
the circle.
Ex. 2. Let PQ be a common direct tangent to the circles
having O and O' as centres.
Let OP and O'Q be radii
to the points of contact, and
let QR be II to 00'. Denote
the radii by r and /. Then
AC = 00' + rr',
BD = 00'rlr'.
.. AC.BD = 00'2(rr')2 =
QR2PR2 = PQ2. (169°, Cor. i)
138 SYNTHETIC GEOMETRY.
Similarly it may be shown that
AD . BC = square on the transverse common tangent.
Exercises.
1. The greater of two chords in a circle is nearer the centre
than the other.
2. Of two chords unequally distant from the centre the one
nearer the centre is the greater.
3. AB is the diameter of a circle, and P, Q any two points on
the curve. AP and BQ intersect in C, and AQ and BP
in C. Then
AP.AC + BQ.BC = AC'.AQ + BC'.BP.
4. Two chords of a circle, AB and CD, intersect in O and
are perpendicular to one another. If R denotes the
radius of the circle and E its centre,
8R2 = AB2 + CD2 + 40E2.
5. Circles are described on the four sides of a quadrangle as
diameters. The common chord of any two adjacent
circles is parallel to the common chord of the other
two.
6. A circle S and a line L, without one another, are touched
by a variable circle Z. The chord of contact of Z passes
through that point of S which is farthest distant
from L.
7. ABC is an equilateral triangle and P is any point on its
circumcircle. Then PA + PB + PC=o, if we consider
the line crossing the triangle as being negative.
8. CD is a chord parallel to the diameter AB, and P is any
point in that diameter. Then
PC2 + PD2 = PA2 + PB2.
CONSTRUCTIVE GEOMETRY. 1 39
SECTION V.
CONSTRUCTIVE GEOMETRY.
1 80°. Problem.— PC& being a given segment, to construct
the segment ABv/2. A b
C^;zj/r.— Draw BC JL to AB and equal to it.
Then AC is the segment ABx/2.
Proof. — Since ABC is rightangled at B,
AC2=AB2 + BC2 = 2AB2, (169, Cor. i)
AC = ABV2.
Cor. The square on the diagonal of a given square is equal
to twice the given square.
181°. Problem.— To construct ABv'S
Constr. — Take BC in line with AB and equal
to it, and on AC construct an equilateral tri
angle ADC. (124°, Cor. I)
BD is the segment ABV3.
Proof.— KBJ) is a ~1, and AD = AC = 2AB.
Also AD2 = AB2hBD2 = 4AB2. (169°, Cor. i)
BD2 = 3AB2, and BD^AB^/S
Cor. Since BD is the altitude of an equilateral triangle
and AB is onehalf the side,
.'. the square on the altitude of an equilateral triangle is
equal to three times the square on the half side.
182°. Problem.— To construct AB^S
Constr. —Draw BC X to AB and equal to twice
AB. Then AC is the segment ABV5.
Pr^^— Since Z.B is a right angle,
AC2 = ABHBC2.
But BC2 = 4AB2;
AC2=5AB2,
and AC = ABV5.
140
SYNTHETIC GEOMETRY.
183°. The three foregoing problems furnish elements of
construction which are often convenient. A few examples
are given.
Ex. I. AB being a given segment, to find a point C in its
, line such that AC' = AB . CB.
A c B Analysis AC2 = AB. CB = AB(AB AC),
AC^ + AC.AB = AB2.
Considering this an algebraic form and solving as a quad
ratic in AC, we have AC = J(ABV5  AB),
and this is to be constructed.
Constr. — Construct AD=ABV5 (by 182°) as in the figure,
c ' A 9 ^ ^^^ ^^t E be the
middle point of BD.
TakeDF = DE.
Then
AF=ABV5AB;
.*. bisecting AF in G,
AG = AC
=MABV5AB),
and the point C is
found.
Again, since ^J^ has
two signs + or ,
Jy^' take its negative sign
and we have AC =  ^AB^S + AB).
Therefore, for the point C, on AD produced take
DF' = DE, and bisect AF' in G'. Then
AG'=(ABV5 + AB);
and since AC is negative we set off AG' from A to C, and
C is a second point.
The points C and C satisfy the conditions,
AC2 = AB . CB and AC2=AB . CB.
A construction effected in this way requires no proof other
than the equation which it represents.
\
CONSTRUCTIVE GEOMETRY. I4I
It is readily proved however. For
AD2=5AB, and also AD2=(AF + FD)2 = (2AC + AB)2,
whence AC2 = AB(AB  AC) = AB . CB.
It will be noticed that the constructions for finding the two
points differ only by some of the segments being taken in
different senses. Thus, for C, DE is taken from DA, and for
C\ added to DA ; and for C, AC is taken in a positive sense
equal to AG, and for C, AC is taken in a negative sense
equal to AG'.
In connection with the present example we remark : —
1. Where the analysis of a problem involves the solution of
a quadratic equation, the problem has two solutions corre
sponding to the roots of the equation.
2. Both of the solutions may be applicable to the wording
of the problem or only one may be.
3. The cause of the inapplicability of one of the solutions
is commonly due to the fact that a mathematical symbol is
more general in its significance than the words of a spoken
language.
4. Both solutions may usually be made applicable by some
change in the wording of the problem so as to generalize it.
The preceding problem may be stated as follows, but
whether both solutions apply to it, or only one, will depend
upon our definition of the word " part." See Art. 23°.
To divide a given segment so that the square tipon one of
the parts is equal to the rectangle on the whole segment and
the other part.
Def. — A segment thus divided is said to be divided into
extreme and mea7i ratio, or in median section.
Ex. 2. To describe a square when the sum of its side and
diagonal is given.
Analysis. — If AB is the side of a square, AB^/2 is its
diagonal, (iSo°)
142
SYNTHETIC GEOMETRY.
yH'
.'. AB(i +J2) is a given segment =S, say. Then
AB = S(V2i).
Constr. — Let EF be the given segment S.
Draw FG JL and = to EF, and with
centre G and radius GF describe a cut
ting EG in H and H'.
EH is the side of the square; whence the
square is easily constructed.
If we enquire what EH' means, we find it to be the side of
the square in which the difference between the side and
diagonal is the given segment S. The double solution here
is very suggestive, but we leave its discussion to the reader.
184°. Problem. — To find a segment such that the rectangle
on it and a given segment shall be equal to a given rectangle.
p G E
"■ Constr. — Let S be the given segment,
and AC the given rectangle.
On DA produced make AP = S, and
draw PBO to cut DC produced in Q.
CQ is the segment required.
Complete the as PEQD, PGBA, and BCQF.
i=iAC=[=:GF = GB.BF = PA.CQ,
S.CQ = [=iAC.
Def. — The segments AP and CQ are reciprocals of one
another with respect to the cuAC as unit.
185°. Problem. — To find the side of a square which is
equal to a given rectangle.
Constr. — Let AC be the rectangle.
Make BE = BC and in line with BA.
On AE describe a semicircle, and pro
duce CB to meet it in F.
^ ^ BF is the side of the required square.
Proof. — Since AE is a diameter and FB a half chord JL to it,
/"^
r\
^
B
CONSTRUCTIVE GEOMETRY.
143
BF2=AB.BE,
BF2=AB.BC.
(176°, Cor. i)
Cor. This is identical with the problem, "To find a geo
metric mean between two given segments," and it furnishes
the means of constructing the segment a, when a=^>Jbc, b and
^ being given. (165°)
Ex. I. To construct an equilateral triangle equal to a given
rectangle. b c
Let AC be the given rectangle, and suppose
PQR to be the required triangle. Then
AB.BC = iPR.QT
= PT.QT.
But QT = PTV3, (181°, Cor.)
PT.QT=PTV3
whence PT2=ABV3iBC. p t r
And PT is the side of a square equal to the rectangle whose
sides are ABAy3 and ^BC, and is found by means of 181°,
127°, and 185°.
Thence the triangle is readily constructed.
Ex. 2. To bisect the area of a triangle by a line parallel to
its base.
Let ABC be the triangle, and assume
PQ as the required line, and complete
the parallelograms AEBC, KFBC, and
let BD be the altitude to AC. Because
PQ is II to AC, BD is ± to PQ. Now
/=7EP = Z=Z7PC, (145°)
E=7FC = e:^EQ, or PQ.BD = AC.BG. (153°, i)
But 2Z=7FQ = £Z:7EC, or 2PQ.bg = AC. BD;
.*. dividing one equation by the other, and reducing to one
line, BD''i = 2BG2;
and therefore BG is onehalf the diagonal of the square
of which BD is the side, and the position of PQ is de
termined.
144
SYNTHETIC GEOMETRY.
1 86°. Problem. — To find the circle which shall pass through
two given points and touch a given line.
Let A, B be the given points and L
the given line.
Constr. — Let the line AB cut L in O.
Take OP = OP', a geometric mean be
tween OA and OB (185°). The circles
through the two sets of three points A, B, P and A, B, P' are
the two solutions.
The proof is left to the reader. (See 176°, Cor. 2.)
187°. Problem. — To find a to pass through two given
points and touch a given 0.
Let A, B be the points and S
the given 0.
Constr, — Through A and B
draw any so as to cut S in
two points C and D. Let the
line CD meet the hne AB in O.
From O draw tangents OP and
OQ to the 0S (114°). Pand
Q are the points of contact for
the 0s which pass through A and B and touch S. There
fore the 0s through the two sets of three points A, B, P and
A, B, Q are the 0s required.
Proof. OB.OA=OC.OD = OQ2=OP2;
therefore the 0s through A, B, P and A, B, Q have OP and
OQ as tangents (176°, Cor. 3). But these are also tangents
to 0S ; therefore P and Q are the points of contact of the
required 0s.
Exercises.
1. Describe a square that shall have twice the area of a
given square.
2. Describe an equilateral triangle equal to a given square.
CONSTRUCTIVE GEOMETRY. 145
3. Describe an equilateral triangle having five times the area
of a given equilateral triangle.
4. Construct ABV7, where AB is a given segment.
5. Construct sJa^ + b^ and sla^b'^, where a and b denote
given line segments.
6. Divide the segment AB in C so that AC2 = 2CB2. Show
that AC is the diagonal of the square on CB. Does
this hold for external division also ?
7. ABCD is a rectangle and DE, a part of DA, is equal
to DC. EF, perpendicular to AD, meets the circle
having A as centre and AD as radius in F. Then DF
is the diagonal of a square equal to the rectangle.
8. In the Fig. of 183°, CE^^sAB . CB,
CD2 = CE2 + 3ED2 = 3AB(AB + CB).
9. Show that the construction of 183° solves the problem,
"To divide a segment so that the rectangle on the parts
is equal to the difference of the squares on the parts."
10. Show that the construction of 183° solves the problem,
" To divide a given segment so that the rectangle on
the whole and one of the parts is equal to the rectangle
on the other part and the segment which is the sum of
the whole and the first part."
11. Construct an equilateral triangle when the sum of its side
and altitude are given. What does the double solution
mean? (See 183°, Ex. 2.)
12. Describe a square in a given acuteangled triangle, so
that one side of the square may coincide with a side of
the triangle.
13. Within a given square to inscribe a square having three
fourths the area of the first.
14. Within an equilateral triangle to inscribe a second equi
lateral triangle whose area shall be one half that of
the first.
1 5. Produce a segment AB to C so that the rectangle on the
sum and difference of AC and AB shall be equal to a
given square.
K
146 SYNTHETIC GEOMETRY.
16. Draw a tangent to a given circle so that the triangle
formed by it and two fixed tangents may be (i) a
maximum, (2) a minimum.
17. Draw a circle to touch two sides of a given square, and
pass through one vertex. Generalize this problem and
show that there are two solutions.
18. Given any two lines at right angles and a point, to find a
circle to touch the lines and pass through the point.
19. Describe a circle to pass through a given point and to
touch a given line at a given point in the line.
20. Draw the oblique lines required to change a given square
into an octagon.
If the side of a square is 24, the side of the result
ing octagon is approximately 10 ; how near is the
approximation ?
21. The area of a regular dodecagon is three times that of
the square on its circumradius.
22. By squeezing in opposite vertices of a square it is trans
formed into a rhombus of onehalf the area of the square.
What are the lengths of the diagonals of the rhombus?
23. P, Q, R, S are the middle points of the sides AB, BC,
CD, and DA of a square. Compare the area of the
square with that of the square formed by the joins AQ,
BR, CS, and DP.
24. ABCDEFGH is a regular octagon, and AD and GE are
produced to meet in K. Compare the area of the tri
angle DKE with that of the octagon.
25. The rectangle on the chord of an arc and the chord of
its supplement is equal to the rectangle on the radius
and the chord of twice the supplement.
26. At one vertex of a triangle a tangent is drawn to its cir
cumcircle. Then the square on the altitude from that
vertex is equal to the rectangle on the perpendiculars
from the other vertices to the tangent.
27. SOT is a centreline and AT a tangent to a circle at the
point A. Determine the angle AOT so that AS = AT.
PART III.
PRELIMINARY.
1 88°. By superposition we ascertain the equality or in
equality of two given linesegments. But in order to express
the relation between the lengths of two unequal segments we
endeavour to find two numerical quantities which hold to one
another the same relations in magnitude that the given seg
ments do.
Let AB and CD be two given segments. If they are com
mensurable (150°, 5) some M.l. can be found with respect to
which the measures of AB and CD (150°, 2) are both whole
numbers. Let in denote the measure of AB and n the
measure of CD with respect to this unitlength.
The numbers in and n hold to one another the same rela
tions as to magnitude that the segments AB and CD do.
The fraction — is called in Arithmetic or Algebra the ratio
n
ofm to 11, and in Geometry it is called the ratio of AB to CD.
Now n has to in the same ratio as unity has to the fraction
— . But if CD be taken as i/J. its measure becomes unity,
n ^
in
while that of AB becomes — •
Therefore the ratio of AB to CD is the measure of AB with
respect to CD as unitlength.
When AB and CD are commensurable this ratio is expres
sible arithmetically either as a whole number or as a fraction ;
147
148 SYNTHETIC GEOMETRY.
but when the segments are incommensurable the ratio can
only be symbolized, and cannot be expressed arithmetically
except approximately.
189°. If we suppose CD to be capable of being stretched
until it becomes equal in length to AB, the numerical factor
which expresses or denotes the amount of stretching neces
sary may conveniently be called the tensor of AB with
respect to CD. (Hamilton.)
As far as two segments are concerned, the tensor, as a
numerical quantity, is identical with the ratio of the segments,
but it introduces a different idea. Hence in the case of com
mensurable segments the tensor is arithmetically expressible,
but in the case of incommensurable ones the tensor may be
symbolically denoted, but cannot be numerically expressed
except approximately.
Thus if AB is the diagonal of a square of which CD is the
side, AB = CDV2 (180°); and the tensor of AB on CD,
Le.^ the measure of AB with CD as unitlength, is that
numerical quantity which is symbolized by sji^ and which
can be expressed to any required degree of approximation by
that arithmetical process known as "extracting the square
root of 2."
190°. That the tensor symbolized by ^2 cannot be ex
pressed arithmetically is readily shown as follows : —
If /^2 can be expressed numerically it can be expressed as
a fraction, — , which is in its lowest terms, and where accord
n
ingly m and n are not both even.
If possible then let ^2 = .
n
Then 211^ = m'^. Therefore m'^ and in are both even and «
is odd. „
m
But if m is even, — is even, and n^ and ;/ are both even.
But n cannot be both odd and even.
Therefore sji cannot be arithmetically expressed.
PRELIMINARY. I49
Illustration of an ificommensurable tensor.
Let BD be equal to AB, and let AC be equal to the
diagonal of a square of which AB is the side.
^ E'F' ^
A B E 'c' 1^ D
Then some tensor will bring AB to AC.
Let BD be divided into 10 equal parts whereof E and F
are those numbered 4 and 5.
Then the tensor i . 4 stretches AB to AE, and tensor i . 5
stretches AB to AF. But the first of these is too small and
the second too great, and C lies between E and F.
Now, let EF be divided into 10 equal parts whereof E', F'
are those numbered i and 2.
Then, tensor i . 41 brings AB to AE', and tensor i . 42
brings AB to AF' ; the first being too small and the second
too great.
Similarly by dividing E'F' into 10 equal parts we obtain
two points ^, yj numbered 4 and 5, which lie upon opposite
sides of C and adjacent to it.
Thus, however far this process be carried, C will always lie
between two adjacent ones of the points last obtained.
But as every new division gives interspaces onetenth of
the length of the former ones, we may obtain a point of
division lying as near C as we please.
Now if AB be increased in length from AB to AD it must
at some period of its increase be equal to AC.
Therefore the tensor which brings AB to AC is a real
tensor which is inexpressible, except approximately, by the
symbols of Arithmetic.
The preceding illustrates the difference between magnitude
and number. The segment AB in changing to AD passes
through every intermediate length. But the commensurable
or numerically expressible quantities lying between i and 2
must proceed by some unit however small, and are therefore
not continuous.
ISO SYNTHETIC GEOMETRY.
Hence a mag7titude is a variable which^ in passing from
one value to another^ passes through every intermediate value.
191°. The tensor of the segment AB with respect to AC, or
the tensor of AB on AC is the numerical factor which brings
AC to AB.
But according to the operative principles of Algebra,
^^.AC=AB,
AB
.*. ^ is the tensor which brings AC to AB.
Hence the algebraic form of a fraction^ when the parts
denote segments^ is interpreted geometrically by the tensor
which brings the denominator to the numerator ; or as the
ratio of the numerator to the denominator.
SECTION I.
PROPORTION AMONGST LINESEGMENTS.
192°. Def — Four linesegments takett in order form a
proportion^ or are in proportion^ when the tensor of the first
on the second is the same as the tensor of the third on the
fourth.
This definition gives the relation
r, (^>
where a, b^ c, and d denote the segments taken in order.
The fractions expressing the proportion are subject to all
the transformations of algebraic fractions (158°), and the re
sult is geometrically true whenever it admits of a geometric
interpretation.
The statement of the proportion is also written
(B)
PROPORTION AMONGST LINESEGMENTS. 151
where the sign : indicates the division of the quantity de
noted by the preceding symbol by the quantity denoted by
the following symbol.
In either form the proportion is read
" rt: is to ^ as ^ is to dP
193°. In the form (b) a and d are called the extremes, and
b and c the means ; and in both forms a and c are called
antecedents and b and d consequents.
In the form (a) a and d^ as also b and c^ stand opposite
each other when written in a cross, as
a\c
b\d'
and we shall accordingly call them the opposites of the
proportion.
194°. I. From form (a) we obtain by crossmultiplication
ad— be,
which states geometrically that
When four segments are in proportion the rectangle
upon one pair of opposites is equal to that upon the other
pair of opposites.
Conversely, let rt(5 and^i'^^'be equal rectangles having for
adjacent sides a, b, and a', b' respectively. Then
ab = a'b',
and this equality can be expressed under any one of the fol
lowing forms, or may be derived from any one of them, viz. ;
a _b' a _a' b _b' b _a'
a' b b' b a' a b' a
in all of which the opposites remain the same. Therefore
2. Two equal rectangles have their sides in proportion,
a pair of opposites of the proportion coming from the same
rectangle.
3. A given proportion amongst four segments may be
written in any order of sequence, provided the opposites
remain the same.
^S2 SYNTHETIC GEOMETRY.
195°. The following transformations are important.
Let ^ = ^, then
a±b c±d / ,r ■ \
1. ~~b~^~T^ {a>btoY  sign)
a c a\c a — c 1 ^ r • \
2. /, = :>==Tr:>=77 {ci>ciox  sign)
b d b + a bd
^ a c
Let ,=^=
b d
3
c
= etc., then
a c
b d
f b + d+f+Qic'
To prove i.
*•■
a c
I'd'
...±.=f±:,and4^ =
c±d
d '
To prove 2.
or
a c
b~d'
, a b
' ' "c'd'
a±c b±d
c d '
a c a±c
b d b±d'
(194°, 3)
To prove 3.
•.•
n __e _
a + c P
irr^—r.^ say,
b 7 7+Q~^+^+/
SIMILAR TRIANGLES.
196°. Def. — I. Two triangles are similar when the angles of
the one are respectively equal to the angles of the other.
{n\ 4)
2. The sides opposite equal angles in the two triangles are
corresponding or homologous sides.
The symbol ^ will be employed to denote similarity, and
will be read "is similar to."
PROPORTION AMONGST LINESEGMENTS. 1 53
In the triangles ABC and A'B'C, if zA=lA' and z.B = ^B',
then also lC=lC and the tri
angles are similar.
The sides AB and A'B' are / / \ //>
homologous, so also are the other / / \ //
pairs of sides opposite equal ^ ' ^ '^—^
angles. ad c
Let BD through B and B'D' through B' make the
^BDA = ^B'D'A'.
Then AABD « AA'B'D' since their angles are respectively
equal. In like manner ADBC « AD'B 'C, and BD and B'D'
divide the triangles similarly.
3. Lines which divide similar triangles similarly are
homologous lines of the triangles, and the intersections of
homologous /i?tes are homologous points.
Cor. Evidently the perpendiculars upon homologous sides
of similar triangles are homologous lines. So also are the
medians to homologous sides ; so also the bisectors of equal
angles in similar triangles ; etc.
197°. Theorem. — The homologous sides of similar triangles
are proportional.
AABC^AA'B'C
having lA. = lA.'
and lR=lB'.
rj,.^ AB BC CA
^^'^ A^=FC = C^ ,
B C
Proof.— V\3,ce A' on A, and let C fall at D. Then, since
lA'=lA, A'B' will lie along AC and B' will fall at some point
E. Now, AA'B'C'^AAED, and therefore ^AED=^B,
and B, D, E, C are concyclic. (107°)
Hence AD.AB = AE.AC, (176°, 2)
or A'C'.AB = A'B'.AC. ,
AB AC /ox
A^B'=A^ (^94,2)
154 SYNTHETIC GEOMETRY.
Similarly, by placing B' at B, we prove that
AB _ BC
A'B' B'C*
AB^BC^CA ,
A'B' B'C C'A'' ^' ' '
Cor. I. Denoting the sides of ABC by a, b, c, and those of
A'B'C by a', b', c',
a _b _c
a~b'~c''
^ a b c a + b + c /,^r° ^\
Cor. 2. v = A' = 7' = vn:i'3:7>' ('95 , 3)
a c a +b +c
i.e., the perimeters of similar triangles are proportional to
any pair of homologous sides.
198°. Theorem. — Two triangles which have their sides pro
b' portional are similar, and have
their equal angles opposite hom
ologous sides. (Converse of 197°.)
AB _ BC _ CA
A'B' B'C C'A''
Then .iA = zj\', ^B = z.B', and
^C=z.C'.
Proof.— On A'C let the AA'DC be constructed so as to
have the ^DA'C'=zA
and ^DC'A'=ZC.
Then AA'DC ^AABC, (196°, Def. i)
^^A AB AC BC . ,.s
^^^ A^==A^' = DC' ^^97)
u . AB AC BC /, X
^"' A^ = A^' = B'C' ^^^P^
A'D = A'B'
and DC' = B'C,
and AA'DC = AA'B'C'. (58°)
ZJ\' = ^A, ^B'=^B,
and ^C'=^C. q.e.d.
PROPORTION AMONGST LINESEGMENTS. 1 55
199°. Theorem. — If two triangles have two sides in each
proportional and the included a
angles equal, the triangles are
similar. D^
then AABC^AA'B'C.
c
Proof. — Place A' on A, and let A'C lie along AB, and
A'B' lie along AC, so that C falls at D and B' at E.
The triangles AED and A'B'C are congruent and therefore
. ., , AB AC
similar, and ^=^
Hence AB . AD = AE . AC ; ( 1 94°)
and .'. B, D, E, C are concyclic. {mT)
Z_AED = aB, and £ADE = zlC, (106°, Cor. 3)
and AABC ^ AAED « AA'B'C. q.e.d.
200°. Theorem. — If two triangles have two sides in each
proportional, and an angle opposite a homologous side in
each equal :
1. If the angle is opposite the longer of the two sides the
triangles are similar.
2. If the angle is opposite the shorter of the two sides the
triangles may or may not be
similar.
I. If BOAB,
AABC^AA'B'C.
Pr^^/— Place A' at A and let B' fall at D, and A'C along
AC. Draw DE  to BC. Then
AABC« AADE, and ^^=§§
156 SYNTHETIC GEOMETRY.
And since B'C'> A'B', the AA'B'C' = AADE and they are
therefore similar. (65°, i)
But AABC^AADE,
AABC^AA'B'C.
2. If BC<AB, B'C'<A'B', and the triangles may or may
not be similar.
Proof.— Since AD = A'B', and DE = B'C', and B'C'<A'B',
.*. the triangles A'B'C and ADE may or may not be
congruent (65°, 2), and therefore may or may not be
similar.
But AABC^AADE,
.*. the triangles ABC and A'B'C may or may not be
similar.
Cor. Evidently, if in addition to the conditions of the
theorem, the angles C and C are both less, equal to, or
greater than a right angle the triangles are similar.
Also, if the triangles are rightangled they are similar.
201°. The conditions of similarity of triangles may be
classified as follows : —
1. Three angles respectively equal. (Def. of similarity.)
2. Three sides proportional.
3. Two sides proportional and the included angles equal.
4. Two sides proportional and the angles opposite the
longer of the homologous sides in each equal.
If in 4 the equal angles are opposite the shorter sides in
each the triangles are not necessarily similar unless some
other condition is satisfied.
By comparing this article with 66° we notice that there is a
manifest relation between the conditions of congruence and
those of similarity.
Thus, if in 2, 3, and 4 of this article the words "propor
tional " and " homologous " be changed to " equal," the
statements become equivalent to 1,2, and 5 of Art. 66°. The
PROPORTION AMONGST LINESEGMENTS. 1 5/
difference between congruence and similarity is the non
necessity of equality of areas in the latter case.
When two triangles, or other figures, are similar, they are
copies of one another, and the smaller may be brought, by a
uniform stretching of all its parts, into congruence with the
larger. Thus the primary idea of similarity is that every
linesegment of the smaller of two similar figures is stretched
to the same relative extent to form the corresponding seg
ments of the larger figure. This means that the tensors of
every pair of corresponding line segments, one from each
figure, are equal, and hence that any two or more line
segments from one figure are proportional to the correspond
ing segments' from the second figure.
Def. — Two linesegments are divided similarly when, being
divided into the same number of parts, any two parts from
one of the segments and the corresponding parts from the
other taken in the same order are in proportion.
202°. Theorejn. — A line parallel to the base of a triangle
divides the sides similarly ; and
Conversely, a line which divides two sides of a triangle
similarly is parallel to the third side. B
DE is II to AC. Then BA and BC are
divided similarly in D and E.
Proof.— Tht triangles ABC and DBE are /^
evidently similar, ^ ^
AB CB ^. . AD CE . ^_o ^
DB = EB'^"^ DB^EB' ^^^^ ' ^^
and AB and CB are divided similarly in D and E. g.e.d.
Conversely, if DE so divides BA and BC that
AD : DB = CE : EB, DE is  to AC.
P.../Since AD^CE ._ AB^CB ^ ^„, ^,^ .^j.^g,,,
ABC and DBE having the angle B common, and the sides
158
SYNTHETIC GEOMETRY.
about that angle proportional, are similar. (i99°)
aBDE=^ and DE is to AC. q.e.d.
Cor. I. Since the triangles ABC and DBE are similar
BA:BD=AC:DE.
203°. Theorem. — Two transversals to a system of parallels
are divided similarly by the parallels.
AA' is II to BB' is  to CC, etc.
Then AD and A'D' are divided similarly.
Proof. — Consider three of the s, AA',
BB', and CC, and draw A'Q  to AD.
Then AP and BQ are 1 — 7 s, and
AB = A'P and BC = PQ. (81°, i)
But A'QC is a triangle and PB' is  to QC.
A'P A'B' . ^,0. ^,. AB BC . o V
^=— (202), 01 ^,g,=^^, (194,3).
Similarly, if DD' be a fourth parallel,
AB
A'B'
BC
B'C
CD
CD'
BC^
B'C
CD
CD"
= etc.
Def, — A set of three or more lines meeting in a point is a
pencil and the lines are rays.
The point is the vertex or centre of the pencil.
Cor. I. Let the transversals meet in O, and let L denote
any other transversal through O.
Then AD, A'D', and L are all divided similarly by
the parallels. But the parallels are transversals to the
pencil.
.*. parallel transversals divide the rays of a pencil
similarly.
Cor. 2. Applying Cor. i of 202",
OA OB OC OD
AA' BB' CC DD"
=etc.
PROPORTION AMONGST LINESEGMENTS. 1 59
204°. Theorem. — The rectangle on any two sides of a tri
angle is equal to twice the rectangle on B
the circumradius (97°, Def.) and the alti
tude to the third side. ^f^ ^^ — ^°
BD is _L to AC and BE is a diameter.
Then BA.BC = BE.BD.
Proof.— lA = lE,
and^ADB=z.ECB=~l,
(106°, Cor. I)
(106°, Cor. 4)
AABD^AEBC, and ^=§,
BA.BC = BE.BD. g.e.d.
Cor. Denoting BD by p and the circumradius by R,
ac= 2/R,
and multiplying by 3, and remembering that pd=2^ (i 53°, 2),
4A'
which (with i7si°, Ex. i) gives the means of calculating the
circumradius of a triangle when its three sides are given .
we obtain
205°. Theorem. — In a concyclic quadrangle the rectangle
on the diagonals is equal to the sum of the rectangles on the
sides taken in opposite pairs.
AC . BD = AB . CD + BC . AD.
Proof.— DvdiVf AE making aAED
= zABC. Then, since ^BCA=^BDA,
the triangles EDA and BCA are similar.
BC.AD=AC.DE.
Again, since z_AEB is supp. to Zj\ED,
and Z.CDA is supp. to ^lABC, therefore triangles BE A and
CD A are similar, and AB . CD = AC . EB.
Adding these results, AB . CD + BC . AD = AC . BD.
This theorem is known as Ptolemy's Theorem.
206°. Def. — Two rectilinear figures are similar when they
l60 SYNTHETIC GEOMETRY.
can be divided into the same number of triangles similar in
pairs and similarly placed.
Thus the pentagons X
and Y can be divided into
the same number of tri
angles.
If then AP « AP',
AQ ^ AQ', AR « AR',
and the triangles are similarly placed, the pentagons are
similar.
The triangles are similarly placed if Z.EAD corresponds to
E'A'D', ^AED to A'E'D', .lDAC to D'A'C, etc.
This requires that the angles A, B, C, etc., of one figure
shall be respectively equal to the angles A', B', C, etc., of the
other figure.
Hence when two rectilinear figures are similar, their angles
taken in the same order are respectively equal, and the sides
about equal angles taken in the same order are pro
portional.
Linesegments, such as AD and A'D', which hold similar
relations to the two figures are similar or homologous lines of
the figures.
207°. Theorem. — Two similar rectilinear figures have any
two linesegments from the one proportional to the homolo
gous segments from the other.
Proof. — By definition AP^^AP', and they are similarly
placed, .. AE : A'E' = AC : A'C.
For like reasons, AD : A'D' = AC : A'C' = AB : A'B'.
AE _ AD _ AC _ AB _
A'E' A'D' A'C A'B' ''
and the same can be shown for any other sets of homologous
linesegments.
Cor. I. All regular polygons of the same species are similar
figures.
PROPORTION AMONGST LINESEGMENTS. l6l
Now, let a^ b, c, ..., a', b\ (f, ..., be homologous sides of two
similar regular polygons, and let r and r' be their circumradii.
Then r and r' are homologous,
r _a _b _c _ _ a^b^c^... / o .
V~a'~'b~'<f~""'V+Y+l^\:: ^'^^'^^
_ perimeter of P
perimeter of P'
But at the limit (148°) the polygon becomes its circumcircle.
.•. the circumferences of any two circles are proportional to
their radii.
Cor. 2. If ^, c' denote the circumferences of two circles and
r and r' their radii,  = ^ = constant.
r r
Denote this constant by 27r, then
C=7.Trr.
It is shown by processes beyond the scope of this work
that TT stands for an incommensurable numerical quantity,
the approximate value of which is 3. 141 5926...
Cor. 3. Since equal arcs subtend equal angles at the centre
(102°, Cor. 2), if s denotes the length of any arc of a circle
whose radius is r, the tensor  varies directly as s varies,
and also varies directly as the angle at the centre varies.
Hence  is taken as the measure of the angle, subtended
by the arc, at the centre. Denote this angle by Q. Then
and when s = r^B becomes the unit angle.
.*. the unit angle is the angle subtended at the centre by an
arc equal in length to the radius.
This unit is called a radian, and the measure of an angle
in radians is called its radian measure. Radian measure will
be indicated by the mark '^.
Cor. 4. When j=^ = a semicircle, Q — ir.
1 62 SYNTHETIC GEOMETRY.
But a semicircle subtends a straight angle at the centre.
.*. tr is the radian measure of a straight angle and  of a J.
Now a straight angle contains i8o°, (4i°)
7r'^=i8o°.
Hence i'^=57°.29578...,
and i° = o'\oi7453... ;
and these multipliers serve to change the expression of a
given angle from radians to degrees or from degrees to
radians.
Cor. 5. Since the area of a circle is equal to onehalf that
of the rectangle on its radius and a segment equal in length
to its circumference, (i49°)
Q) = \cr=\.2Trr.r (Cor. 2)
= 7rr2.
.'. the area of a is tt times that of the square on its radius.
208°. Theorem. — The bisectors of the vertical angle of a
triangle each divides the base into parts which are propor
g tional to the conterminous
sides.
BD and BD' are bi
sectors of Z.B. Then
AD_AD^_AB
DC CD' BC*
/*r^^— Through C draw
EE'lltoAB. Then
EBE'='~1(45°), and^E = ^ABD = ^DBC,
BC = EC = CE'. (88°, 3)
But ABD and ABD' are triangles having EE'  to the
common base AB.
AB_AD ^ AB AD' , o ^ v
ECDC ^"^ CE' = CD" ^^°^ ' ^^'^
^_ AD_AD' AB ,
DC~Ciy = BC ^^'^'^•
Cor. D and D' divide the base internally and externally in
PROPORTION AMONGST LINESEGMENTS. 163
the same manner. Such division of a segment is called har
monic division.
.*. the bisectors of any angle of a triangle divide the oppo
site side harmonically.
209°. Theorem.— K line through the vertex of a triangle
dividing the base into parts which are proportional to
the conterminous sides is a bisector of the vertical angle.
(Converse of 208°.)
Let the line through B cut AC internally in F. Then, AD
being the internal bisector ^=^^ (208°), and  = ^^ by
, ,, . . AF AD
hypothesis,.. FC^DC
But AD is < AF while DC is > FC.
.'. the relation is impossible unless F and D coincide, i.e.^
the line is the bisector AD.
Similarly it may be proved that if the line divides the base
externally it is the bisector AD'.
210°. Theorem. — The tangent at any point on a circle and
the pei*pendicular from that point upon t he di ameter divide
the diameter harmonically.
AB is divided harmonically in
M and T. a
Proof.LQ,VT=^LVWY=~^, (110°)
ACPM^ACTP,
r
and
CM^CP
CP CT'
CB+CM CT+CB
or
CM_CB
CB CT
AM AT
CBCM CtCB' MB BT'
,\ AB is divided harmonically in M and T.
(195°, 2)
q.e.d.
211". The following examples give important results.
164
SYNTHETIC GEOMETRY.
Ex. I. L, M, and N are tangents which touch the circle at
A, B, and P.
AX and BY are _Ls on
N, PC is X on AB, and
PO and PR are ±s upon
L and M.
Let N meet the chord
of contact of L and M in
T. Then the triangles
TAX, TPC, TBY are all
TA.TB _TP2
PC2'
(176°, Cor. 2)
(A)
similar,
But
AX. BY
Then
(114°, Cor. i)
1^
and
and
Similarly
Ex.
TA_TP_TB.
AX PC BY '
TA.TB = TP2,
AX.BY = PC2
Again, let L and N intersect in V.
VP=VA,
/.V0P = ^VXA = ~1,
^QVP = z.XVA.
AVXA = AVQP,
AX = PQ.
BY = PR, .. PO.PR = PC^ (B)
AD is a centreline and DO a perpendicular to it,
and AQ is any line from A to
the line DO.
Let AQ cut the circle in P.
Then AADQ^AAPB,
AD^AQ
AP AB'
or AD.AB = AP.AO.
But the circle and the point D
being given, AB . AD is a given
constant.
AP . AQ = a constant.
Conversely, if Q moves so that the cziAP.AQ remains
constant, the locus of Q is a line ± to the centreline
through A. ,
PROPORTION AMONGST LINESEGMENTS. 165
Now, let the dotted lines represent rigid rods of wood or
metal jointed together so as to admit of free rotation about
the points A, C, P, U, V, and Q, and such that UPVQ is a
rhombus (82°, Def. i), and AU=AV, and AC = CP, AC
being fixed.
PQ is the right bisector of UV, and A is equidistant from
U and V. Therefore A, P, Q are always in line.
Also, PUO is an isosceles triangle and UA is a line to the
base, therefore UA2UP2 = AP . AQ (174°).,. But, UA and
UP being constants, AP . AQ is constantl" ' '''
And AC being fixed, and CP being equal to AC, P moves
on the circle through A having C as centre.
.'. Q describes a line _L to AC.
This combination is known as Peaucelller's cellj and is
interesting as being the first successful attempt to describe a
line by circular motions only.
Ex. 3. To construct an isosceles triangle of which each
basal angle shall be double the vertical angle.
Let ABC be the triangle required, and let AD
bisect the Z.A.
Then ^B=z_BAD=^DAC, and ^C is common
to the triangles ABC and DAC. Therefore these
triangles are similar, and the ACAD is isosceles
and AD = AC.
Also, AABD is isosceles and AD = DB=AC.
BA:AC = AC:DC,
or BC:BD = BD:DC.
BC.DC = BD2.
And BC is divided into extreme and mean ratio at D (183°,
Ex. i). Thence the construction is readily obtained.
Cor. I. The isosceles triangle ADB has each of its basal
angles equal to onethtrd its vertical angle.
Cor. 2. JJ\BC = 36°, ^BAC = 72°, ZBDA= 108°. Hence
(i) Ten triangles congruent with ABC, placed side by side
with their vertices at B, form a regular decagon. (132°)
1 66
SYNTHETIC GEOMETRY.
(2) The bisectors AD and CE of the basal angles of the
AABC meet its circumcircle in two points which, with
the three vertices of the triangle, form the vertices of a
regular pentagon.
(3) The ^BDA = the internal angle of a regular pentagon.
The following Mathematical Instruments are im
P portant : —
I. Proportional Compasses.
This is an instrument primarily for the purpose
of increasing or diminishing given linesegments
in a given ratio ; ?>., of multiplying given line
segments by a given tensor.
If AO = BO and QO = PO, the triangles AOB,
POQ are isosceles and similar, and
AB:PQ = OA:OP.
B Hence, if the lines are one or both capable of
rotation about O, the distance AB may be made to vary at
pleasure, and PQ will remain in a constant ratio to AB.
The instrument usually consists of two brass bars with
slots, exactly alike, and having the point of motion O so
arranged as to be capable of being set at any part of the slot.
The points A, B, P, and Q are of steel.
2. The Sector.
This is another instrument which pri
marily serves the purpose of increasing or
diminishing given linesegments in given
ratios.
This instrument consists of two rules
equal in length and jointed at O so as to
be opened and shut like a pair of com
passes. Upon each rule various lines are
drawn corresponding in pairs, one on each rule.
Consider the pair OA and OB, called the "line of lines."
Each of the lines of this pair is divided into 10 equal parts
PROPORTION AMONGST LINESEGMENTS. 1 6/
which are again subdivided. Let the divisions be numbered
from o to lo along OA and OB, and suppose that the points
numbered 6 are the points P and Q. Then GAB and OPQ
are similar triangles, and therefore PQ : AB=OP : OA. But
OP = j%AO. .. PQ = i%AB.
And as by opening the instrument AB may be made equal to
any segment not beyond the compass of the instrument, we
can find PQ equal to ^q of any such given segment.
The least consideration will show that the distance 55 is
AB, 33 is f^^AB, etc. Also that 33 is f of 77, 55 is f of
77, etc. Hence the instrument serves to divide any given
segment into any number of equal parts, provided the num
ber is such as belongs to the instrument.
The various other lines of the sector serve other but very
similar purposes.
3. TAe Pantagraph or Eidograph.
Like the two preceding in
struments the pantagraph pri y^
marily increases or diminishes
segments in a given ratio, but
unlike the others it is so ar
ranged as to be continuous in
its operations, requiring only
one setting and no auxiliary
instruments.
It is made of a variety of
forms, but the one represented
in the figure is one of the most convenient.
AE, AB, and BF are three bars jointed at A and B. The
bars AE and BF are attached to the wheels A and B respec
tively, which are exactly of the same diameter, and around
which goes a very thin and fle^xible steel band C.
The result is that if AE and BF are so adjusted as to be
parallel, they remain parallel however they be situated with
respect to AB. E, F are two points adjustable on the bars
6S
SYNTHETIC GEOMETRY.
AE and BF, and D is a point in line with EF, around which
the whole instrument can be rotated.
Now let EG KM be any figure traced by the point E ; then
F will trace a similar figure FHLN.
Evidently the triangles DAE and DBF remain always
similar however the instrument is transformed. Therefore
DF is in a constant ratio to DE, viz., the ratio DB : DA.
Now, when E comes to G, F comes to some point H in
fine with GD, and such that DH : DG = DB : DA.
.*. the triangles EDG and FDH are similar, and FH is [[
to EG, and has to it the constant ratio DB : DA. Similarly
HL is II to GK and has to it the same constant ratio, etc.
.'. the figures are similar, and the ratio of homologous lines
in GMandHN is AD :DB.
The three points E, D, and F being all adjustable the ratio
can be changed at pleasure.
Altogether the Pantagraph is a highly important instrument,
and when so adjusted that E, D, and F are not in line its
results offer some interesting geometrical features.
4 :
i 'c
> 1
c
) A
4. The Diagonal Scale.
This is a divided scale in
which, by means of similar tri
angles, the difficulty of reading
off minute divisions is very much
diminished.
Its simpler form is illustrated in
°' ^ the figure.
A scale divided to fortieths of an inch is, on account of the
closeness of the divisions, very difficult to read.
In the scale represented OA is \ inch. The distance AB
is divided into 10 equal parts by the horizontal parallel lines
numbered i, 2, 3, etc. Then QBO' is a triangle whereof the
horizontal lines are all parallel to the base. Hence it is
readily seen from the proportionality of the homologous sides
of the similar triangles formed that the intercept on the hori
PROPORTION AMONGST LINESEGMENTS. 1 69
zontal line i, between GO' and OB, is tgO'E, that is
j^ inch.
Similarly the intercept on the horizontal line 2 is ^ inch,
on 3, £'q inch, etc.
Hence from j?^ to g^ is one inch and sevenfortieths.
In a similar manner diagonal scales can be made to divide
,any assumed unitlength into any required number of minute
parts.
The chief advantages of such scales are that the minute
divisions are kept quite distinct and apparent, and that errors
are consequently avoided.
Exercises.
1. ABCD is a square and P is taken in BC so that PC is
onethird of BC. AC cuts thie diagonal BD in O, and
AP cuts it in E. Then OE is onetenth of DB.
2. If, in I, OE is oneeighth of DB, how does P divide BC ?
3. If BP is one nth of BC, what part of DB is OE ?
4. Given three linesegments to find a fourth, so that the
four may be in proportion.
5. The rectangle on the distances of a point and its chord of
contact from the centre of a circle is equal to the
square on the radius of the circle.
6. OD and DO are fixed lines at right angles and O is a
fixed point. A fixed circle with centre on OD and
passing through O cuts OQ in P. Then OP . OQ is a
constant however OQ be drawn.
7. To divide a given segment similarly to a given divided
segment.
8. To divide a given segment into a given number of equal
parts.
9. Two secants through A cut a circle in B, D, and C, E
respectively. Then the triangles ABE and ACD are
similar. So also are the triangles ABC and AED.
10. Two chords are drawn in a circle. To find a point on
lyo SYNTHETIC GEOMETRY.
the circle from which perpendiculars to the chords are
proportional to the lengths of the chords.
11. ABC is a triangle and DE is parallel to AC, D being on
AB and E on CB. DC and AE intersect in O. Then
BO is a median.
12. If BO, in II, cuts DE in P and AC in O, BO is divided
harmonically by P and Q.
13. A and B are centres of fixed circles and AX and BY are
parallel radii. Show that XY intersects AB in a
fixed point.
14. In the triangle ABC, BD bisects the ^B and cuts AC in
D. Then BD2=AB. BC AD . DC. (Employ the
circumcircle.)
15. ABC is rightangled at B and BD is the altitude on AC.
(i) The As ADB and BDC are each similar to ABC.
(2) Show by proportion that AB2 = AD . AC,
and BD2=AD.DC.
16. If R and r denote the radii of the circumcircle and in
circle of a triangle, 2Rr(rt \d + c) = abc.
17. In an equilateral triangle the square on the side is equal
to six times the rectangle on the radii of the circum
circle and incircle.
18. OA, OB, OC are three lines. Draw a line cutting them
so that the segment intercepted between OA and OC
may be bisected by OB.
19. What is the measure of an angle in radians when its
measure in degrees is 68° 1 7' .'*
20. How many radians are in the angle of an equilateral A ?
21. The earth's diameter being 7,960 miles, what is the dis
tance in miles between two places having the same
longitude but differing 16° in latitude.''
22. Construct a regular pentagon, a regular decagon, a
regular polygon — of 1 5 sides, of 30 sides, of 60 sides.
23. ABODE is a regular pentagon.
(i) Every diagonal is divided into extreme and mean
ratio by another diagonal.
PROPORTION AMONGST LINESEGMENTS. I/I
(2) The diagonals enclose a second regular pentagon.
24. Compare the side and the areas of the two pentagons of
23 (2).
25. If one side of a rightangled triangle is a mean propor
tional between the other side and the hypothenuse, the
altitude from the right angle divides the hypothenuse
into extreme and mean ratio.
26. A variable line from a fixed point A meets a fixed circle
in P, and X is taken on AP so that AP . AX = a con
stant. The locus of X is a circle.
27. If two circles touch externally their common tangent is a
mean proportional between their diameters.
28. Four points on a circle are connected by three pairs of
lines. If a, oj denote the perpendiculars from any fifth
point on the circle to one pair of lines, /3, ^^ to another
pair, and 7, y^ to the third pair, then aai=/3j3i = 77i.
(Employ 204°.)
29. A line is drawn parallel to the base of a trapezoid and
bisecting the nonparallel sides. Compare the areas
of the two trapezoids formed.
30. Draw two lines parallel to the base of a triangle so as to
trisect the area.
31. ABC is rightangled at B, and AP is the perpendicular
from A to the tangent to the circumcircle at B. Then
AP.AC = AB.
SECTION II.
FUNCTIONS OF ANGLES.— AREAL RELATIONS.
212°. De/. — When an element of a figure undergoes change
the figure is said to vary that element.
If a triangle changes into any similar triangle it varies its
magnitude while its form remains constant ; and if it changes
1/2 SYNTHETIC GEOMETRY.
into another form while retaining the same area, it varies its
form while its area remains constant, etc.
Similar statements apply to other figures as well as triangles.
When a triangle varies its magnitude only, the tensors or
ratios of the sides taken two and two remain constant. Hence
the tensors or ratios of the sides of a triangle taken two and
two determine the form of the triangle but not its magni
tude ; i.e.^ they determine the angles but not the sides.
{n% 3 ; 197° ; 198° ; 201°)
A triangle, which, while varying its size, retains its form, is
sometimes said to remain similar to itself, because the tri
angles due to any two stages in its variation are similar to
one another.
213°. In the rightangled triangle the ratios or tensors of
the sides taken in pairs are important functions of the angles
and receive distinctive names.
^' '' The AOPM is rightangled at M,
and the :lPOM is denoted by e.
PM
Then, ~~ is the sine of ^, and is
contracted to sin B in writing.
^^ is the sine of the ^OPM, but as ^OPM is the comple
ment of ^, this tensor is called the cosine of ^, and is written cos Q.
'' is the tangent of B contracted to tan B.
OM *
Cor. I. V ™ = PM..OP, ., tan<» = liil^.
OM OP OM' cos»
PM^.oM==op^ .. e)%(^y
Cor. 2.
or sin^f cos'^^=i.
214°. LetOP' = OP be drawn so that aP'OM' = POM = ^,
and let P'M' be _L on OM.
Then z.P'OM is the supplement of ^, and the triangles
P'OM' and POM are congruent.
FUNCTIONS OF ANGLES. — AREAL RELATIONS. 1 73
sinP'OM = ^ = ™^sin..
i.e., an angle and its supplement have the same sine.
2. CosP'OM = 5^. But in changing from OM to OM',
on the same line, OM vanishes and then reappears upon the
opposite side of O.
Therefore OM and OM' have opposite senses (156°), and
if we consider OM positive, OM' is negative.
0M'=  OM, and hence
an angle and its supplement have cosines which are equal in
numerical value but opposite in sign.
215°, Theorem. — The area of a parallelogram is the pro
duct of two adjacent sides multiplied by the sine of their
included angle. (152°,!)
AC is a I 7 and BP is X upon AD.
Then BP is the altitude, and the
area = AD.BP. (153°, i)
But BP = ABsin^BAP.
£=7 = AB . AD sin ^BAP=rt<5 sin ^.
Cor. I. Since the area of a triangle is onehalf that of the
parallelogram on the same base and altitude,
.'. the area of a triangle is onehalf the product of any two
sides multiplied by the sine of the included angle. Or
2A = ^<^sin C = ^i:sin A=<f^sin B.
216°. Theorem. — The area of any quadrangle is onehalf
the product of the diagonals multiplied by the sine of the
angle between them.
ABCD is a quadrangle of which AC
and BD are diagonals.
Let^AOB = ^ = ^COD.
Then ^BOC=z.AOD = supp. of ^.
AAOB =^0A . OB sin B,
ABOC=iOB . CO sin B, ACOD =^0C . DO sin B,
74
SYNTHETIC GEOMETRY.
ADOA= DO . OA sin 6, and adding,
Od.=:AC.BDsin^.
(compare 162°)
217°. BD being the altitude to AC in the AABC, we have
B from 172°, 2,
But
r2.
AD = AB cos A=r cos A,
a^ = b^+C'2bccos\.
When B comes to B' the lK becomes
(214°, 2)
D' A DC
obtuse, and cos A changes sign.
If we consider the cosine with respect to its magnitude
only, we must write + before the term 2bccosK, when A be
comes obtuse. But, if we leave the sign of the function to be
accounted for by the character of the angle, the form given
is universal.
Cor. I. ABCD is a parallelogram. Consider the AABD,
then '^T)'^ = a'^\b^2abcosd.
Next, consider the AABC. Since
^ABC is the supplement of ^, and
BC = AD = /^,
AO=d^ + b'^ + 2abcosd.
and writing these as one expression,
a'^>rb'^±2abcosd^
gives both the diagonals of any 1 — 7 , one of whose angles is 0.
Cor. 2. DE = rt:cos^(CEbeingJ_to AD), CE = rtsin^.
AE = ^ + ^cos^;
and tanCAE = CE^_^zsin^
AE
b + aco^ d'
which gives the direction of the diagonal.
218". Def. — The ratio of any area X to another area Y is
the measure of X when Y is taken as the unitarea, and is
accordingly expressed as — . (Compare i88°.)
I. Let X and Y be two similar rectangles. Then X = ab
FUNCTIONS OF ANGLES.— AREAL RELATIONS. 175
and Y = a'd\ where a and /; are adjacent sides of the nnX and
a' and d' those of the cziY.
X _a d
But because the rectangles are similar, — = .
X^^ ^' ^'
Y a"''
/.<?., the areas of similar rectangles are proportional to the
areas of the squares upon homologous sides.
2. Let X and Y be two similar triangles. Then
X = ^ad sin C, Y = ha' d' sin C,
Y c^b' rt'^'
because the triangles are similar, (i97°)
i.e.^ the areas of similar triangles are proportional to the
areas of the squares upon homologous sides.
3. Let X denote the area of the pentagon ABCDE, and Y
that of the similar pentagon p^
A'B'CD'E'. Then A^\ R
P^AD^ R^AC^
P' A'D'2' R' A'C'^' / p \ Q
Q' D'C'2* ^
But (X)
P' R' Q' P' + Q' + R'
, AD _ AC _ DC
ana aTD'A'C'D'C"
X_ DC^
Y D'C'^'
And the same relation may be proved for any two similar
rectilinear figures whatever.
.'. the areas of any two similar rectilinear figures are pro
portional to the areas of squares upon any two homologous
lines.
176 SYNTHETIC GEOMETRY.
4. Since two circles are always similar, and are the limits
of two similar regular polygons,
.*. the areas of any two circles are proportional to the areas
of squares on any homologous chords of the circles, or on
linesegments equal to any two similar arcs.
5. When a figure varies its magnitude and retains its form,
any similar figure may be considered as one stage in its
variation.
Hence the above relations, i, 2, 3, 4, may be stated as
follows : —
The area of any figure with coiista7it form varies as the
square upon any one of its linesegments.
Exercises.
Two triangles having one angle in each equal have
their areas proportional to the rectangles on the sides
containing the equal angles.
Two equal triangles, which have an angle in each equal,
have the sides about this angle reciprocally propor
tional, i.e.^ a : a' = b' : b.
The circle described on the hypothenuse of a rightangled
triangle is equal to the sum of the circles described on
the sides as diameters.
If semicircles be described outwards upon the sides of a
rightangled triangle and a semicircle be described in
wards on the hypothenuse, two crescents are formed
whose sum is the area of the triangle.
AB is bisected in C, D is any point in /::m^
AB, and the curves are semicircles. ^ q
Prove that P 45 = fR. A c d
If ^, b denote adjacent sides of a parallelogram and also
of a rectangle, the ratio of the area of the parallelogram
to that of the rectangle is the sine of the angle of the
parallelogram.
FUNCTIONS OF ANGLES! — AREAL RELATIONS. 1 7/
7. The sides of a concyclic quadrangle are a, b^ c^ d. Then
the cosine of the angle between a and b is
{d'^b^c'd'')l2{^ab^cd).
8. In the quadrangle of 7, \is denotes onehalf the perimeter,
area = ^{{sd){s b)(s  c){s ~d)}.
9. In any parallelogram the ratio of the rectangle on the
sum and differences of adjacent sides to the rectangle
on the diagonals is the cosine of the angle between the
diagonals.
10. U a, b be the adjacent sides of a parallelogram and 6 the
angle between them, one diagonal is double the other
when cos^=l(^ + ^\
11. If one diagonal of a parallelogram is expressed by
/ J ^^^'—^ — H, the other diagonal is n times as long.
12. Construct an isosceles triangle in which the altitude is a
mean proportional between the side and the base.
13. Three circles touch two lines and the middle circle
touches each of the others. Prove that the radius of
the middle circle is a mean proportional between the
radii of the others.
14. In an equilateral triangle describe three circles which
shall touch one another and each of w hich shall touch
a side of the triangle.
15. In an equilateral triangle a circle is described to touch
the incircle and two sides of the triangle. Show that
its radius is onethird that of the incircle.
PART IV.
SECTION I.
GEOMETRIC EXTENSIONS.
220°. Let two lines L and M passing through the fixed
points A and B meet at P.
When P moves in the direction of the arrow, L and M
approach towards parallelism, and the
angle APB diminishes. Since the
^ • . *" ~^ lines are unlimited (21°, 3) P may re
cede from A along L until the segment AP becomes greater
than any conceivable length, and the angle APB becomes
less than any conceivable angle.
And as this process may be supposed to go on endlessly, P
is said to "go to infinity^' or to "be at infinity," and the
z_APB is said to vanish.
But lines which make no angle with one another are parallel,
.'. Parallel lines meet at infinity^ and lines which meet at
injitiity are parallel.
The symbol for " infinity " is 00 .
The phrases " to go to infinity," " to be at infinity," must
not be misunderstood. Infinity is not a place but a property.
Lines which meet at 00 are lines so situated that, having the
same direction they cannot meet at any finite point, and
therefore cannot meet at all, within our apprehension, since
every point that can be conceived of is finite.
178
GEOMETRIC EXTENSIONS.
1/9
The convenience of the expressions will appear throughout
the sequel.
Cor. Any two lines in the same plane meet : at a finite
point if the lines are not parallel, at infinity if the lines are
parallel.
221°. L and M are lines intersecting in O, and P is any
point from which PB and PA are 
respectively to L and M. A third
and variable line N turns about P
in the direction of the arrow.
1. AX. BY = a constant (184')
= U say.
When N comes to parallelism
with L, AX becomes infinite and
BY becomes zero.
.. 00.0 is indefinite since U may have any value we
please.
2. The motion continuing, let N come into the position N'.
Then AX' is opposite in sense to AX, and BY' to BY. But
AX increased to 00 , changed sign and then decreased ab
solutely, until it reached its present value AX', while BY
decreased to zero and then changed sign.
/. a magnitude changes sign when it passes through zero
or infinity.
3. It is readily seen that, as the rotation continues, BY' in
creases negatively and AX' decreases, as represented in one
of the stages of change at X" and Y". After this Y" goes off
to 00 as X" comes to A. Both magnitudes then change sign
again, this time BY" by passing through 00 and AX''' by pass
ing through zero.
.Since both segments change sign together the product or
rectangle remains always positive and always equal to the
constant area U.
l8o SYNTHETIC GEOMETRY.
222°. A line in the plane admits of one kind of varia
tion, rotation. When it rotates about a fixed finite point it
describes angles about that point. But since all the lines of
a system of parallels meet at the same point at infinity, rota
tion about that point is equivalent to translation, without
rotation, in a direction orthogonal to that of the line.
Hence any line can be brought into coincidence with any
other line in its plane by rotation about the point of intersection.
223". If a line rotates about a finite point while the point
simultaneously moves along the line, the point traces a curve
to which the line is at all times a tangent. The line is then
said to envelope the curve, and the curve is called the en
velope of the line.
The algebraic equation which gives the relation between
the rate of rotation of the line about the point and the rate of
translation of the point along the line is the intrinsic equation
to the curve.
224°. A linesegment in the plane admits of two kinds of
variation, viz., variation in length, and rotation.
If one endpoint be fixed the other describes some locus
depending for its character upon the nature of the variations.
The algebraic equation which gives the relation between
the rate of rotation and the rate of increase in length of the
segment, or radius vector, is the polar equation of the locus.
When the segment is invariable in length the locus is a
circle.
225°. A line which, by rotation, describes an angle may
rotate in the direction of the hands of a clock or in the con
trary direction.
If we call an angle described by one rotation positive we
must call that described by the other negative. Unless con
venience requires otherwise, the direction of rotation of the
hands of a clock is taken as negative.
GEOMETRIC EXTENSIONS.
I8l
An angle is thus counted from zero to a circumangle either
positively or negatively.
The angle between AB and A'B' is the
rotation which brings AB to A'B', and
is either + a or — j3, and the sum of these A
two angles irrespective of sign is a cir
cumangle.
When an angle exceeds a circumangle the excess is taken
in Geometry as the angle.
Ex. QA and QB bisect the angles CAB and ABP extern
ally ; to prove that z_P = 2^Q.
The rotation which brings CP
to AB is 20, AB to BP is +2^9,
.. ^P = 2(/3a).
Also, the rotation which brings
AQ to AB is a, and AB to BO
is +/3, .. ^Q=/3a.
z.(CP.BP) = 2Z(AQ.BQ).
This property is employed in the working of the sextant.
226°. Let AB and CD be two diameters at right angles.
The rectangular sections of the plane
taken in order of positive rotation and
starting from A are called respectively
the first, second, third, and fourth
quadrants, the first being AOC, the ^\
second COB, etc.
The radius vector starting from co
incidence with OA may describe the
positive zJ\OP, or the negative Z.AOP'.
Let these angles be equal in absolute value, so that the
AMOPs AMOP', PM being ± on OA.
Then PM=P'M, since in passing from P to P', PM
passes through zero.
sinAOP' =
P'M
OP
PM
OP
= sinAOP.
82
SYNTHETIC GEOMETRY.
and
cosAOP'=OM^=0^=cosAOP.
.'. the sine of an angle changes sign when the angle does,
but the cosine does not.
227°. As the angle AOP increases, OP passes through the
several quadrants in succession.
When OP lies in the ist Q., sin AOP and cos AOP are
both positive ; when OP lies in the 2nd g., sin AOP is posi
tive and cos AOP is negative ; when OP lies in the 3rd O.,
the sine and cosine are both negative ; and, lastly, when OP
lies in the 4th ^., sin AOP is negative and the cosine
positive.
Again, when P is at A, i.AOP = o, and PM==o, while
OM = OP. .•. sin = and cos 0=1.
When P comes to C, PM = OP and 0M=o, and denoting
a right angle by \ (207°, Cor. 4)
sin 2= I, and cos^ = o.
When P comes to B, PM=o and OM=  1,
sin 7r = o, and cos7r= — I.
Finally when P comes to D, PM= OP and OM=o.
sin ^'^=  I, and cos ^^^ = 0.
•7. ' ?
These variations of the sine and cosine for the several
quadrants are collected in the following table : —
ISt(?.
2nde.
3rd<>.
4th (>.
Sine,
Cosine, ....
+
+
+

+
Sine,
Cosine, ....
From To
I
From To
I
T
From To
 r
 1
From To
 I
I
GEOMETRIC EXTENSIONS. 183
228°. ABC is a triangle in its circumcircle whose diameter
we will denote by d. ^ ^^ ^b
Let CD be a diameter.
Then ^D = Z.A, (106°, Cor. i)
and Z.CBD=~ ^^
CB = CDsinCDB = ^sinA = ^.
and from symmetry,
<!/= ^ _ b ^ c
sin A sin B sin C*
Hence the sides of a triangle are proportional to the sines of
the opposite angles ; and the diameter of the circumcircle is
the quotient arising from dividing any side by the sine of the
angle opposite that side.
PRINCIPLE OF ORTHOGONAL PROJECTION.
229°. The orthogonal projection {i6y°, 2) of PO on L is
P'O', the segment intercepted between
the feet of the perpendiculars PP' and
QQ'.
Now P'Q' = PQcos(PQ . P'Q').
.'. the projection of any segment on a
given line is the segment multiplied by the cosine of the
angle which it makes with the given line.
From left to right being considered as the + direction
along L, the segment PQ lies in the ist Q., as may readily be
seen by considering P, the point from which we read the
segment, as being the centre of a circle through Q.
Similarly QP lies in the 3rd Q., and hence the projection
of PO on L is + while that of OP is .
When PQ is J_ to L, its projection on L is zero, and wheTi
II to L this projection is PO itself.
Results obtained through orthogonal projection are univer
sally true for all angles, but the greatest care must be
1 84
SYNTHETIC GEOMETRY.
exercised with regard to the signs of angular functions
concerned.
Ex. AX and OY are fixed lines at right angles, and AQ is
any line and P any point.
Required to find the ±PQ in terms of
AX, PX, and the z_A.
Take PO as the positive direction, and
project the closed figure POAXP on the
line of PO. Then
pr.PQ + pr.QA + pr.AX + pr.XP = o. (i68')
Now, pr.PQ is PQ, and pr.QA = o ; AX lies in ist Q.^ and
XP in the 3rd Q.
Moreover z.(AX . PQ), i.e., the rotation which brings AX to
PQ in direction is ZN, and its cosine is +.
cos4AX.PQ)=+sinA.
Also, pr.XP is XPcos^XPQ= XPcosA.
PO = XPcosAAXsinA.
SIGNS OF THE SEGMENTS OF DIVIDED
LINES AND ANGLES.
230°. AOB is a given angle and ^AOB= Z.BOA.
Let OP divide the Z.AOB internally, and OQ divide it ex
ternally into parts denoted respectively
by a, /3, and a!^.
If a is the ^AOP and /3 the ^POB, a
and /3 are both positive. But if we
write Q for P, a! = /AOQ, and
/3' = ^Q0B, and a and /S' have contrary signs.
On the other hand, if a is ^AOP and ^ the ^BOP, a and /8
have contrary signs, while replacing P by Q gives a and ^
with hke signs.
The choice between these usages must depend upon con
venience : and as it is more symmetrical with a twoletter
GEOMETRIC EXTENSIONS. 1 85
notation to write AGP, BOP, AGO, BOO, than AGP, FOB,
etc., we adopt the convention that internal division of an
angle gives segments with opposite signs, while external
division gives segments with like signs.
In like manner the internal division of the segment AB
gives parts AP, BP having unlike signs, while external divi
sion gives parts AQ, BQ having hke signs.
Def. — A set of points on a line is called a 7ajige, and the
line is called its axis.
By connecting the points of the range with any point not
on its axis we obtain a corresponding pencil. (203°, Def)
Cor. To any range corresponds a pencil for every vertex,
and to any vertex corresponds a range for every axis, the
axis being a transversal to the rays of the pencil.
If the vertex is on the axis the rays are coincident ; and if
the axis passes through the vertex the points are coincident.
231°. BY is any line dividing the angle B, and CR, AP are
perpendiculars upon BY. B
Then AAPY^ACRY,
and AP is AB sin ABY,
and CR is BC sin CBY,
AY^AB sin ABY
CY CB • sin CBY'
Therefore a line through the vertex of a tri
angle' divides the base into segments which
are proportional to the products of each conterminous side
multiplied by the sine of the corresponding segment of the
vertical angle.
Cor. I. Let BY bisect /.B, then ^^=.
YC a
AY=^(^AY), and AY = — .
Thence YC = ^.
a + c
1 86 SYNTHETIC GEOMETRY.
Which are the segments into which the bisector of the ^B
divides the base AC.
Cor. 2. In the AABY,
BY2 = ABHAY2AB.AY.cosA. (217°)
But cos A = ^'±^'":'' (217^), and AY=^,
ibc " a + c
whence by reduction
which is the square of the length of the bisector.
Cor. 3. When AY = CY, BY is a median, and
AB^sinYBC
CB sinABY"
.'.a median to a triangle divides the angle through which it
passes into parts whose sines are reciprocally as the con
terminous sides.
232°. In any range, when we consider both sign and mag
nitude, the sum AB + BC + CD + DE + EA=o,
however the points may be arranged.
For, since we start from A and return to A, the translation
in a + direction must be equal to that in a  direction.
That this holds for any number of points is readily seen.
Also, in any pencil, when we consider both sign and mag
nitude, the sum zA0B + ^B0Chz.CODf^D0A=o.
For we start from the ray OA and end with the ray OA,
and hence the rotation in a f direction is equal to that in a
 direction.
RANGES AND PENCILS OF FOUR.
533°. Let A, B, C, P be a range of four, then
AB.CPlBC.AP + CA.BP = o.
GEOMETRIC EXTENSIONS. 1 8/
Proof.— AP = AC + CP, and BP = BC + CP.
.'. the expression becomes
BC(AC + CA) + (AB + BC + CA)CP,
and each of the brackets is zero (232°). .'. etc.
234°. Let O . ABCP be a pencil of four. Then
sinAOB.sinCOP + sinBOC.sinAOP + sinCOA.sinBOP^o.
P^^^y:_/^AOB=OA . OB sin AOB,
also AAOB=^AB.;^,
where p is the common altitude to all
the triangles.
AB ./ = OA . OB . sin AOB.
Similarly, CP./=OC. OP. sin COP. '^ ^ c\
AB . CP .^2== OA . OB . OC . OP sin AOB . sin COP.
Now, /" and OA . OB . OC . OP appear in every homologous
product, .•.(AB.CP + BC.AP + CA.BP)/^
= OA . OB . OC . OP(sin AOB . sin COP
+ sin BOC . sin AOP + sin COA . sin BOP).
But the bracket on the left is zero (233°), and OA. OB.OC.OD
is not zero, therefore the bracket on the right is zero. q.e.d.
235''. From P let perpendiculars PA', PB', PC be drawn to
OA, OB, and OC respectively. Then
sinAOP = ^, sinBOP=^^, etc.,
and putting these values for sin AOP, etc., in the relation of
234°, we have, after multiplying through by OP,
C'P.sinAOB + A'P.sinBOC + B'P.sinCOA = o.
Or, let L, M, and N be any three concurrent lines, /, ///, 7i
the perpendiculars from any point P upon L, M, and N
respectively, then
>^ /sin MN+;;zsinNL + ;;sin LM=o.
where MN denotes the angle between M and N, etc.
SYNTHETIC GEOMETRY.
Let four rays be disposed in the order OA,
P OB, OC, OP, and let OP be perpendicular
to OA.
Denote /.AOC by A, and ^AOB by B.
Then 234° becomes
sinBcosA + sin(AB)sin2sinAcosB=o,
or, sin(AB)=sinAcosBcosAsinB.
Similarly, by writing the rays in the same order and making
^BOP a n, and denoting ^AOB by A and ^BOC by B, we
obtain sin (A + B) = sin A cos B + cos A sin B.
Also, by writing the rays in the order OA, OP, OB, OC,
and denoting z_AOP by A and Z.BOC by B, we obtain
(i) when ^AOB=n,
cos (A  B)= cos A cos B + sin A sin B ;
(2) when zAOC=n,
cos (A + B) = cosAcosBsinAsinB ;
which are the addition theorems for the sine and cosine.
Ex. 2. ABC is a triangle and P is any point. Let PX,
p PY, PZ be perpendiculars upon BC,
K^. ^ A CA, AB, and be denoted by P«, P^, Pc
1 \ ''""v// I'espectively.
I /^/ I Then (235°)
X B c D PQsinA + PYsinB + PZsinC = o.
But if AD is±to BC, AD = ^sin C = QX.
(PX(^sinC)sinA + PYsinB +PZsinC = o,
or .2:(P„sinA) = <^sinAsinC.
Similarly, S(PaSin A) = rsin Bsin A
= ^sin C sin B,
2(P„sin A)=^{abcs\n^K sin2B sin^C}.
Hence the function of the perpendicular
P,, sin A + Pj sin B + Pc sin C
is constant for all positions of P. This constancy is an
important element in the theory of trilinear coordinates.
GEOMETRIC EXTENSIONS.
189
237". A, B, C being a range of three, and P any point not
on the axis,
AB.CP4BC.AP + CA.BP
AB.BC.CA.
Proof.— LtX P() be X to AC. Then
AO = AC + CQ, BQ = BC + CQ,
and the expression becomes
(AB + BC + CA)(PQ2 + CQ) + BC.CA(BCAC)
= BC . CA(BC + CA) = BC . CA . BA
= AB.BC.CA.
Exercises.
1. A number of stretched threads have their lower ends fixed
to points lying in line on a table, and their other ends
brought together at a point above the table. What is
the character of the system of shadows on the table
when {a) a point of light is placed at the same height
above the table as the point of concurrence of the
threads ? {b) when placed at a greater or less height t
2 . If a line rotates uniformly about a point while the point
moves uniformly along the line, the point traces and
the line envelopes a circle.
3. If a radius vector rotates uniformly and at the same time
lengthens uniformly, obtain an idea of the curve traced
by the distal endpoint.
4. Divide an angle into two parts whose sines shall be in a
given ratio. (Use 231°, Cor. 3.)
5. From a given angle cut off a part whose sine shall be to
that of the whole angle in a given ratio.
6. Divide a given angle into two parts such that the product
of their sines may be a given quantity. Under what
condition is the solution impossible ?
7. Write the following in their simplest form : —
sin(7r^), sin (2+^), sin {tt (^) },
C0s(27r + ^), cos j27r(^)}, COS { ^  (J)^) }.
IQO
SYNTHETIC GEOMETRY.
8. Make a table of the variation of the tangent of an angle
in magnitude and sign.
9. OM and ON are two lines making theiLMON = w, and
PM and PN are perpendiculars upon OM and ON
respectively. Then OP sin a; = MN.
A transversal makes angles A', B', C with the sides BC,
CA, AB of a triangle. Then
sin A sin A' + sin B sin B' + sin C sin C = o.
OA, OB, OC, OP being four rays of any length whatever,
AAOB.ACOP+ABOC.AAOP+ACOA.ABOP=o.
If r be the radius of the incircle of a triangle, and r^ be
that of the excircle to side a^ and if p^ be the altitude
to the side a^ etc.,
.^ .(sin A + sinB + sinC)
sni A
^\ (sin A + sinB + sinC),
sm A
10
II.
J 2,
13
A
and
+ + =
^+^
^3 P\ Pi /;
+ ^ = i
(Use 235°.)
The base AC of a triangle is trisected at M and N, then
BN = i(3BC + 6BA22AC).
SECTION II.
CENTRE OF MEAN POSITION.
A, B, C, D are any points in line, and perpendiculars
AA', BB', etc., are drawn to any fixed
line L. Then there is, on the line,
evidently some point, O, for which
AA' + BB' + CC'+DD' = 40N ;
and ON is less than A A' and greater
than DD'.
The point O is called the centre o/jnean position, or simply
the 7ne(m centi'e, of the system of points A, B, C, D.
CENTRE OF MEAN POSITION. IQI
Again, if we take multiples of the perpendiculars, as a.AA',
b . BB', etc., there is some point O, on the axis of the points,
for which
^?.AA' + /5.BB' + r.CC' + ^. DI)' = (rt + /A+6+^/)ON.
Here again ON lies between AA' and DD'.
O is then called the nieati centre of the system of points for
the system of multiples.
Def.— For a range of points with a system of multiples we
define the mean centre by the equation
2(^?.A0) = o,
where 2(/t . AO) is a contraction for
^.AO + ^.BO + 6.CO + ...,
and the signs and magnitudes of the segments are both
considered.
The notion of the mean centre or centre of mean position
has been introduced into Geometry from Statics, since a
system of material points having their weights denoted by «,
b, c, ..., and placed at A, B, C, ..., would "balance" about
the mean centre O, if free to rotate about O under the action
of gravity.
The mean centre has therefore a close relation to the
" centre of gravity " or "mass centre" of Statics.
239°. Theorem. — If P is an independent point in the line of
any range, and O is the mean centre, ^ ^
S(^ . AP) = S(«) . OP. "a b S c p
/•r^^/ AP = AO + OP, BP = BO + OP, etc.,
rt.AP = ^.AO + rt.OP, ^.BP = <^.BO + ^.OP, etc.
2(^ . AP) = Z(^ . AO) + 2(^) . OP.
But, if O is the mean centre,
2(<z. AO)=o, by definition,
2(rt.AP)=S(.?)OP.
Ex. The mean centre of the basal vertices of a triangle
192 SYNTHETIC GEOMETRY.
when the multiples are proportional to the opposite sides is
the foot of the bisector of the vertical angle.
240°. Let A, B, C, ... be a system of points situated any
where in the plane, and let AL, BL, CL, ..., AM, BM, CM,
..., denote perpendiculars from A, B, C, ... upon two Hnes L
and M.
Then we define the mean centre of the system of points for
a system of multiples as the point of intersection of L and M
when 2(^z.AL) = o,
and 2(rt:.AM)=o.
If N be any other line through this
centre, S(^.AN)=o.
For, let A be one of the points.
Then, since L, M, N is a pencil of three
and A any point, (235°)
AL . sin MON + AM . sin NOL+AN . sin LOM =0,
also BL . sin MON + BM . sin NOL + BN . sin LOM = 0,
and multiplying the first b}' <?, the second by b^ etc., and adding,
2(^.AL)sinMON + 2(^.AM)sinNOL + 2(rt.AN)sinLOM=o.
But 2:(^. AL) = S(rt'. AM) = o, by definition,
2:(rt.AN) = o.
241". 'J'heore7n. — If O be the mean centre of a system of
points for a system of multiples, and L any line whatever,
20?.AL) = 2(rt).OL.
Proof. — Let M be  to L and pass through O. Then
AL = AM + ML, .. ^.AL=^.AM + ^?.ML,
BL = BM + ML, ,. /;.BL = ^.BM + /5.ML,
adding, ^{a . AL) = Y.{a . AM) + ^{d) . ML.
But, since M passes through O,
S(«.AM)=:o, and ML = OL,
S(^.AL)=S(^).OL.
CENTRE OF MEAN POSITION. 1 93
242°. Theorem. — The mean centre of the vertices of a
triangle with multiples proportional to
the opposite sides is the centre of the
incircle.
Proof. — Take L along one "of the sides,
as BC, and let/ be the ± from A. Then A
2(^.AL) = ^./
and S(.i:).OL = (« + <^+<:).OL,
■•■(^39°) OL = ^^=f = .; (.53', Ex..)
i.e.^ the mean centre is at the distance r from each side, and
is the centre of the incircle.
Cor. I. If one of the multiples, as a, be taken negative,
OL= ~^f =z:A^ _^' (j53° Ex. 2)
/>., the mean centre is beyond L, and is at the distance r' from
each side, or it is the centre of the excircle to the side a.
Cor. 2. If any line be drawn through the centre of the in
circle of a triangle, and a, j8, 7 be the perpendiculars from the
vertices upon it, aa + d^ + cy=o,
and if the line passes through the centre of an excircle, that
on the side ^ for example, aa = b^ + cy.
Exercises.
If a line so moves that the sum of fixed multiples of the
perpendiculars upon it from any number of points is
constant, the line envelopes a circle whose radius is
The mean centre of the vertices of a triangle, for equal
multiples, is the centroid.
The mean centre of the vertices of any regular polygon, for
equal multiples, is the centre of its circumcircle.
N
194 SYNTHETIC GEOMETRY.
243°. Theorem. — If O be the mean centre of a system of
points for a system of multiples, and P any independent point
in the plane,
Proof. — Let O be the mean centre, P the
independent point, and A any point of the
system. Let L pass through O and be per
pendicular to OP, and let AA' be perpen
dicular to OP. Then
AP2=AOHOP220P.OA',
and «.AP2 = ^.A02 + «.OP220P.^.OA'.
Similarly <5. BP2 = <^. 30^ + ^. OP220P.^. OB',
2(^.AP2) = 2(«.A02)l2(^).0P220P.2(^.0A').
But 2(^.0A') = 2(«.AL)=o, (241°)
2(^. AP2) = 2(^. A02) + 2(rt). 0P2. q.e.d.
Cor. In any regular polygon of ;^ sides th the sum of the
squares on the joins of any point with the vertices is greater
than the square on the join of the point with the mean centre
of the polygon by the square on the circumradius.
For making the multiples all unity,
2(AP2) = /z^ + «OP2,
i2(AP2) = OP2 + r2.
71
Ex. Let a, b, c be the sides of a triangle, and a, j8, 7 the
joins of the vertices with the centroid. Then (242°, Ex. 2)
2(AP2)=2(A02)f30P2.
ist. Let P be at A, ^^.^^^^a^ + zS^lT'+Sa',
2nd. „ P „ B, r^ + ^^^a^ + iSHv' + S^',
3rd. „ P „ C, «2 + <^^ = a2 + /32f7H3r,
whence Vz^+(^^ + c^=3(a+/3H7).
Ex. If ABCDEFGH be the vertices of a regular octagon
taken in order, AC2 + ADHAE2tAF2 + AG2=i4r2.
CENTRE OF MEAN POSITION. 195
244°. Let O be the centre of the incircle of the AABC and
let P coincide with A, B, and C in succession.
ist. bc"^ cl^ = ^{a . AO^) + 2(rt)A02,
2nd. ac^ +t«2_2(^.A02) + S(rt)B02,
3rd. aB'^ba'^ =2(^.AO2) + 2(.0CO2.
Now, multiply the ist by a, the 2nd by b^ the 3rd by c, and
add, and we obtain, after dividing by {a^rb^rc\
^{a. AO^) =abc.
Cor. I, For any triangle, with O as the centre of the incircle,
the relation 2(^. AP2)=S(rt. A02)+2(rt)OP2
becomes 'Z{a.AV^)=abc+2s. OP^,
and, if O be the centre of an excircle on side ^, for example,
S(^. AP2)= abc+2{sa)0V\
where a denotes that a alone is negative.
Cor. 2. Let P be taken at the circumcentre, and let D be
the distance between the circumcentre and the centre of the
incircle. Then AP = BP = CP = R.
2s'K^=^abc + 2sT>\
But abc=^/S^, (204°, Cor.)
and s=^, (153°, Ex. i)
D2=R22Rr.
Cor. 3. If Dj be the distance between the circumcentre and
the centre of an excircle to the side a, we obtain in a similar
manner D ^ = R^ + 2 Rr^.
Similarly D2^ = RH2Rr2,
D32 = R2 + 2Rr3.
245°. Ex. To find the product OA. OB . DC, where O is
the centre of the incircle.
Let P coincide with A. Then (244°)
bc'' + ^c=abc+2s.K0\
196 SYNTHETIC GEOMETRY.
Similarly B0^=£%1^\ and C02 = ^%l£),
AO^. BO^. CO^:
s s
_ a^^c^s(s  a)(s  /?){s  c)
=^A'=i6RV,
and OA.OB.OC = 4R^2.
246°. If S(rt . AP^) becomes constant, /^, we have
^ = 2(^.A02) + 2(rt)0P^
and 2(^z . AO^) being independent of the position of P, and
therefore constant for variations of P, OP is also constant,
and P describes a circle whose radius is
p2_ >^S(^.AO^)
.'. If a point so moves that the sum of the squares of its
joins with any number of fixed points, each multiplied by a
given quantity, is constant, the point describes a circle
whose centre is the mean point of the system for the given
multiples.
Exercises,
1. If O', O", O'" be the centres of the escribed circles,
AO'.BO".CO"'=4R«5'.
2. A0'.B0'.C0' = 4R^i'.
3. s.OL = {s a)0'L + {s ^)0"L + {s c)0"'L,
where L is any line whatever.
4. If P be any point,
5. (sa). O'O^ + {sd). 0"02 + {s~ c)0"'0'' = labc.
6. If D is the distance between the circumcentre and the
centroid, Ti" =\{c)P:  a^  V"  c").
OF COLLIN.EARITY AND CONCURRENCE. I9;
SECTION III.
OF COLLINEARITY AND CONCURRENCE.
247°. Def. I. — Three or more points in line are collinear,
and three or more lines meeting in a point are concurrent.
Def. 2. — A tetragram or general quadrangle is the figure
formed by four lines no three of
which are concurrent, and no two
of which are parallel.
Thus L, M, N, K form a tetra l/ ^^^^/c
gram. A, B, C, D, E, F are its six
vertices. AC, BD are its internal ^^
diagonals, and EF is its external diagonal.
248°. The following are promiscuous examples of collinear
ity and concurrence.
Ex. I. AC is a 1 7 , and P is any point. Through P, GH
is drawn 1 to BC, and EF  k N o
to AB.
The diagonals EG, HF,
and DB of the three / 7s
AP, PC, and AC are concur
rent.
EG and HF meet in some D H c L
point O ; join BO and complete the ZZZ70KDL, and make
the extensions as in the figure.
We are to prove that D, B, O are collinear.
/'r^<?/:— /ZZ7KG = Z=7GM, and ^Z7FL = £i:7FN, (145°)
Ei:7KG = ^=Z7GF + ZZ:7BM,
and z=:7FL = £:Z7GF + z:=7NB.
Hence EI::7KB = CZ7BL,
B is on the line DO, (i45°j Cor. 2)
and D, B, O are collinear.
198
SYNTHETIC GEOMETRY.
Ex. 2. The middle points of the three diagonals of a tetra
gram are collinear.
ABCDEF is the tet
ragram, P, Q, R the
middle points of the
diagonals.
Complete the paral
lelogram AEDG, and
through B and C draw
lines II to AG and AE
respectively, and let
them meet in T.
Then, from Ex. i,
IH passes through F. Therefore EI HP is a triangle, and
the middle points of EI, EH, and EF are collinear.
(84°, Cor. 2)
But these are the middle points of AC, BD, and EF re
spectively. .'. P, Q, R are collinear.
Ex. 3. Theorem. — The circumcentre, the centroid, and the
orthocentre of a triangle are col
linear.
Proof.— L^\. YD and ZD be
the right bisectors of AC and AB.
Join BY, CZ, and through E,
A Y H c the intersection of these joins,
draw DE to meet the altitude BH in O.
Then D is the circumcentre and E is the centroid. Since
DY is II to BH, the triangles YDE and BOE are similar.
But BE = 2EY, (85°, Cor.)
OE = 2DE,
and as D and E are fixed points, O is a fixed point.
.*. the remaining altitudes pass through O.
249°. 77/^<7r^;«.— Three concurrent lines perpendicular to
the sides of a triangle at X, Y, Z divide the sides so that
BX2 + CY2 + AZ2 = CXHAY2lBZ2;
OF COLLINEARITY AND CONCURRENCE. 199
and, conversely, if three lines perpendicular to the sides of a
triangle divide the sides in this manner, B
the lines are concurrent.
Proof.— \.^\, OX, OY, OZ be the lines. z^
Then BX2CX2 = B02C02,(i72°,i) ^j
Similarly CY^ AY2 = C02A02,
AZ2BZ2 = A02BO^ A
BX2 + CY2 + AZ2CX"^AY2BZ2 = o. q.eul
Conversely, let X, Y, Z divide the sides of the triangle in
the manner stated, and let OX, OY, perpendiculars to BC
and CA, meet at O. Then OZ is _L to AB.
Proof.— 1{ possible let OZ' be J_ to AB, Then, by the
theorem, BXH CY^ + AZ'^  CX^  AY^  BZ'2=o,
and by hyp. BX2 + CYHAZ2CX2AY2BZ2=o,
AZ'2AZ2=BZ'2BZ2.
But these differences have opposite signs and cannot be equal
unless each is zero. .*. Z' coincides with Z.
Exercises.
1. When three circles intersect two and two, the common
chords are concurrent.
Let S, Si, S2 be the circles, and A, B, C their centres.
Then (113°) the chords are perpendicular to the sides
of the AABC at X, Y, and Z. And if r, r^, r^ be the
radii of the circles,
BX2CX2 = ri2r22, etc., etc.,
and the criterion is satisfied.
.*. the chords are concurrent.
2. The perpendiculars to the sides of a triangle at the points
of contact of the escribed circles are concurrent.
3. When three circles touch two and two the three common
tangents are concurrent.
4. If perpendiculars from the vertices of one triangle on the
sides of another be concurrent, then the perpendiculars
200 SYNTHETIC GEOMETRY.
from the vertices of the second triangle on the sides
of the first are concurrent.
5. When three perpendiculars to the sides of a triangle are
concurrent, the other three at the same distances from
the middle points of the sides are concurrent.'
6. Two perpendiculars at points of contact of excircles are
concurrent with a perpendicular at a point of contact of
the incircle.
250°. Theorem. — When three points X, Y, Z lying on the
sides BC, CA, and AB of a triangle are collinear, they divide
the sides into parts which fulfil the relation
f. BX.CY.AZ ^
{(X) ■ = I :
^ ' CX.AY.BZ
and their joins with the opposite vertices divide the angles
into parts which fulfil the relation
,,N sin BAX . sin CBY . sin ACZ _
^^ sinCAX.sinABY.sinBCZ *
Proof of {a).— On the axis of X, Y, Z draw the perpendicu
B lars AP, BQ, CR.
On account of similar /\s,
^=?S ^=CR AZ^AP
CX CR' AY AP' BZ BQ'
BX.CY.AZ _
CX.AY.BZ '
Proof of {b). BX=ABAX^B AsinBAX
■^•^^^ CX ACAX CAsinCAX'
sinBAX_CA BX
sinCAX BA'CX'
Similarlv si" CBY _AB CY sinACZ _BC AZ
^' sinABY CB'AY' sin BCZ AC " BZ*
sin BAX . sin CBY . sin ACZ _ BX . CY. AZ _
SinCAX.sinABY.sinBCZ CX.AY.BZ ' q.e.d.
The preceding functions which are criteria of collinearity
OF COLLINEARITY AND CONCURRENCE. 201
will be denoted by the symbols
It is readily seen that three points on the sides of a triangle
can be collinear only when an even number of sides or angles
(2 or o) are divided internally, and from 230° it is evident that
the sign of the product is + in these two cases.
Hence, in applying these criteria, the signs may be dis
regarded, as the final sign of the product is determined by
the number of sides or angles divided internally.
The converses of these criteria are readily proved, and the
proofs are left as an exercise to the reader.
Ex. If perpendiculars be drawn to the sides of a triangle
from any point in its circumcircle, the feet of the perpendicu
lars are collinear.
X, Y, Z are the feet of the perpendicu
lars. If X falls between B and C, ^OBC
is < a ~~, and therefore Z.OAC is > a ""],
and Y divides AC externally ;
.'. it is a case of collinearity.
Now, BX = OBcosOBC,
and AY=OAcosOAC.
But, neglecting sign, cosOBC=cosOAC,
BX^OB
AY OA'
CY^OC
BZ OB
BX
CX
and similarly.
AZ
CX"
OA
OC
Vex/
and X, Y, Z are collinear.
Def. — The line of collinearity of X, Y, Z is known as
" Simson's line for the point O."
251°. Theorem. — When three lines through the vertices of a
triangle are concurrent, they divide the angles into parts
202 SYNTHETIC GEOMETRY.
which fulfil the relation
, V s in BAX . sin CBY . sin ACZ _ _
^^' sinCAX.sinABY.sinBCZ
and they divide the opposite sides into parts which fulfil the
^^ CX.AY.BZ
To prove {a). — Let O be the point of concurrence of AX, BY,
and CZ, and let OP, OQ, OR be perpendiculars on the sides.
sin BAX OR
Then
sinCAX OQ'
sinCBY _OP
sinABY OR*
sinACZ _OQ
sinBCZ OP'
sinBAX\
A QY
, . , . /sin D/\A.\
..multiplying, (inrCAx)=''
the negative sign resulting from the three angles being
divided internally. (230°)
To prove (<5).— From B and C let BE and CF be per
pendiculars upon AX.
Then, from similar As BEX and CFX,
BX_BE_ ABsinBAX
CX CF ACsinCAX*
e. ., , CY_BCsinCBY AZ_CAsinACZ
j)imiiariy, AY ~BA sinABY' BZ~CBsinBCZ'
^)=i, from (^). ^^^
The negative sign results from the three sides being divided
internally. (230°)
It is readily seen that three concurrent lines through the
vertices of a triangle must divide an odd number of angles
and of sides internally, and that the resulting sign of the
product is accordingly negative.
Hence, in applying the criteria, the signs of the ratios may
be neglected.
OF COLLINEARITY AND CONCURRENCE. 203
The remarkable relation existing between the criteria for
coUinearity of points and concurrence of lines will receive an
explanation under the subject of Reciprocal Polars.
Exercises.
252°. I. Equilateral triangles ABC, BCA', CAB' are de
scribed upon the sides AB, BC, CA of any triangle.
Then the joins A A', BB', CC are concurrent.
Proof.— S'mce AC' = AB, AB'=AC,
and ^CAC'=^BAB',
ACAC'=AB'AB, and aAC'C = ^ABB'.
i^ sin ACZ _sin ACC'_AC'_AB (oi9,°\
" sin ABY~ sin ACC~ AC ~ AC' ^ ^
c ., , sin BAX BC sin CBY CA
^'"^'^^'^^^ ihTBCZ^BA' ilKCAX^CB'
/sin BAX \_
VsinCAX/ '
and hence the joins AA', BB', CC are concurrent.
2. The joins of the vertices of a A with the points of con
tact of the incircle are concurrent.
3. The joins of the vertices of a A with the points of
contact of an escribed are concurrent.
4. ABC is a A, rightangled at B, CD is = and _L to CB,
and AE is = and _L to AB. Then EC and AD inter
sect on the altitude from B.
5. The internal bisectors of two angles of a A and the
external bisector of the third angle intersect the
opposite sides collinearly.
6. The external bisectors of the angles of a A intersect the
opposite sides collinearly.
7. The tangents to the circumcircle of a A, at the vertices
of the A, intersect the opposite sides collinearly.
8. If any point be joined to the vertices of a A, the lines
through the point perpendicular to those joins intersect
the opposite sides of the A collinearly.
204
SYNTHETIC GEOMETRY.
9, A cuts the sides of a A in six points so that three
of them connect with the opposite vertices concurrently.
Show that the remaining three connect concurrently
with the opposite vertices.
[o. Is the statement of Ex. i true when the As are all de
scribed internally upon the sides of the given A ?
[I. If L is an axis of symmetry to the congruent As ABC
and A'B'C, and O is any point on L, A'O, B'O, and
CO intersect the sides BC, CA, and AB collinearly.
253°. Theorem. — Two triangles which have their vertices
connecting concurrently have their corresponding sides inter
secting collinearly. (Desargue's Theorem.)
\z ABC, A'B'C are two As
having their vertices connect
ing concurrently at O, and
their corresponding sides in
tersecting in X, Y, Z. To
prove that X, Y, Z are col
linear.
Proof. — To the sides of
AA'B'C draw perpendiculars
AP, AP', BQ, BQ', CR, CR'.
Then, from similar As,
bx^bq;
CX CR'
CY^CR'
AY AP'
AZ_AF
BZ BQ'
A P\ BO\CR'
AP . BQ . CR*
But t^=«4^LAA;^;,
sm AA'C"
with similar expressions for the other ratios.
OF COLLINEARITY AND CONCURRENCE. 205
Also, since AA', BB', CC are concurrent at O, they divide
the angles A' B', C so that
sin AA^B\ sin BB'C. sin CC^A ^ ^ ^
sin AA'C sin BB'A'. sin CC'B' '
( — V j = I, and X, Y, Z are collinear.
The converse of this theorem is readily proved, and will be
left as an exercise to the reader.
Ex. A', B', C are points upon the sides BC, CA, AB re
spectively of the AABC, and AA', BB', CC are concurrent
in O. Then
1. AB and A'B', BC and B'C, CA and CA' meet in three
points Z, X, Y, which are collinear.
2. The lines AX, BY, CZ form a triangle with vertices A",
B", C", such that AA", BB", CC" are concurrent in O.
OF RECTILINEAR FIGURES IN PERSPECTIVE.
254°. Def. — AB and A'B' are two segments and AA' and
BB' meet in O. a b
Then the segments AB and A'B' are said to
be in perspective at O, which is called their
centre of perspective.
The term perspective is introduced from
Optics, because an eye placed at O would see
A' coinciding with A and B' with B, and the
segment A'B' coinciding with AB.
By an extension of this idea O' is also a
centre of perspective of AB and B'A'. O is o
then the external centre of perspective and O' is the internal
centre.
Def. — Two rectilinear figures of the same number of sides
are in perspective when every two corresponding sides have
the same centre of perspective.
206 SYNTHETIC GEOMETRY.
Cor. I. From the preceding definition it follows that two
rectilinear figures of the same species are in perspective when
the joins of their vertices, in pairs, are concurrent.
Cor. 2. When two triangles are in perspective, their ver
tices connect concurrently, and their corresponding sides
intersect collinearly. (253°)
In triangles either of the above conditions is a criterion of
the triangles being in perspective.
Def. — The line of collinearity of the intersections of corre
sponding sides of triangles in perspective is called their axis
of perspective ; and the point of concurrence of the joins of
corresponding vertices is the centre of perspective.
255°. Let AA', BB', CC be six points which connect con
currently in the order written.
These six points may be connected in four different ways
so as to form pairs of triangles having the same centre of
perspective, viz.,
ABC, A'B'C; ABC, A'B'C ; AB'C, A'BC; A'BC, AB'C.
These four pairs of conjugate triangles determine four axes
of perspective, which intersect in six points ; these points are
centres of perspective of the sides of the two triangles taken
in pairs, three X, Y, Z being external centres, and three X',
Y', Z' being internal centres. (254°)
The points, the intersections which determine them, and
the segments of which they are centres of perspective are
given in the following table : —
OINT.
Determined by
Intersection of
Centre of
Perspective to
X
BCB'C
BC'B'C
Y
CAC'A'
CA'C'A
Z
ABA'B'
AB'A'B
X'
BC'B'C
BCB'C
Y'
CA'C'A
CA~C'A'
Z'
AB'A'B
ABA'B'
OF INVERSION AND INVERSE FIGURES. 207
And the six points lie on the four lines thus,
XYZ, X'Y'Z, X'YZ', XY'Z'.
Exercises
1. The triangle formed by joining the centres of the three
excircles of any triangle is in perspective with it.
2. The three chords of contact of the excircles of any tri
angle form a triangle in perspective with the original.
3. The tangents to the circumcircle of a triangle at the three
vertices form a triangle in perspective with the original.
SECTION IV.
OF INVERSION AND INVERSE FIGURES.
256°. Def. — Two points so situated upon a centreline of a
circle that the radius is a geometric mean (169°, Def) between
their distances from the centre are called inverse points with
respect to the circle.
Thus P and Q are inverse points if
CP.CQ = CB2=R2, a
R being the radius.
The 0S is the circle of inversion
or the inverting 0, and C is the centre of inversion.
Cor. From the definition : —
1. An indefinite number of pairs of inverse points may lie on
the same centreline.
2. An indefinite number of circles may have the same two
points as inverse points.
3. Both points of a pair of inverse points lie upon the same
side of the centre of inversion.
4. Of a pair of inverse points one lies within the circle and
one without.
208 SYNTHETIC GEOMETRY.
5. P and Q come together at B ; so that any point on the
circle of inversion is its own inverse.
6. When P comes to C, Q goes to co ; so that the inverse of
the centre of inversion is any point at infinity.
257°. Problem. — To find the circle to which two pairs of
coUinear points may be inverse points.
P^,_— .^Q c p! ~^Q' P> Qj P') Q' ^6 the four
collinear points, of which
PQ and P'Q' are respec
tively to be pairs of in
verse points.
Through P, Q and
\v through P', Q' describe
any two circles S, S' to intersect in two points U and V. Let
the connector UV cut the axis of the points in C, and let
CT be a tangent to circle S'. Then C is the centre and CT
the radius of the required circle.
Proof.— CT2 = CP'. CQ' = CU. CV = CQ. CP.
Cor. If the points have the order P, P', Q, Q' the centre C
is real and can be found as before, but it then lies within
both circles S and S', and no tangent can be drawn to
either of these circles ; in this case we say that the radius
of the circle is imaginary although its centre is real.
In the present case P and Q, as also P' and Q', lie upon
opposite sides of C, and the rectangles CP. CQ and CP'. CQ'
are both negative. But R^ being alwa;^s positive (163°)
cannot be equal to a negative magnitude.
When the points have the order P, P', Q', Q, the circle of
inversion is again real.
Hence, in order that the circle of inversion may be real,
each pair of points must lie wholly without the other, or one
pair must lie between the others.
OF INVERSION AND INVERSE FIGURES. 209
Exercises.
1. Given a and a point without it to find the inverse point.
2. Given a and a point within it to find the inverse point.
3. Given two points to find any to which they shall be
inverse.
4. In 3 the is to have a given radius.
5. In 3 the is to have a given centre on the line of the
points.
258°. Theorem. — A which passes through a pair of
inverse points with respect to another cuts the latter
orthogonally. (i 15°, Defs. i, 2)
And, conversel}^, a which cuts another orthogonally
determines a pair of inverse points on any centreline of the
latter.
1. P and O are inverse to 08.
Then CP.C0 = CT2,
CT is tangent to 0S'.
(176°, Cor. 3)
And .*. S' cuts S orthogonally
since the radius of S is perpen
dicular to the radius of S' at its endpoint.
2. Conversely, let S' cut S orthogonally. Then ZCTC
is a ""I, and therefore CT is tangent to S' at the point T.
Hence CT^^CP.CQ,
and P and Q are inverse points to 08.
Cor. I. A through a pair of points inverse to one another
with respect to two 0s cuts both orthogonally.
Cor. 2. A which cuts two 0s orthogonally determines
on their common centreline a pair of points which are inverse
to one another with respect to both 0s.
Cor. 3. If the 08 cuts the 0s 8' and 8" orthogonally, the
tangents from the centre of 8 to the 0s 8' and 8" are radii of
S and therefore equal.
2IO
SYNTHETIC GEOMETRY.
.*. (178°) a which cuts two 0s orthogonally has its
centre on their radical axis.
Cor. 4, A having its centre on the radical axis of two
given 0s, and cutting one of them orthogonally, cuts the
other orthogonally also.
259°. Let P, O be inverse points to circle S and D any
point on it.
Then
CP.CQ=CD2,
.. CP:CD = CD:CQ.
Hence the triangles CPD
and CDQ are similar, and
PD and DQ are homolo
gous sides.
PD2_CD2_ CP.CQ _CP
QD2 CQ2 CQ2 CQ
.'. the squares on the joins of any point on a circle with a
pair of inverse points with respect to the circle are propor
tional to the distances of the inverse points from the centre.
Cor. I. If P and Q are fixed, PD^: QD^ is a fixed ratio.
.'. the locus of a point, for which the squares on its joins to
two fixed points have a constant ratio, is a circle having the
two fixed points as inverse points.
Cor. 2. When D comes to A and B we obtain
CP_PD2_PA2_PB2
CQ QD'* 0A2 QB2'
Pp^PA^PB
QD QA QB
Hence DA and DB are the bisectors of the ^PDQ, and
the segments PB and BQ subtend equal angles at D.
Hence the locus of a point at which two adjacent segments
of the same line subtend equal angles is a circle passing
through the common endpoints of the segments and having
their other endpoints as inverse points.
OF INVERSION AND INVERSP: FIGURES. 211
Cor. 3. Let P', Q' be a second pair of inverse points. Then
Z.BDP' = ^BDQ', and ^BDP = ^BDQ,
z.PDP' = ^QDQ';
or the segments PP' and QO' subtend equal angles at D.
Hence the locus of a point at which two nonadjacent seg
ments of the same line subtend equal angles is a circle having
the endpoints as pairs of inverse points.
Cor. 4. Since AP : AQ = PB : BQ, (Cor. i)
.*. P and Q divide the diameter AB in the same manner in
ternally and externally, and B and A divide the segment PQ
in the same manner internally and externally.
.*. from 208°, Cor., P, Q divide AB harmonically, and A, B
divide PO harmonically.
Hence, when two segments of the same line are such that
the endpoints of one divide the other harmonically, the circle
on either segment as diameter has the endpoints of the other
segment as inverse points.
Exercises.
1. If a variable circle passes through a pair of inverse points
with respect to a fixed circle, the common chord of the
circles passes through a fixed point.
2. To draw a circle so as to pass through a given point and
cut a given circle orthogonally.
3. To draw a circle to cut two given circles orthogonally.
4. On the common centreline of two circles to find a pair of
points which are inverse to both circles.
Let C, C be the centres of the circles S and S'. Take
any point P, without both circles," and find its inverses
P' and P" with respect to both circles. (257°, Ex. i)
The circle through P, P', and P" cuts the common
centreline CC in the required points Q and Q'.
5. To describe a circle to pass through a given point and cut
two given circles orthogonally.
212 SYNTHETIC GEOMETRY.
6. To determine, on a given line, a point the ratio of whose
distances from two fixed points is given.
7. To find a point upon a given line from which the parts of
a given divided segment may subtend equal angles.
8. A, B, C, D are the vertices of a quadrangle. Gn the
diagonal BD find a point at which the sides BA and
BC subtend equal angles.
9. To draw a circle to pass through a given point and touch
two given lines.
260°. Def. — One figure is the inverse of another when every
point on one figure has its inverse upon the other figure.
Theorem. — The inverse of a circle is a circle when the
centre of inversion is not on the figure to be inverted.
s
T
Let O be the centre of inversion and S be the circle to be
inverted ; and let A', B', C be the inverses of A, B, C respec
tively. To prove that the L^Ch! = ~^
Pr^t?/— OA. OA'=OB. OB' = OC. 0C'=R2, (256°)
A0A'C5=?A0CA, and AOB'C'^AOCB.
(i) aOC'A'=z.OAC, and (2) ZOC'B' = z.OBC.
And ^B'CA' = ^OCA'OCB'=^OAC^OBC
= z.ACB=n,
since ACB is in a semicircle.
.*. as C describes S, its inverse, C, describes the circle S'
on A'B' as diameter. q.e.d.
OF INVERSION AND INVERSE FIGURES. 21 3
Cor. I. The point O is the intersection of common direct
tangents.
Cor. 2. •.• CD'. OD = OT. OT'. ^^.^ = RK ?^',
where R is the radius of the circle of inversion ;
.*. the centre of a circle and the centre of its inverse are not
inverse points, unless OD = OT, i.e., unless the centre of
inversion is at 00 .
Cor. 3. When the circle to be inverted cuts the circle of
inversion, its inverse cuts the circle of inversion in the same
points. (256°, Cor. 5)
261°. Theorem. — A circle which passes through the centre
of inversion inverts into a line.
Let O be the centre of inversion
and S the circle to be inverted, and
let P and P' be inverses of Q and Q'.
Proof.— '$AnQ.Q
OP.OO = OP'. 00' = R2,
OP:OP' = OQ':OQ,
and the triangles OPP' and OQ'Q
are similar, and ^OPP' = Z.OQ'Q = "~> since OQ'Q is in a
semicircle. And as this is true however OP' be drawn, PP'
is a line ± to OP, the common centreline of the circle of
inversion and the circle to be inverted. q.e.d.
Cor. I. Since inversion is a reciprocal process, the
inverse of a line is a circle through the centre of inversion
and so situated that the line is J_ to the common centre
line of the two circles.
Cor. 2. Let I be the circle of inversion, and let PT and PT'
be tangents to circles I and S respectively. Then,
PT2 = OP2OT2=OP2OP.OQ = OP.PQ=PT'2,
PT = Pr,
.*. when a circle inverts into a line with respect to another
circle, the line is the radical axis of the two circles. (178°, Def.)
214
SYNTHETIC GEOMETRY.
Cor. 3. If a circle passes through the centre of inversion
and cuts the circle of inversion, its inverse is their common
chord.
Cor. 4. A centreline is its own inverse.
Cor. 5. Considering the centre of inversion as a pointcircle,
its inverse is the line at 00 .
262°. A circle which cuts the circle of inversion orthogon
ally inverts into itself.
Since circle S cuts circle I orthogon
ally OT is a tangent to S, and hence
OP.OO = OT2,
.'. P inverts into Q and O into P, and
the arc TQV inverts into TPV and vice
versa. q.e.d.
Cor. Since I cuts S orthogonally, it is evident that I inverts
into itself with respect to S.
263°. A circle, its inverse, and the circle of inversion have
t^ a common radical axis.
Let I be the circle of inver
'^^^^^ I sion, and let the circle S' be
the inverse of S.
o^^ \J I \ The tangents TT' and VV
v^^^~ff_ X meet at O (260°, Cor. i), and
T, T' are inverse points. D,
the middle point of TT' is on the radical axis of S and S',
and the circle with centre at D and radius DT cuts S and S'
orthogonally. But this circle also cuts circle I orthogonally
(258°). .*. D is on the radical axes of I, S and S'.
Similarly D', the middle point of the tangent VV, is on the
radical axes of I, S and S'.
.*. the three circles I, S, and S' have a common radical
axis passing through D and D'. q.e.d.
OF INVERSION AND INVERSE FIGURES. 21 5
Remarks. — This is proved more simply by supposing one
of the circles to cut the circle of inversion. Then its inverse
must cut the circle of inversion in the same points, and the
common chord is the common radical axis.
The extension to cases of nonintersection follows from the
law of continuity.
264°. Theorem. — The angle of intersection of two lines or
circles is not changed in magnitude by inversion.
Let O be the centre of in
version, and let P be the point
of intersection of two circles S
and S', and Q its inverse.
Take R and T points near P, o
and let U and V be their in
verses. Then
0U.0R = 0Q.0P = 0V.0T = R2,
AOQU^AORP,
and ^OQU = ^ORP = ^RPX^ROP.
Similarly ^OQV = ^TPX^TOP,
^UQV = ^RPT^ROT.
But at the limit when R and T come to P the angle between
the chords RP and PT becomes the angle between the circles
(115°, Def. I ; 109°, Def. i). And, since ^ROT then vanishes,
we have ultimately ^UQV = I.RPT.
Therefore S and S', and their inverses Z and Z', intersect at
the same angle.
Cor. I. If two circles or a line and a circle touch one
another their inverses also touch one another.
Cor. 2. If a circle inverts into a line, its centrelines invert
into circles having that line as a common diameter. For,
since the circle cuts its centrelines orthogonally, their in
verses must cut orthogonally. But the centreline is the only
line cutting a circle orthogonally.
2l6
SYNTHETIC GEOMETRY.
Exercises.
1. What is the result of inverting a triangle with respect to
its incircle ?
2. The circle of selfinversion of a given circle cuts it or
thogonally.
3. Two circles intersect in P and O, and AB is their common
centreline. What relation holds between the various
parts when inverted with P as centre of inversion ?
4. A circle cuts two circles orthogonally. Invert the system
into two circles and their common centreline.
5. Three circles cut each other orthogonally. If two be in
verted into lines, their intersection is the centre of the
third.
265°. The two following examples are important.
Ex. I. Any two circles cut their common centreline, and a
circle which cuts them orthogonally in two sets of points
which connect concurrently on the lastnamed circle.
S and S' are the
two given circles and
Z a circle cutting
them orthogonally.
Invert S and S'
and their common
centre line with re
spect to a circle
which cuts S and S' orthogonally and has its centre at some
point O on Z. S and S' invert into themselves, and their
centreline into a circle through O cutting S and S' ortho
gonally, i.e.^ into circle Z.
.•. A' is the inverse of A, B' of B, etc , and the points AA',
BB', CC, DD' connect concurrently at O.
Ex. 2. The ninepoints circle of a triangle touches the in
circle and the excircles of the triangle.
OF INVERSION AND INVERSE FIGURES. 21/
Let ABC be any triangle having its side AB touched by
the incircle I at b h ^
T, and by the ex
circle to the side
c at T'. Take
CH = CAand CD
= CB, and join DH
and HA.
From the sym
metry of the fig
ure it is evident
that HD touches
both the circles I and S. Let E and F be the middle points
of AB and AC, and let EF cut HA in G.
From 135°, Ex. i, AT = BT'=jrt;,
whence ET = ET' = (^/^).
But, since EF bisects HA, EG = ^BH = (^^),
ET = ET' = EG,
and the circle with E as centre and EG as radius cuts I and
S orthogonally, and, with respect to this circle, the circles I
and 8 invert into themselves.
Now, PF :HC = DF :DC = BCCF:BC,
PF = ii^(BCCF) = ^— ,
BC 2a
EP = EFPF = ^j^,
2 2a
EP.EF = f''^h— ) = K^^)2 = EG2.
^2 2a'2
.'. P inverts into F, and the line HD into the circle through
E and F, and by symmetry, through the middle point of BC.
But this is the ninepoints circle (116°, Ex. 6). And since
HD touches I and S, the ninepoints circle, which is the
inverse of HD, touches the inverses of I and S, t.e., I and S
themselves.
And, similarly, the ninepoints circle touches the two
remaining excircles.
2l8 SYNTHETIC GEOMETRY.
SECTION V.
OF POLE AND POLAR.
266°. Def. — The line through one of a pair of inverse points
perpendicular to their axis is the polat' of the other point with
respect to the circle of inversion, and the point is the pole of
the line.
The circle is called, in this relation, Xh^ polar circle, and its
centre is called the polar cejitre.
From this definition and from the nature of inverse points
we readily obtain the following : —
Cor. I. The polar of the polar centre is a line at infinity.
But, since the point which is the inverse of the centre may go
to 00 along any centreline, all the hnes obtained therefrom
are polars of the centre. And as a point has in general but
o/te polar with respect to any o?ie circle, we speak of the polar
of the centre as being the line at infinity, thus assuming that
there is but one line at infinity.
Cor. 2. The polar of any point on the circle is the tangent
at that point ; or, a tangent to the polar circle is the polar to
the point of contact.
Cor. 3. The pole of any line lies on that centreline of the
polar circle which is perpendicular to the former line.
Cor. 4. The pole of a centreline of the polar circle lies
at 00 on the centreline which is perpendicular to the
former.
Cor. 5. The angle between the polars of two points is
equal to the angle subtended by these points at the polar
centre.
267°. Theorem. — If P and Q be two points, and P lies on
the polar of Q, then O lies on the polar of P.
OF POLE AND POLAR. 219
OP and OQ are centrelines of the polar circle
and PE, ± to OQ, is the polar of Q.
To prove that QD, ± to OP, is the
polar of P.
Proof.— The As ODQ and OEP
are similar.
OE:OP = OD:OQ,
and.. OE.OQ = OP.OD.
But E and Q are inverse points with
respect to circle I,
P and D are inverse points,
and .'. DQ is the polar of P.
Def. — Points so related in position that each lies upon the
polar of the other are conjugate points, and lines so related
that each passes through the pole of the other are conjugate
lines.
Thus P and Q are conjugate points and L and M are con
jugate lines.
Cor. I. If Q and, accordingly, its polar PV remain fixed
while P moves along PE, L, which is the polar of P, will
rotate about Q, becoming tangent to the circle when P comes
to U or V, and cutting the circle when P passes without.
Similarly, if Q moves along L, M will rotate about the
point P.
Cor. 2. As L will touch the circle at U and V, UV is the
chord of contact for the point Q.
.*. for any point without a circle its chord of contact is its
polar.
Cor. 3. For every position of P on the line M, its polar
passes through Q.
.'. collinear points have their polars concurrent, and con
current lines have their poles collinear, the point of concur
rence being the pole of the line of collinearity.
220 SYNTHETIC GEOMETKY.
Exercises.
1. Given a point and a line to find a circle to which they are
pole and polar.
2. In Ex, I the circle is to pass through a given point.
3. In the figure of 267° trace the changes,
(a) when P goes to 00 along M ;
(6) when P goes to 00 along OD ;
(c) when P moves along UV, what is the locus of D ?
4. From any point on a circle any number of chords are
drawn, show that their poles all lie on the tangent at
the point.
5. On a tangent to a circle any number of points are taken,
show that all their polars with respect to the circle pass
through the point of contact.
268°. Theorem. — The point of intersection of the polars of
two points is the pole of the join of the points.
A Let the polars of B and of C pass through A.
Then A lies on the polar of B, and therefore B
lies on the polar of A (267°). For similar
B . . c reasons C lies on the polar of A.
.*. the polar of A passes through B and C and is their join.
q.e.d.
Cor. Let two polygons ABCD... and abc... be so situated
that a is the pole of AB, b of BC, c
of CD, etc.
Then, since the polars of a and b
meet at B, B is the pole of ab ;
similarly C is the pole of hc^ etc.
kE .'.if two polygons are such that
the vertices of one are poles of the sides of the other, then,
reciprocally, the vertices of the second polygon are poles of
the sides of the first, the polar circle being the same in
each case.
OF POLE AND POLAR.
221
Def. I. Polygons related as in the preceding corollary are
polar reciprocals to one another.
Def. 2. — When two polar reciprocal As become coincident,
the resulting A is selfreciprocal or selfconjngate^ each vertex
being the pole of the opposite side.
I^^f' 3 — The centre of the with respect to which a A is
selfreciprocal is the polar centre of the A? and the O itself
is the polar circle of the A
269°. The orthocentre of a triangle is its polar centre
. Let ABC be a selfconjugate A a
Then A is the pole of BC, and B of
AC, and C of AB.
Let AX, J_ to BC, and BY, ± to
AC, meet in O. Then O is the ortho
centre. (88°, Def)
Now, as AX is \_ to BC, and as A
is the pole of BC, the polar centre
lies on AX. For similar reasons it
lies on BY. (266°, Cor. 3)
.*. O is the polar centre of the AABC. q.e.d.
Cor. I. With respect to the polar of the A? the on
AO as diameter inverts into a line _L to AO (261°). And as
A and X are inverse points, this line passes through X ;
therefore BC is the inverse of the on OA as diameter.
Similarly, AC is the inverse of the on OB as diameter,
and AB of the on OC as diameter.
Cor. 2. As the on AO inverts into BC, the point D is
inverse to itself, and is on the polar of the A (256°, 5)
.'. OD is the polar radius of the A
Cor. 3. If O falls within the A^ it is evident that the on
OA as diameter will not cut CB. In this case the polar
centre is real while the polar radius is imaginary. (257°, Cor.)
Hence a A which has a real polar circle must be obtuse
angled.
222 SYNTHETIC GEOMETRY.
Cor. 4. The on BC as diameter passes through Y since
Y is a ~~1
But B and Y are inverse points to the polar 0.
.*. the polar cuts orthogonally the on BC as diameter.
(258°)
Similarly for the circles on CA and AB.
.'. the polar of a A cuts orthogonally the circles having
the three sides as diameters.
Cor. 5. The^A0Z=^B,^BOZ = zA, and ^OAC=^(C)
And CX = OC sin AOZ = OC sin B, also =^cosC,
0C=^ — — .cosC= ^cos C,
sm B
where d is the diameter of the circumcircle (228°) to the
triangles AOC or BOC or AOB or ABC, these being all
equal. (116°, Ex. 4)
Similarly OA = ^cosA, OB = ^cosB.
But OX = OCcosB=^cosBcosC,
R2 = OX. OA=^2cosAcosBcosC.
In order that the righthand member may be +, one of the
angles must be obtuse.
Cor. 6. R2 = OC.OZ = OC(OC + CZ) = OC2 + OC.CZ,
and OC=<^cosC, and CZ = ^sinB = ^, (228°)
R2=^2(i_sin2C)rt^cosC
= ^2_^(^2 + ^2 + ^2). (217°)
If O is within the triangle, d"^ < \{a^ ^ b'^ ■\ c^) and R is
imaginary.
Exercises.
1. If two triangles be polar reciprocals, the inverse of a side
of one passes through a vertex of the other.
2. A rightangled triangle has its rightangled vertex at the
centre of a polar circle. What is its polar reciprocal .'*
3. In Fig. of 269°, if the polar circle cuts CY produced in C,
prove that CY=YC'.
OF POLE AND POLAR. 223
4. If P be any point, ABC a triangle, and A'B'C its polar
reciprocal with respect to a polar centre O, the per
pendiculars from O on the joins PA, PB, and PC
intersect the sides of A'B'C coUinearly.
270°. Theore77t.—\i two circles intersect orthogonally, the
endpoints of any diameter of either are conjugate points
with respect to the other.
Let the circles S and S ' in
tersect orthogonally, and let
PQ be a diameter of circle S'.
Then P' is inverse to P, and
P'Q is ± to CP.
.'. P'Q is the polar of P with
respect to circle S.
.'. Q lies on the polar of P, and hence P lies on the polar
of O, and P and Q are conjugate points (267° and Def.).
q.e.d.
Cor. I. PQ2 = CP2 + CQ22CP.CP' (172°, 2)
= CP2 + C022R2
= CP2R2 + CQ2R2 = T2 + T'2,
where T and T' are tangents from P and Q to the circle S.
.*. the square on the join of two conjugate points is equal
to the sum of the squares on the tangents from these points
to the polar circle.
Cor. 2. If a circle be orthogonal to any number of other
circles, the endpoints of any diameter of the first are conju
gate points with respect to all the others. And when two
points are conjugate to a number of circles the polars of either
point with respect to all the circles pass through the other
point.
271°. Theorhn. — The distances of any two points from a
polar centre are proportional to the distances of each point
from the polar of the other with respect to that centre.
(Salmon)
224
SYNTHETIC GEOMETRY.
NN' is the polar of P and MM' is the polar of Q.
PO:00 = PM:ON.
Then
MM'
Proof.— h^t Om be 1 to
and O;/ be II to NN'. Then
OP'. OP = OQ'. OQ
OP_OQ'_M;;^
OQ OP' N;^"
But the triangles OP;;z and 00;/
are similar,
OQ Q« N;2
_ P^+M;;g _PM ,
Q;/ + N« ON" ^
Cor. I. A, B are any two points and L and M their polars,
and P the point of contact of any tangent N.
AX and BY are _L upon N, and PH and PK are J_ upon L
and M respectively. Then
M = ?2and AX_AO
PK R PH R'
BY. AX AO.BO ;_ ^,^„,
PK.pH = ^^=^'^^°^^''"''
AX.BY = ^.PH.PK.
If A and B are on the circle, L and M become tangents
having A and B as points of contact, and AO = BO = R.
AX.BY = PH.PK. (See 211°, Ex. i)
Exercises.
If P and Q be the endpoints of any diameter of the polar
circle of the AABC, the chords of contact of the point
P with respect to the circles on AB, BC, and CA as
diameters all pass through Q.
Two polar reciprocal triangles have their corresponding
vertices joined. Of what points are these joins the
polars ?
OF POLE AND POLAR.
225
A, B, C are the vertices of a triangle and L, M, N the
corresponding sides of its reciprocal polar. If T be a
tangent at any point P, and AT is ± to T, etc.,
AT.BT.CT AG. BO. CO
PL.PM.PN = R3 =^ '^"''^^'
If A, B, C are on the circle,
AT . BT . CT = PL . PM . PN.
In Ex. 3, if A', B', C be the vertices of the polar reciprocal,
AT . BT . CT _ A^O . B^O . CO
AT.BT.CT R3
The righthand expression is independent of the position
ofT.
If ABC, A'B'C be polar reciprocal triangles whose sides
are respectively L, M, N and L', M', N', and if AM' is
the ± from A to M', etc.,
AM'. BN'. CL' = AN'. BL'. CM',
and A'M . B'N . C'L = A'N . B'L . CM.
etc.,
272°. Theorem. — Triangles which are polar reciprocals to
one another are in perspective. ^'
Let ABC and A'B'C be polar recipro
cals. Let AP, AP' be perpendiculars on ^
A'B' and A'C, BQ and BO' be perpen
diculars on B'C and B'A', etc.
Then (271°) AP'_AO BQ'^BO
men (271) ^^, CR CO
AP'. BQ'. CR' _
AP.B(2.CR '
But AP' = AA'sinAA'P', AP = AA'sin AA'P', .
. AP'^sinAA'P'
AP sin AA'P'
and similarly for the other ratios. Hence AA', BB', CC
divide the angles at A, B, and C, so as to fulfil the criterion
of 251°.
.*. AA', BB', and CC are concurrent, and the triangles are
in perspective. (254°, Cor. 2)
P
226 SYiNTHETIC GEOMETRY.
SECTION VI.
OF THE RADICAL AXIS.
273°. Dcf. I. — The line perpendicular to the common
centreline of two circles, and dividing the distance be
tween the centres into parts such that the difference of
their squares is equal to the difference of the squares on the
conterminous radii, is the radical axis of the two circles.
Cor. I. When two circles intersect, their radical axis is the
secant line through the points of intersection.
Cor. 2. When two circles touch, their radical axis is the
common tangent at their point of contact.
Cor. 3. When two circles are mutually exclusive without
contact, their radical axis lies between them.
Cor. 4. When two circles are equal and concentric, their
radical axis is any line whatever, and when unequal and con
centric it is the line at 00 .
Def. 2. — When three or more circles have a common
radical axis they are said to be coaxal.
274°. If several circles pass through the same two points
they form a coaxal system.
For (273°, Cor. i) the line through the points is the radical
axis of all the circles taken in pairs, and is therefore the
common radical axis of the system.
Such circles are called circles of the common point species^
contracted to f./. circles.
Let a system of ^./.circles S, S^, S2, ..., pass through the
common points P and O, and let L'L be the right bisector
ofPO.
Then the centres of all the circles of the f./.system lie
on L'L and have M'M for their common radical axis.
OF THE RADICAL AXIS.
227
Hence from any point C in M'M the tangents to all the
circles are equal to one another. (178°)
Let CT be one of these tangents. The circle Z with C as
centre and CT as radius cuts all the <;./. circles orthogonally.
Similarly, a system of circles Z, Zj, Zg, ... may be found
with centres lying on M'M such that each one of the system
cuts orthogonally every one of the r./. circles.
Since the centre of any circle of this new system is obtained
by drawing a tangent from any one of the circles, as Sg, of
the <:.^. species, to meet MM', it follows that no circle of this
new system can have its centre lying between P and Q. As
T approaches P the dependent circle Z contracts until it be
comes the pointcircle P, when T comes to coincidence with P.
Hence P and O are limiting forms of the circles having
their centres on M'M and cutting the c.p.chdes orthogon
ally. The circles of this second system are consequently
called limithig point circles^ contracted to /./.circles.
From the way in which /./.circles are obtained we see that
from any point on L'L tangents to circles of the /./.system
are all equal, and hence that L'L is the radical axis of the
/./.circles. Thus the two systems of circles have their radi
228 SYNTHETIC GEOMETRY.
cal axes perpendicular, and every circle of one system cuts
every circle of the other system orthogonally.
Hence P and Q are inverse points with respect to every
circle of the /./.system, and with respect to any circle of
either system all the circles of the other system invert into
themselves.
If P and Q approach L, the c.p.cirdes separate and the
/./.circles approach, and when P and Q coincide at L the
circles of both species pass through a common point, and the
two radical axes become the common tangents to the respec
tive systems.
If this change is continued in the same direction, P and Q
become imaginary, and two new limiting points appear on
the line L'L, so that the former /./.circles become ^./.circles,
and the former ^./.circles become /./.circles.
Thus, in the systems under consideration, two limiting
points are always real and two imaginary, except when they
all become real by becoming coincident at L.
Cor. I. As the r./.circles and the /^.circles cut each other
orthogonally, the endpoints of a diameter of any circle of one
species are conjugate points with respect to every circle of
the other species. But a circle of either species may be found
to pass through any given point (259°, Ex. 5). .'. the polars
of a given point with respect to all the circles of either species
are concurrent.
Cor. 2. Conversely, if the polars of a variable point P with
respect to three circles are concurrent, the locus of the point
is a circle which cuts them all orthogonally.
For let Q be the point of concurrence. Then P and O are
conjugate points with respect to each of the circles. Hence
the circle on PQ as diameter cuts each of the circles ortho
gonally. (270°)
Cor. 3. If a system of circles is cut orthogonally by two
circles, the system is coaxal.
For the centres of the cutting circles must be on the radical
OF THE RADICAL AXIS.
229
axis of all of the other circles taken in pairs ; therefore they
have a common radical axis.
Cor. 4. If two circles cut two other circles orthogonally, the
common centreline of either pair is the radical axis of the
other pair.
Cor. 5. Two /./.circles being given, a circle of any required
magnitude can be found coaxal with them. But if the circles
be of the ^./.species no circle can be coaxal with them whose
diameter is less than the distance between the points.
Exercises.
1. Given two circles of the /./.species to find a circle with a
given radius to be coaxal with them.
2. Given two circles of either species to find a circle to pass
through a given point and be coaxal with them.
3. To find a point upon a given line or circle such that tan
gents from it to a given circle may be equal to its
distance from a given point.
4. To find a point whose distances from two fixed points may
be equal to tangents from it to two fixed circles.
275°. Theorem. — The difference of the squares on the tan
gents from any point to two circles is equal to twice the
rectangle on the distance between the centres of the circles
and the distance of the point from their radical axis.
Let P be the point, S and
S' the circles, and LI their
radical axis. Let PQ be J_
to AB.
PT^ — PT'^
= PA2PB2(r2r'^),
where r, r' are radii of S
and S'.
But, 273°, Def I,
r2r'2 = AI=^
230 SYNTHETIC GEOMETRY.
and PA'^PB2 = AQ2QB2, (172°, i)
PT2PT'2 = AQ2QB2(AI2IB2)
= AB(AQ  QB)  AB(AI  IB)
= 2AB.IQ = 2AB.PL. q.e.d.
This relation is fundamental in the theory of the radical axis.
Cor. I. When P is on the radical axis PL=o, and the
tangents are equal, and when P is not on the radical axis the
tangents are not equal.
Cor. 2. The radical axis bisects all common tangents to the
two circles.
Cor. 3. If P lies on the circle S', PT' = o, and
PT2 = 2AB.PL,
.*. the square of the tangent from any point on one circle to
another circle varies as the distance of the point from the
radical axis of the circles.
Cor. 4. If C is the centre of a circle S" passing through P
and coaxal with S', PT2 = 2AC . PL.
Now, if P could at any time leave this circle we would have
PT2_pr'2 = 2Ac.pl,
where PT" is the tangent from P to the circle S"
p'r2_p'p2_ p'p"2
which is impossible finless PT"=o.
Hence the locus of a point, which so moves that the square
on the tangent from it to a given circle varies as the distance
of the point from a given line, is a circle, and the line is the
radical axis of this circle and the given circle.
Cor. 5. Let PT' = ^ . PT, where ^ is a constant. Then
PT2Pr = (i/&2)px2 = 2Ab.pl,
pp2 2AB. PL
As PT2 varies as PL, P lies on a circle coaxal with S and S'.
.'. the locus of a point from which tangents to two given
circles are in a constant ratio is a circle coaxal with the two.
OF THE RADICAL AXIS. 23 1
Exercises.
1. In Cor. 5 what is the position Y)f the locus for k = o^ /C'=t,
^ = > I, /& negative?
2. What is the locus of a point whose distances from two
fixed points are in a constant ratio t
3. P and Q are inverse points to the circle I, and a line
through P cuts circle I in A and B. PQ is the internal
or external bisector of the ^AOB, according as P is
within or without the circle.
4. P, Q are the limiting points of the //^.circles S and S', and
a tangent to S' at T cuts S p^^ ^g
in A and B.
Then, considering P as a
pointcircle, tangents from
any point on S to P and S'
are in a constant ratio.
.. AP:AT = BP:BT, and PT is the external bi
sector of lAPB. If S' were enclosed by S, BT would
be an internal bisector.
5. The points of contact of a common tangent to two l.p.
circles subtend a right angle at either limiting point.
276°. Theorejn. — The radical axes of three circles taken in
pairs are concurrent.
Let Si, S2, S3 denote the circles, and let L be the radical
axis of Si and S2, M of S2 and S3, and N of S3 and S^
L and M meet at some point O, from which 0Ti = 0T2,
and OT2=OT3, where OT^ is the tangent from O to S^, etc.,
0Ti = 0T3, and O is on N,
.*. L, M, and N are concurrent at O.
Def. — The point of concurrence of the three radical axes of
three circles taken in pairs is called the radical centre of the
circles.
Cor. I. If Si, S2 are cut by a third circle Z, the common
chords of Si, Z and S2, Z intersect on the radical axis of Si
and So.
232 SYNTHETIC GEOMETRY.
Hence to find the radical axis of two given circles S^ and
S2, draw any two circles Z and Z^ cutting the given circles.
The chords Sj, Z and 83, Z give one point on the radical axis
and the chords S^, Zj and S2, Z^ give a second point.
Cor. 2. If three circles intersect each other, their three
common chords are concurrent. (See 249°, Ex. i.)
Cor. 3. If a circle touches two others, the tangents at the
points of contact meet upon the radical axis of the two.
Cor. 4. If a circle cuts three circles orthogonally, its centre
is at their radical centre and its radius is the tangent from
the radical centre to any one of them.
Cor. 5. If in Cor. 4 the three circles are coaxal, any num
ber, of circles may be found to cut them orthogonally, and
hence they have no definite radical centre, as any point upon
the common radical axis of the three becomes a radical
centre.
Cor. 6. If in Cor. 4 the three circles mutually intersect one
another, the radical centre is within each circle (Cor. 2), and
no tangent can be drawn from the radical centre to any one
of the circles. In this case the circle which cuts them all or
thogonally has a real centre but an imaginary radius.
277°. Theorem. — li ^.ny three lines be drawn from the ver
tices of a A to the opposite sides, the polar centre of the A
is the radical centre of the circles having these lines as
A R z B diameters.
ABC is a A and O its ortho
centre, and AP, BQ, CR are lines
from the vertices to the opposite
sides.
Q •.• ^BXA=n,
o the circle on AP as diameter
passes through X, and OX. OA is equal to the square on the
tangent from O to the circle on AP.
OF THE RADICAL AXIS. 233
Similarly OY. OB is the square of the tangent from O to
the circle on BQ as diameter, and similarly for OZ.OC.
But as O is the polar centre of AABC, (269°)
OX. OA = OY. OB = OZ. OC
.*. the tangents from O to the three circles on AP, BQ, and
CR are equal, and O is their radical centre. q.e.d.
Cor. I. Let P, Q, R be collinear.
Then the polar centre of AABC is the radical centre of
circles on AP, BQ, and CR as diameters.
Again, in the AAQR AP, QB, and RC are lines from the
vertices to the opposite sides.
.'. the polar centre of AAQR is the radical centre of circles
on AP, BQ, and CR as diameters.
Similarly the polar centres of the As BPR and CPQ are
radical centres to the same three circles.
But these As have not a common polar centre, as is readily
seen. Hence the same three circles have four different
radical centres. And this is possible only when the circles
are coaxal. (276°, Cor. 5)
.*. the circles on AP, BQ, and CR are coaxal.
.'. if any three collinear points upon the sides of a A be
joined with the opposite vertices, the circles on these joins as
diameters are coaxal.
Cor. 2. Since ARPC is a quadrangle or tetragram (247°,
Def. 2), and AP, BQ, CR are its three diagonals,
.'. the circles on the three diagonals of any quadrangle are
coaxal.
Cor. 3. The middle points of AP, BQ, and CR are col
linear. But ARPC is a quadrangle of which AP and CR are
internal diagonals, and BQ the external diagonal.
.*. the middle points of the diagonals of a complete quad
rangle, or tetragram, are collinear. (See 248°, Ex. 2)
Cor. 4. The four polar centres of the four triangles deter
mined by the sides of a tetragram taken in threes are collinear
234 SYNTHETIC GEOMETRY.
and lie upon the common radical axis of the three circles
having the diagonals of the tetragram as diameters.
278°. Theorem. — In general a system of coaxal circles
inverts into a coaxal system of the same species.
(i.) Let the circles be of the r./. species.
The common points become two points by inversion, and
the inverses of all the circles pass through them. Therefore
the inverted system is one of ^./.circles.
Cor: I. The axis of the system (LL' of Fig. to 274") inverts
into a circle through the centre of inversion (261°, Cor. i), and
as all the inverted circles cut this orthogonally, the axis of
the system and the two common points invert into a circle
through the centre and a pair of inverse points to it.
(258°, Conv.)
Cor. 2. If one of the common points be taken as the centre
of inversion, its inverse is at 00 .
The axis of the system then inverts into a circle through
the centre of inversion, and having the inverse of the other
common point as its centre, and all the circles of the system
invert into centrelines to this circle.
(2.) Let the circles be of the /./.species.
Let the circles S and S' pass through the limiting points,
and be thus ^./.circles.
Generally S and S' invert into circles which cut the in
verses of all the other circles orthogonally. (264°)
.'. the intersections of the inverses of S and S' are limiting
points, and the inverted system is of the /./.species.
Cor. 3. The axis of the system (MM' of Fig. to 274°) be
comes a circle through the centre and passing through the
limiting points of the inverted system, thus becoming one of
the r./. circles of the system.
Cor. 4. If one of the limiting points be made the centre of
OF THE RADICAL AXIS.
235
inversion, the circles S and S' become centrelines, and the
/./.circles become concentric circles.
Hence concentric circles are coaxal, their radical axis
being at 00 .
Exercises.
1. What does the radical axis of (i, 278^^) become?
2. What does the radical axis of (2, 278°) become ?
3. How would you invert a system of concentric circles into
a common system of /./.circles ?
4. How would you invert a pencil of rays into a system of
^./.circles.
5. The circles of 277° are common point circles.
279". Theorem. — Any two circles can be inverted into
equal circles.
Let S, S' be the circles
having radii r and r', and
let C, C be the equal
circles into which S and
S' are to be inverted ; and
let the common radius be p.
But, since P and Q and also P' and Q' are inverse points,
OP.OO = OP'.OQ',
OQ2 r , ,
and (275°, Cor. 5) O lies on a circle coaxal with S and S'.
And with any point on this circle as a centre of inversion S
and S' invert into equal circles.
Cor. I. Any three noncoaxal circles can be inverted into
equal circles.
\l6
SYNTHETIC GEOMETRY.
For, let the circles be S, S', S", and let Z denote the locus
of O for which S and S' invert into equal circles, and Z' the
locus of O for which S and S" invert into equal circles. Then
Z' is coaxal with S and S". And, as S, S', and S" are not
coaxal, Z and Z' intersect in two points, with either of which
as centre of inversion the three given circles can be inverted
into equal circles.
Cor. 2. If S, S', and S" be /./.circles, Z and Z' being
coaxal with them cannot intersect, and no centre exists with
which the three given circles can be inverted into equal circles.
But if S, S' and S" be ^/.circles, Z and Z' intersect in the
common points, and the given circles invert into centrelines
of the circle of inversion, and having each an infinite radius
these circles may be considered as being equal. (278°, Cor. 2)
Cor. 3. In general a circle can readily be found to touch
three equal circles. Hence by inverting a system of three
circles into equal circles, drawing a circle to touch the three,
and then reinverting we obtain a circle which touches three
given circles.
If the three circles are coaxal, no circle can be found to
touch the three.
280°. Let the circles S and S', with centres A and B and
radii r and r, be cut
by the circle Z with
centre at O and radius
OP = R. Let NLbe
the radical axis of S
and S'.
Since AP is J. to
the tangent at P to
the circle S, and OP
is JL to the tangent
at P to the circle Z,
the ^APO = ^ is the angle of intersection of the circles S
OF THE RADICAL AXIS. 53/
and Z (115°, Def. i). Similarly BQO0 is the angle of
intersection of the circles S' and Z. Now
PP' = 2rcos^ = ROP',
and QQ' = 2r'cos0 = ROQ',
OP'  OQ' = 2(r'cos  r cos ^).
But R.OP'R.OQ' = OT2Or2 (where OT is the
tangent from O to S, etc.) =2AB . OL, (275°)
R = _ ^ .OL.
rcos<i>rcosd
Cor. I. When 6 and <p are constant, R varies as OL.
.*. a variable circle which cuts two circles at constant
angles has its radius varying as the distance of its centre from
the radical axis of the circles.
Cor. 2. Under the conditions of Cor. i ON varies as OL,
and .*. ^, is constant.
ON
.•.a variable circle which cuts two circles at a constant
angle cuts their radical axis at a constant angle.
Cor. 3. When 0L = o, /coscp^rcosd,
and r : r' = cos0 : cos^.
.'. a circle with its centre on the radical axis of two other
circles cuts them at angles whose cosines are inversely as the
radii of the circles.
Cor. 4. If circle Z touches S and S', d and <p are both zero
ot both equal to tt, or one is zero and the other is tt.
AB
.'. when Z touches S and S' R= . , . OL, where the
variation in sign gives the four possible varieties of contact.
Cor. 5. When d = <f) = 2' Z cuts S and S' orthogonally, and
OL=o, and the centre of the cutting circle is on the radical
axis of the two.
238
SYNTHETIC GEOMETRY.
SECTION VII.
CENTRE AND AXES OF SIMILITUDE OR
PERSPECTIVE.
The relations of two triangles in perspective have been
given in Art. 254°. We here propose to extend these rela
tions to the polygon and the circle.
Let O, any point, be connected Avith the vertices A,
B, C, ... of a polygon, and on
OA, OB, OC, ... let points a, b,
c, ... be taken so that
OA : 0^ = 0B : Ol?=OC : Or...
and
OA : 0^' = 0B : 0^' = 0C : O^:'...
Then, since OAB is a A and ab
is so drawn as to divide the sides
proportionally in the same order,
.. ab is II to AB. (202°, Conv.)
Similarly,
be is II to BC, cd to CD, etc.,
b'c' is II to BC, (fd' to CD, etc.,
AO AB ^ /\Oab « /Spa'b',
AOl^C^ /\Obc^ AOb'c', ...
.*. the polygons ABC..., abc..., and a'b'c'... are all similar
and have their homologous sides parallel.
Def.—The. polygons ABCD... and abed... are said to be
similarly placed, and O is their ^aY^r;/^?/ centre of similitude;
while the polygons ABCD... and a'b'e'd'... are oppositely
placed, and O is their internal centre of similitude.
Hence, when the lines joining any point to the vertices of
a polygon are all divided in the same manner and in the
same order, the points of division are the vertices of a second
similarly,
and
CENTRE OF SIMILITUDE OR PERSPECTIVE. 239
polygon similar to the original, and so placed that the
homologous sides of the two polygons are parallel.
282°. When two similar polygons are so placed as to have
their homologous sides parallel, they are in perspective, and
the joins of corresponding vertices concur at a centre of
similitude.
Let ABCD..., abed... be the polygons.
Since they are similar, AB : ab=^^C : bc=CD : cd... (207"),
and by hypothesis AB is  to ab, BC to bc^ etc.
Let ha and ^b meet at some point O.
Then GAB is a A and ab\s\\ to AB.
OB_AB_BC_ .^
Wb ab be '
.". C<; passes through O, and similarly D^^ passes through O,
etc.
By writing alb'd ... for abe... the theorem is proved for the
polygon a'b'e'd', which is oppositely placed to ABCD...
Cor. I. If Aa and V>b meet at 00, <?^ = AB, and hence
^(r=BC, etc., and the polygons are congruent.
Cor. 2. The joins of any two corresponding vertices as A,
C ; a, c ', a\ e' are evidently homologous lines in the polygons
and are parallel.
Similarly any line through the centre O, as XxOx' is
homologous for the polygons and divides them similarly.
283°. Let the polygon ABCD... have its sides indefinitely
increased in number and diminished in length. Its limiting
form (148°) is some curve upon which its vertices lie. A
similar curve is the limiting form of the polygons abed... as
also oi a'b'c'd'..., since every corresponding pair of limiting or
vanishing elements are similar.
Hence, if two points en a variable radius vector have the
ratio of their distances from the pole constant, the loci of the
240 • SYNTHETIC GEOMETRY.
points are similar curves in perspective, and having the pole
as a centre of perspective or simihtude.
Cor. I. In the limiting form of the polygons, the line BC
becomes a tangent at B, and the line be becomes a tangent at
b. And similarly for the line be'.
.'. the tangents at homologous points on any two curves in
perspective are parallel.
284°. Since abed. . . and a'b'e'd'. . . are both in perspective with
ABCD... and similar to it, we see that two similar polygons
may be placed in two different relative positions so as to be
in perspective, that is, they may be similarly placed or oppo
sitely placed.
In a reguLir polygon of an even number of sides no dis
tinction can be made between these two positions ; or, two
similar regular polygons are both similarly and oppositely
placed at the same time when so placed as to be in per
spective.
Hence two regular polygons of an even number of sides
and of the same species, when so placed as to have their sides
respectively parallel, have two centres of perspective, one due
to the polygons being similarly placed, the external centre ;
and the other due to the polygons being oppositely placed,
the internal centre.
Cor. Since the limiting form of a regular polygon is a circle
(148°), two circles are always similarly and oppositely placed
at the same time, and accordingly have always two centres of
perspective or similitude.
285°. Let S and S' be two circles with centres C, C and
radii r, r' respectively, and let O and O' be their centres of
perspective or similitude.
Let a secant line through O cut S in X and Y, and S' in X'
and Y'.
CENTRE OF SIMILITUDE OR PERSPECTIVE. 241
Then O is the centre of similitude due to considering the
circles S and S' as being
similarly placed.
Hence X and X', as also
Y and Y', are homologous
points, and (283°, Cor. i)
the tangents at X and X'
are parallel. So also the
tangents at Y and Y' are
parallel.
Again O' is the centre
of similitude due to con
sidering the circles as
being oppositely placed,
and for this centre Z and Y' as also U and U' are homologous
points ; and tangents at Y' and Z are parallel, and so also
are tangents at U and U'.
Hence YZ is a diameter of the circle S and is parallel to
Y'Z' a diameter of the circle S'.
Hence to find the centres of similitude of two given
circles : — Draw parallel diameters, one to each circle, and
connect their endpoints directly and transversely. The
direct connector cuts the common centreline in the external
Q
242 SYNTHETIC GEOMETRY.
centre of similitude, and the transverse connector cuts it in
the internal centre of simihtude.
286°. Since OX : OX' = OY : OY', if X and Y become coin
cident, X' and Y' become coincident also.
.'. a line through O tangent to one of the circles is tangent
to the other also, or O is the point where a common tangent
cuts the common centreline. A similar remark applies to O'.
When the circles exclude one another the centres of
similitude are the intersections of common tangents of the
same name, direct and transverse.
When one circle lies within the other (2nd Fig.) the com
mon tangents are imaginary, although O and O' their points
of intersection are real.
287°. Since AOCY^AOC'Y', .. OC:OC = r:?'',
and since AO'CZ^ AO'C'Y', .. O'C : 0'C' = r : r\
.'. the centres of similitude of two circles are the points
which divide, externally and internally, the join of the centres
of the circles into parts which are as the conterminous radii.
The preceding relations give
OC = /!. CC, and 0'C=;^ . CC.
r J' r \r
OC is < r according as CC is < r' — r,
and O'C is ^ r according as CC is % r' + r.
Hence
1. O lies within the circle S when the distance between the
centres is less than the difference of the radii, and O' lies
within the circle S when the difference between the centres is
less than the sum of the radii.
2. When the circles exclude each other without contact both
centres of similitude lie without both circles.
3. When the circles touch externally, the point of contact
is the internal centre of similitude.
4. When one circle touches the other internally, the point
of contact is the external centre of similitude.
CENTRE OF SIMILITUDE OR PERSPECTIVE. 243
5. When the circles are concentric, the centres of similitude
coincide with the common centre of the circles, unless the
circles are also equal, when one centre of similitude becomes
any point whatever.
6. If one of the circles becomes a point, both centres of
similitude coincide with the point.
288°. Def. — The circle having the centres of similitude of
two given circles as endpoints of a diameter is called the
circle of similitude of the given circles.
The contraction of s. will be used for circle of similitude.
Cor. I. Let S, S' be two circles and Z their of s.
Since O and O' are
two points from which
tangents to circles S and
S' are in the constant
ratio of r to r', the circle o"
Z is coaxal with S and
S' (275°, Cor. 5). Hence
any two circles and their Q) of s. are coaxal.
Cor. 2. From any point P on circle Z,
PT :TC = PT':T'C,
and.. ^TPC = ^T'PC'.
Hence, at any point on the Q) of s. oi two circles, the two
circles subtend equal angles.
Cor. 3. OC = CC. ,^. and O'C = CC. ^. ^287°)
^ r'r r' + r ' ' ^
whence 00'= CC.
The Q) of s.\s a line, the radical axis, when the given
circles are equal {r=r').
The (z) of s. becomes a point when one of the two given
circles becomes a point (r or r' = o).
The ofs. is a point when the given circles are con
centric (CC'=o).
244 SYNTHETIC GEOMETRY.
289°. Def. — With reference to the centre O (Fig. of 285°),
X and y, as also X' and Y, are called antihomologoiis points.
Similarly with respect to the centre O', U' and Z, as also U
and Y', are antihomologous points.
Let tangents at X and Y' meet at L. Then, since CX is 
to CX', ^CXY=^C'X'Y' = z.C'Y'X'. But ^LXY is comp. of
^CXY and Z.LY'X' is comp. of Z.C'Y'X'.
ALXY' is isosceles, and LX = LY'.
L is on the radical axis of S and S'.
Similarly it may be proved that pairs of tangents at Y and
X', at U and Y', and at U' and Z, meet on the radical axis of
S and S', and the tangent at U passes through L.
.'. tangents at a pair of antihomologous points meet on the
radical axis.
Cor. I. The join of the points of contact of two equal
tangents to two circles passes through a centre of similitude
of the two circles.
Cor. 2. When a circle cuts two circles orthogonally, the
joins of the points of intersection taken in pairs of one from
each circle pass through the centres of similitude of the two
circles.
290°. Since OX:OX' = r:r',
OX.OY':OX'. OY' = r:r'.
But OX'. OY' = the square of the tangent from O to the circle
S' and is therefore constant.
OX . 0Y' = , . 0T'2 = a constant.
r
.'. X and Y' are inverse points with respect to a circle
whose centre is at O and whose radius is OT' /^,.
B^/. — This circle is called the circle of antisimilihide^ and
will be contracted to ofans.
Evidently the circles S and S' are inverse to one another
with respect to their ofans.
CENTRE OF SIMILITUDE OR PERSPECTIVE. 245
For the centre 0' the product OU . OY' is negative, and the
of ans. corresponding to this centre is imaginary.
Cor. I. Denoting the distance CC by d^ and the difference
between the radii {r' — r) by 5, we have
R' = ^^'— 52'
where R = the radius of the of ans. Hence
1. When either circle becomes a point their of ans.
becomes a point.
2. When the circles S and S' are equal, the of ans. be
comes the radical axis of the two circles.
3. When one circle touches the other internally the of
ans. becomes a pointcircle. (d=d.)
4. When one circle includes the other without contact the
of ans. is imaginary. (d<d.)
Cor. 2. Two circles and their circle of antisimilitude are
coaxal. (263°)
Cor. 3. If two circles be inverted with respect to their
circle of antisimilitude, they exchange places, and their radi
cal axis being a line circle coaxal with the two circles
becomes a circle through O coaxal with the two.
The only circle satisfying this condition is the circle of
similitude of the two circles. Therefore the radical axis
inverts into the circle of similitude, and the circle of simili
tude into the radical axis.
Hence every line through O cuts the radical axis and the
circle of similitude of two circles at the same angle.
291°. D^f — When a circle touches two others so as to
exclude both or to include both, it is said to ifouc/i them
similarly^ or to have contacts of like kind with the two.
When it includes the one and excludes the other, it is said to
touch them dissimilarly^ or to have contacts of unlike kinds
with the two.
246
SYNTHETIC GEOMETRY.
292°. Theorem. — When a circle touches two other circles,
its chord of contact passes through their external centre of
similitude when the contacts are of like kind, and through
their internal centre of similitude when the contacts are of
unlike kinds.
Proof. — Let circle Z touch circles S and S' at Y and X'.
Then CYD and C'X'D are lines. (113°, Cor. i)
Let XYX'Y' be the secant through Y and X'. Then
^CXY = aCYX = ^DYX' = ^DX'Y = ^CX'Y'.
.*. CX and C'X' are parallel, and X'X passes through the
external centre of similitude O. (285")
Similarly, if Z' includes both S and S', it may be proved
that its chord of contact passes through O.
Again, let the circle W, with centre E, touch S' at Y' and
S at U so as to include S' and exclude S, and let UY' be the
chord of contact. Then
^CVU = i.CUV = ^EUY' = _EY'U,
.*. EY' and CV are parallel and VY' connects them trans
versely ; .*. VY' passes through O'. q.e.d.
Cor. I. Every circle which touches S and S' similarly is
cut orthogonally by the external circle of antisimilitude of S
and S'.
CENTRE OF SIMILITUDE OR PERSPECTIVE. 247
Cor. 2. If two circles touch S and S' externally their points
of contact are concyclic. (n6°, Ex. 2)
But the points of contact of either circle with S and S' are
antihomologous points to the centre O.
.'. if a circle cuts two others in a pair of antihomologous
points it cuts them in a second pair of antihomologous points.
Cor. 3. If two circles touch two other circles similarly, the
radical axis of either pair passes through a centre of simili
tude of the other pair.
For, if Z and Z' be two circles touching S and S' externally,
the external circle of antisimilitude of S and S' cuts Z and Z'
orthogonally (Cor. i) and therefore has its centre on the
radical axis of Z and Z'.
Cor. 4. If any number of circles touch S and S' similarly,
they are all cut orthogonally by the external circle of anti
similitude of S and S', and all their chords of contact and all
their chords of intersection with one another are concurrent
at the external centre of antisimilitude of S and S'.
293°. 71ieorein.—\{ the circle Z touches the circles S and
S', the chord of contact of Z and the radical axis of S and S'
are conjugate lines with respect to the circle Z.
Proof. — Let Z touch S and S' in Y and X' respectively.
The tangents at Y and X' meet at a point P on the radical
axis of S and S'. (178 )
But P is the pole of the chord of contact YX'.
.*. the radical axis passes through the pole of the chord of
contact, and reciprocally the chord of contact passes through
the pole of the radical axis (267°, Def ) and the lines are
conjugate. q.c.d.
248
SYNTHETIC GEOMETRY.
AXES OF SIMILITUDE.
294°. Let Sj, S2, S3 denote three circles having their centres
A5 B, C and radii r^, ?^2y ^z^ ^"<^ let X, X', Y, Y', Z, Z' be their
six centres of similitude.
Now X, Y, Z are
three points on the
sides of the AABC,
BX^r,
CX r.l
and
_'3
r^y
CY
AY
AZ
BZ
Vex/ '
and X, Y, Z are col
linear.
Similarly it is proved that the triads of points XY'Z', YZ'X',
ZX'Y' are collinear.
Def. — These lines of coUinearity of the centres of similitude
of the three circles taken in pairs are the axes of similitude
of the circles. The line XYZ is the external axis, as being
external to all the circles, and the other three, passing be
tween the circles, are internal axes.
Cor. I. If an axis of similitude touches any one of the
circles it touches all three of them. (286°)
Cor. 2. If an axis of similitude cuts any one of the circles
it cuts all three at the same angle, and the intercepted chords
are proportional to the corresponding radii.
Cor. 3. Since XYX'Y' is a quadrangle whereof XX', YY',
and ZZ' are the three diagonals, the circles on XX', YY', and
ZZ' as diameters are coaxal. (277°, Cor. 2)
.'. the circles of similitude of three circles taken in pairs
are coaxal.
AXES OF SIMILITUDE OR PERSPECTIVE. 249
Cor. 4. Since the three circles of similitude are of the c.p.
species, two points may be found from which any three circles
subtend equal angles. These are the common points to the
three circles of similitude. (288°, Cor. 2)
Cor. 5. The groups of circles on the following triads of
segments as diameters are severally coaxal,
AX, BY, CZ ; AX, YZ', Y'Z ; BY, Z'X, ZX'; CZ, XY', X'Y.
295°. Any two circles Z and Z', which touch three circles
Si, S2, S3 similarly, cut their circles of antisimilitude ortho
gonally (292°, Cor. i), and therefore have their centres at the
radical centre of the three circles of antisimilitude.
(276°, Cor. 4)
But Z and Z' have not necessarily the same centre.
.*. the three circles of antisimilitude of the circles S^, S2, and
S3 are coaxal, and their common radical axis passes through
the centres of Z and Z'.
296°. Theorem. — If two circles touch three circles similarly,
the radical axis of the two is an axis of similitude of the
three ; and the radical centre of the three is a centre of
similitude of the two.
Proof. — The circles S and S'
touch the three circles A, B,
and C similarly.
I. Since S and S' touch A
and B similarly, the radical axis
of S and S' passes through a
centre of similitude of A and B.
(292°, Cor. 3)
Also, the radical axis of S
• and S' passes through a centre of similitude of B and C, and
through a centre of similitude of C and A.
,*. the radical axis of S and S' is an axis of similitude of
the three circles A, B, and C.
250 SYNTHETIC GEOMETRY.
2. Again, since A and B touch S and S', the radical axis of
A and B passes through a centre of simiHtude of S and S'.
For similar reasons, and because A, B, and C touch S and
S' similarly, the radical axes of B and C, and of C and A,
pass through the same centre of similitude of S and S'. But
these three radical axes meet at the radical centre of A, B,
and C.
.', the radical centre of A, B, and C is a centre of simili
tude of S and S'. q.e.d.
297°. Problem. — To construct a circle which shall touch
three given circles.
In the figure of 296°, let A, B, and C be the three given
circles, and let S and S' be two circles which are solutions
of the problem.
Let L denote one of the axes of similitude of A, B, and C,
and let O be their radical centre. These are given when the
circles A, B, and C are given.
Now L is the radical axis of S and S' (296°, i), and O is
one of their centres of similitude.
But as A touches S and S' the chord of contact of A passes
through the pole of L with respect to A (293°). Similarly the
chords of contact of B and C pass through the poles of L
with respect to B and C respectively. And these chords are
concurrent at O. (292°)
Hence the following construction : —
Find O the radical centre and L an axis of simiHtude of A,
B, and C. Take the poles of L with respect to each of these
circles, and let them be the points p, q, r respectively.
Then Op, Og, Or are the chords of contact for the three
given circles, and three points being thus found for each of
two touching circles, S and S', these circles are determined.
(This elegant solution of a famous problem is due to M.
Gergonne.)
Cor. As each axis of similitude gives different poles with
respect to A, B, and C, while there is but one radical centre
AXES OF SIMILITUDE OR PERSPECTIVE. 25 I
O, in general each axis of similitude determines two touching
circles ; and as there are four axes of similitude there are
eight circles, in pairs of twos, which touch three given circles.
Putting i and e for internal and external contact with the
touching circle, we may classify the eight circles as follows :
(See 294°)
Axes of Similituul.
A
B
c
X Y
Z
i
e
i
e
:]'
X Y'
Z'
e
i
i
e
:]^>
X' Y
Z'
i
e
e
i
:]^
X' Y'
Z
i
e
i
e
:.}.pr.
PART V.
ON HARMONIC AND ANHARMONIC RATIOS—
HOMOGRAPHY. INVOLUTION, ETC.
SECTION I.
GENERAL CONSIDERATIONS IN REGARD TO
HARMONIC AND ANHARMONIC DIVISION.
298°. Let C be a point dividing a segment AB. The posi
tion of C in relation to A and B is determined by the ratio
, ^ , AC : BC. For, if we know this ratio, we
^ D B know completely the position of C with
respect to A and B. If this ratio is negative, C lies between
A and B ; if positive, C does not lie between A and B. If
AC:BC= — I, C is the internal bisector of AB ; and if
AC : BC= + 1, C is the external bisector of AB, i.e., a point
at CO in the direction AB or BA.
Let D be a second point dividing AB. The position of D
is known when the ratio AD : BD is known.
Be/. — If we denote the ratio AC : BC by w, and the ratio
AD : BD by ;/, the two ratios 7/1 : n and n : m, which are
reciprocals of one another, are called the two anhannonic
ratios of the division of the segment AB by the points C and
D, or the harmonoids of the range A, B, C, D.
252
HARMONIC AND ANHARMONIC DIVISION. 253
Either of the two anharmonic ratios expresses a relation
between the parts into which the segment AB is divided by
the points C and D.
Evidently the two anharmonic ratios have the same sign,
and when one of them is zero the other is infinite, and vice
versa.
These ratios may be written : —
AC . AD AC.BD AC . ED
BC ' BD BC.AD AD . BC*
AD . AC AD^C AI:K^
BD ' BC BD.AC AC.BD*
The last form is to be preferred, other things being convenient,
on account of its symmetry with respect to A and B, the
endpoints of the divided segment.
299°. The following results readily follow.
T , AC.BD, , T^t, AC 1 AD , ,.1
1. Let ^^^^S7^ be +. Then  and — ^ have like signs,
AD . riC x>v^ dU
and therefore C and D both divide AB internally or both
externally. (298")
In this case the order of the points must be some one of
the following set, where AB is the segment divided, and the
letters C and D are considered as being interchangeable :
CDAB, ACDB, CABD, ABCD.
T , AC.BD, T,, AC J AD ,
2. Let ^^g^ be . Then — and ^ have opposite
signs, and one point divides AB internally and the other
externally.
The order of the points is then one of the set CADB, ACBD.
3. When either of the two anharmonic ratios is ±1, these
ratios are equal.
^ ^' ^?= + ■• '^''^" ^=nG' ^"^ ^ ^"^ ° ^^«
both internal or both external.
254 synthetic geometry.
., acbc adbd ab ab
^^^° bc=Td'"'bc=bd'
and C and D coincide.
Hence, when C and D are distinct points, the anharmonic
ratio of the parts into which C and D divide AB cannot be
positive unity.
^ . . AC.BD ^ T,, „ AC AD
5 L^^AD7BC=^ ^^^^BC=BD
And since C and D are now one external and one internal
(2), they divide the segment AB in the same ratio internally
and externally, disregarding sign. Such division of a line
segment is called harmonic. (208°, Cor. i)
Harmonic division and harmonic ratio have been long em
ployed, and from being only a special case of the more
general ratio, this latter was named "anharmonic" by
Chasles, " who was the first to perceive its utility and to
apply it extensively in Geometry."
300°. Def. — When we consider AB and CD as being two
segments of the same line we say that CD divides AB, and
that AB divides CD.
Now the anharmonic ratios in which CD divides AB are
AC.BD , AD . BC
AD.BC ^" AC.BD'
And the anharmonic ratios in which AB divides CD are
CA^DB , CB.DA
CB.DA CA.DB*
But the anharmonic ratios of these sets are equal each to
each in both sign and magnitude.
.'. thea7iharmonic ratios in which CD divides AB are the
same as those in which AB divides CD.
Or, any two segments of a common tine divide each other
eqiiianharmonically.
301°. Four points A, B, C, D taken on a line determine six
segments AB, AC, AD, BC, BD, and CD.
HARMONIC AND ANHARMONIC DIVISION. 255
These may be arranged in three groups of two each, so that
in each group one segment may be considered as dividing
the others, viz., AB, CD ; BC, AD ; CA, BD.
Each group gives two anharmonic ratios, reciprocals of one
another ; and thus the anharmonic ratios determined by a
range of four points, taken in all their possible relations, are
six in number, of which three are reciprocals of the other
three.
These six ratios are not independent, for, besides the
reciprocal relations mentioned, they are connected by three
relations which enable us to find all of them when any one is
given.
P,^„_AC.BD, p BA.CD, ^ CB.AD, ^
^^""'^ ADTBC ^^ ^' BD7CA ^^ ^' CDTAB ^^ ^^
Then P, O, R are the anharmonic ratios of the groups
ABCD, BCAD, and CABD, each taken in the same order.
But in any range of four (233°) we have
AB.CD + BC.AD + CA.BD=o.
And dividing this expression by each of its terms in succes
sion, we obtain Q + ^^ = R + ^ = P +  = i.
From the symmetry of these relations we infer that any
general properties belonging to one couple of anharmonic
ratios, consisting of any ratio and its reciprocal, belong
equally to all.
Hence the properties of only one ratio need be studied.
The symbolic expression {ABCD} denotes any one of the
anharmonic ratios, and may be made to give all of them by
reading the constituent letters in all possible orders.
Except in the case of harmonic ratio, or in other special
cases, we shall read the symbol in the one order of alternat
ing the letters in the numerator and grouping the extremes
and means in the denominator. Thus
{ABCD} denotes ^§^
It is scarcely necessary to say that whatever order may be
256 SYNTHETIC GEOMETRY.
adopted in reading the symbol, the same order must be em
ployed for each when comparing two symbols.
302°. Theorem. — Any two constituents of the anharmonic
symbol may be interchanged if the remaining two are inter
changed also, without affecting the value of the symbol.
Proof. {ABCD} = AC.BD : AD.BC.
Interchange any two as A and C, and also interchange the
remaining two B and U. Then
{CDAB}=CA.DB:CB.DA
= AC.BD : AD.BC.
Similarly it is proved that
[ABCD} = {BADC} = {CDAB} = {DCBA}. q.e.d.
303°. If interchanging the first two letters, or the last two,
without interchanging the remaining letters, does not alter
the value of the ratio, it is harmonic.
For, let {ABCD} = {ABDC}.
^, AC.BD_AD^BC
AD.BC AC.BD'
or, multiplying across and taking square roots,
AC. BD=± AD.BC.
But the positive value must be rejected (299°, 4), and the
negative value gives the condition of harmonic division.
304". Let ABCD be any range of four and O any point not
on its axis.
The anharmonic ratio of the pencil
O . ABCD corresponding to any given
ratio of the range is the same function
of the sines of the angles as the given
ratio is of the corresponding segments.
ry,^^ sin AOC. sin BOD , , AC.BD
Thus .  — • tt^^ corresponds to , ,^ ^ ;
smAOD.smBOC ^ AD.BC
r, symbolically, 0{ABCD} corresponds to {ABCD}.
HARMONIC AND ANHARMONIC DIVISION. 257
To prove that the corresponding anharmonic ratios of the
range and pencil are equal.
AC_ AAOC _ OA.OCsinAOC _OA sinAOC
BC ABOC OB.OCsinBOC OB ' sin BOC
o. ., , BD OB sin BOD
Similarly, aD = OA' ilKAOD'
AC . BD _ sin AOC . sin BOD
AD.BC sinAOD.sinBOC*
Hence, symbolically
{ABCD}=0{ABCD} ;
and, with necessary formal variations, the anharmonic ratio
of a range may be changed for that of the corresponding
pencil, and vice versa, whenever required to be done.
Cor. I. Two angles with a common vertex divide each
other equianharmonically. (300°)
Cor. 2. If the anharmonic ratio of a pencil is +1, two rays
coincide, and if  1, the pencil is harmonic. (299°, 4, 5)
Cor. 3. A given range determines an equianharmonic pen
cil at every vertex, and a given pencil determines an equian
harmonic range on every transversal.
Cor. 4. Since the sine of an angle is the same as the sine
of its supplement (214°, i), any ray may be rotated through a
straight angle or reversed in directionwithout affecting the ratio.
Corollaries 2, 3, and 4 are of special importance.
305. Theorem, — If three pairs of corresponding rays of two
equianharmonic pencils intersect collinearly, the fourth pair
intersect upon the line of collinearity.
Proof.— \.Q.\.
0{ABCD}=0'{ABCD'},
and let the pairs of corresponding rays
OA and O'A, OB and O'B, OC and O'C
intersect in the three collinear points A,
B, and C. Let the fourth corresponding
rays meet the axis of ABC in D and D'
respectively. Then {ABCD} = ^ABCD'}, (304°)
R
258 SYNTHETIC GEOMETRY.
AD.BC AD'.BC'^" AD AD"
which is possible only when D and D' coincide.
.*. the fourth intersection is upon the axis of A, B, and C,
and the four intersections are collinear. q.e.d.
Cor. If two of the corresponding rays as OC and 0"C
become one line, these rays may be considered as intersecting
at all points on this line, and however A and B are situated
three corresponding pairs of rays necessarily intersect col
linearly.
.*. when two equianharmonic pencils have a pair of cor
responding rays in common, the remaining rays intersect
collinearly.
306°. Theoretn. — If two equianharmonic ranges have three
pairs of corresponding points in perspective, the fourth
points are in the same perspective.
Proof.—
{ABCD} = {A'B'C'D'},
and A and A', B and B', and C and C
are in perspective at O. Now
0{ABCD} = 0{A'B'C'D'},
and we have two equianharmonic pen
cils of which three pairs of correspond ^'
ing rays meet collinearly at A, B, and C. Therefore OD'
and OD meet at D, or D and D' are in perspective at O.
Cor. If two of the corresponding points, as C and C", be
come coincident, these two points are in perspective at
every centre, and hence three corresponding pairs of points
are necessarily in perspective.
.*. when two equianharmonic ranges have a pair of cor
responding points coincident, the remaining pairs of cor
responding points are in perspective.
HARMONIC RATIO. 259
SECTION II.
HARMONIC RATIO.
307". Harmonic ratio being a special case of anharmonic
ratio (299°, 5), the properties and relations of the latter
belong also to the former.
The harmonic properties of a divided segment may ac
cordingly be classified as follows : —
1. The dividing points alternate with the end points of the
divided segment.
For this reason harmonic division is symbolized by writing
the letters in order of position, as, {APBO}, where A and B
are the end points of the segment and P and Q the dividing
points (301°). A— P— B— O.
2. The dividing points P and O divide the segment extern
ally and internally in the same ratio, neglecting sign. (299°, 5)
3. If one segment divides another harmonically, the second
also divides the first harmonically. (300°)
4. A harmonic range determines a harmonic pencil at every
vertex, and a harmonic pencil determines a harmonic range
on every transversal. (304°^ Cor. 3)
5. If one or more rays of a harmonic pencil be reversed in
direction the pencil remains harmonic. (304°) Cor. 4)
6. Two harmonic pencils which have three pairs of corre
sponding rays intersecting collinearly have all their corre
sponding rays intersecting collinearly. (305°)
7. Two harmonic ranges which have three pairs of corre
sponding points in perspective have all their corresponding
points in perspective. (306°)
8. If two harmonic pencils have a corresponding ray from
€ach in common, all their corresponding rays intersect col
linearly. (305'* Cor.)
9. If two harmonic ranges have a corresponding point
from each in common, all their corresponding points are in
perspective. (306°, Cor.)
26o SYNTHETIC GEOMETRY.
aoS"". Let APBO be a harmonic range. Then
 I , . . AP:PB = AQ:BQ,
^ ^ ^ ^ :. AP:AO = ABAP:AOAB.
Taking AP, AB, AG as three magnitudes, we have the
statement : —
The first is to the third as the difference between the first
and second is to the difference between the second and the
third. And this is the definition of three quantities in
Harmonic Proportion as given in Arithmetic and Algebra.
Exercises.
1. When three line segments are in harmonic proportion the
rectangle on the mean and the sum of the extremes is
equal to twice the rectangle on the extremes.
2. The expanded symbol {APBQ}=i gives AP : AG
=  BP : BQ. Why the negative sign ?
3. Prove from the nature of harmonic division that when P
bisects AB, Q is at 00 .
4. Prove that if OP bisects z_AOB internally GO bisects it
externally ; O {APBQ} is equal to — i.
5. Trace the changes in the value of the ratio AC : BC as G
moves from  00 to + 00 .
309*. In the harmonic range APBQ, P and Q are called
conjugate points, and so also are A and B.
Similarly in the harmonic pencil O.APBQ, OP and OQ
are conjugate rays, and so also are OA and OB.
Ex. I. Given three points of a harmonic range to find the
fourth.
Let A, P, B be the three given points.
By (259°, Ex. 7) find any point O at which the segments
AP and PB subtend equal angles. Draw 00 the external
bisector of the ^AOB. Q is the fourth point.
For OP and OQ are internal and external bisectors of the
,z_AOB. (208", Cor. i)
HARMONIC RATIO.
261
Ex. 2. Given three rays to find a fourth so as to make the
pencil harmonic. o
Let OA, OP, OB be the three rays. 0^
'On OA take any two equal distances ^^
OD and DE.
Draw DF i to OB, and draw OQ  to
EF. OQ is the fourth ray required.
For since OD = DE, EF = FG. And OQ meets EF at 00
are harmonic and hence O . APBQ are
Then EFGoo
harmonic.
Cor. In the symbolic expression for a harmonic ratio a pair
of conjugates can be interchanged without destroying the
harmonicism.
{APBQ} = {BPAQ} = {BQAP} = {AQBP},
for {APBQ} gives AP . BQ : AG . BP=  i,
and {BPAQ} gives BP . AG : BQ . AP,
and being the reciprocal of the former its value is  i also.
And similarly for the remaining symbols.
HARMONIC PROPERTIES OF THE TETRAGRAM
OR COMPLETE QUADRANGLE.
310°. Let ABGD be a quadrangle, of which AC, BD, and
EF are the three diagonals. (247°, Def 2)
Also let the line EO cut two sides
in G aijd H, and the line FO cut
the other two sides in K and L.
Then .
I. AEDisa Awhereof AC, EH,
and DB are concurrent lines from
the vertices to the opposite sides.
.. AB.EC.DH=AH.DC.EB.
(251°, b)
Also, AED is a A and FCB is a
transversal.
AB.EC.DF = AF.DC
262 SYNTHETIC GEOMETRY.
and dividing the former equality by the latter,
DH__AH
DF AF'
and AHDF is a harmonic range.
2. Again, {AHDF} = E{AHDF} = E{LOKF} = E{BGCF}.
(307°, 4)
LOKF and BGCF are harmonic ranges.
3. 0{AHDF}=0{CEDK} = F{CEDK} = F{GEHO}
= F{BEAL}, (307°, 4, 5)
but {CEDK} = {DKCE}, etc. (309°, Cor.)
DKCE, HOGE, ALBE are harmonic ranges.
4. If AC be produced to meet EF in I, AOCI is a harmonic
range.
.'. all the lines upon which four points of the figure lie are
divided harmonically by the points.
And the points E, F, and O at which four hnes concur are
vertices of harmonic pencils.
Exercises.
1. A line  to the base of a A has its points of intersection
with the sides connected transversely with the end
points of the base. The join of the vertex with the
point of intersection of these connectors is a median,
and is divided harmonically.
(Let F go to 00 in the last figure.)
2. ABC is a A and BD is an altitude. Through any point
O on BD, CO and OA meet the sides in F and E
respectively. Show that DE and DF make equal
angles with AC.
3. The centres of two circles and their centres of similitude
form a harmonic range.
4. In the Fig. of 310° the joins DI, IB, BH, and LD are all
divided harmonically.
HARMONIC RATIO. 263
311°. Let APBQ be a harmonic range and let C be the
middle point of AB. Then
AP__BP_PB
AQ BQ BQ
T.,,, • CB + CP_CBCP
^^^^ 'CB + CQ"CQCB»
CB+CP _ CQ+CB
°^ CBCP CQCB'
, CB CO
whence ^=^.
.*. CP . CQ = CB2, Qj. p aj^fj Q aj.g inverse points to the circle
having C as centre and CB as radius.
.'. I. The diameter of a circle is divided harmonically by
any pair of inverse points.
And a circle having a pair of co7ijugates of a harmonic
range as endpoints of a diameter has the other pair of
conjugates as inverse points.
Again, let EF be any secant through P meeting the polar
of P in V.
A circle on PV as diameter passes through Q and P, and
therefore cuts S orthogonally. (258°)
Hence also the circle S cuts the circle on PV orthogonally,
and E and F are inverse points to the circle on PV.
.'. EPFV is a harmonic range.
.*. 2. A line is cut harmonically by a pointy a circle^ and
the polar of the point with respect to the circle.
Ex. P, Q are inverse points, and from Q a line is drawn
cutting the circle in A and B. The join PB cuts the circle in
A'. Then AA' is ± to PQ.
312°. Let P be any point and L its polar with respect to
the circle Z, And let PCD and PBA be any two secants.
Then
I. PCED and PBFA are harmonic ranges having P a cor
responding point in each. Therefore AD, FE, and BC arc
concurrent. And BC and AD meet on the polar of P. (309°, 9)
264
SYNTHETIC GEOMETRY.
2. Again, since {PCED}={PDEC}, (309°, Cor.)
.*. PDEC and PBFA are harmonic ranges having P a cor
responding point in each. Therefore DB, EF, and CA are
concurrent, and AC and DB meet on the polar of P.
.*. If from any point two secants be dj'aivft to a circle, the
connectors of their points of intersection with the circle meet
upon the polar of the first point.
3. Since O is on the polar of P, P is on the polar of O.
But since Q is a point from which secants are drawn satis
fying the conditions of 2, Q is on the polar of O.
.. PQ is the polar of O.
Now ABCD is a concyclic quadrangle whereof AC and
BD are internal diagonals and PQ the external diagonal.
.*. In any concyclic qicadrangle t}ie external diagonal is the
polar of the point of intersection of the internal diagoiials,
with respect to the circumcircle.
4. Since Q is on the polar of P and also on that of O,
therefore PO is the polar of O, and POQ is a triangle self
conjugate with respect to the circle.
5. Let tangents at the points A, B, C, D form the circum
scribed quadrangle USVT.
HARMONIC RATIO.
265
Then S is the pole of AB, and T of DC.
.*. ST is the polar of P, and S and T are points on the
line QO.
Similarly U and V are points on the line PC.
But XY is the external diagonal of USVT, and its pole is
0. the point of intersection of DB and AC.
.*. X and Y are points on the line PQ.
Hence, If tatigents be drazun at the vertices of a coney cite
quadrangle so as to form a circninscribed quadrangle^ the in
ternal diagonals of the two quadrangles are concurrent, and
their external diagonals are seginents of a coinmon line; and
the point of concurrence and the line are pole and polar with
respect to the circle.
Exercises.
1. UOVP and SOTQ are harmonic ranges.
2. If DB meets the line PQ in R, lOR is a selfconjugate
triangle with respect to the circle.
3. To find a circle which shall cut the sides of a given
triangle harmonically.
4. QXPY is a harmonic range.
313°. Let S be a circle and A'P'B'Q' a harmonic range.
Taking any point O on the p
•circle and through it projecting
rectilinearly the points A'P'B'Q'
we obtain the system APBO,
which is called a harmonic sys q,^^ \"""T\~~ \ ^~A^
tern of points on the circle.
Now, taking O' any other point
on the circle, O'. APBQ is also V \ \ I / y^'
harmonic. For
^AOP=_AO'P,
aPOB = ^PO'B, etc.
.'. Def — Four points on a circle
formaharmonicsystem when their A p b' q'
joins with any fifth point on the circle form a harmonic pencil.
266 SYNTHETIC GEOMETRY.
Cor. I. Since sinz.AOP = i^^, sin^POB = £^, etc., (228°)
a a
.. (304°), neglecting sign, AP . BQ = AQ . PB,
.•. When four poiiits form a harmonic system on a circle^ the
rectangles on the opposite sides of the nor7nal quadrangle
which they determine are equal.
Cor. 2. If O comes to A, the ray OA becomes a tangent
at A.
.*. When fottr poi7its form a harmonic system on a circle^
the tangent at any one of them and the chords from the point
of contact to the others form a harmo7iic pencil.
314°. Let the axis of the harmonic range APBQ be a tan
gent to the circle S.
Through A, P, B, and
Q draw the tangents A^,.
B^, Vp, and Q^.
These four tangents
form a harmonic system
of tangents to the circle S
Let L be any other tan
gent cutting the four tan
gents of the system in A'
P', B', and (^.
Then, considering A«, P^, B(5, etc., as fixed tangents, and
A'P'B'Q' as any other tangent.
lA.OV = lA.O'V\ z.POB = ^P'OB', etc., (116°, Ex. i>
.. the pencils O.APBQ and O . A'P'B'Q' are both har
monic, and A'P'B'Q' is a harmonic range.
.*. Whetifour tangents form a harjnonic system to a circle,
they intersect any other tangent in points which foi'i7i a har
7no7iic range.
Cor. I. If the variable tangent coincides with one of the
fixed tangents, the point of contact of the latter becomes one
of the points of the range.
HARMONIC RATIO. 26/
.'. When four ta7igenis form a harmo7iic system to a circle,
each tangent is divided harmonically by its point of contact
and its intersections with the other tangents.
Exercises.
1. Tangents are drawn at A, A' the endpoints of a diameter,
and two points P, B are taken on the tangent through
A such that AB = 2AP. Through P and B tangents are
drawn cutting the tangent at A' in P' and B'. Then
2A'B' = A'P', and AA', PB', and BP' are concurrent.
2. Four points form a harmonic system on a circle. Then the
tangents at one pair of conjugates meet upon the secant
through the other pair.
3. If four tangents form a harmonic system to a circle, the
point of intersection of a pair of conjugate tangents lies
on the chord of contact of the remaining pair.
4. If four points form a harmonic range, their polars with
respect to any circle form a harmonic pencil ; and
conversely.
SECTION III.
OF ANHARMONIC PROPERTIES.
315°. Let A, E, C and D, B, F be two sets of three col
linear points having their ^
axes meeting in some
point R.
Join the points alter
nately, as ABCDEFA.
Then AB and DE, BC and
EF, CD and FA meet in
P, Q, O. To show that these points are collinear.
268 SYNTHETIC GEOxMETRY.
0{ECQF} = C{EOQF} (referred to axis EF)
= C{RDBF} (referred to axis DR)
A{RDBF}
= A{EDPF} (referred to axis DE)
= 0{EDPF}
= 0{ECPF}. (by reversing rays, etc.)
. .'. the pencils O. ECQF and O. ECPF are equianharmonic,
and having three rays in common the fourth rays must be in
common, /.<?., they can differ only by a straight angle, and
therefore O, P, Q are collinear.
(Being the first application of anharmonic ratios the work
is very much expanded.)
.' . If six lines taken in order intersect alternately in two
sets of three collinear points^ they intersect in a third set of
three collijiear points.
Cor. I. ABC and DEF are two triangles, whereof each has
one vertex lying upon a side of the other:
If AB and DE are taken as corresponding sides, A and F
are noncorresponding vertices. But, if AB and EF are
taken as corresponding sides, A and D are noncorresponding
vertices.
Hence the intersections of AB and EF, of ED and CB,
and of AD and CF are collinear.
.*. If two triangles have each a vertex lying upon a side of
the other, the remaining sides and the joins of the remaining
noncorresponding vertices intersect collinearly.
Cor. 2. Joining AD, BE, CF, ADBE, EBFC, and ADFC
are quadrangles, and P, O, O are respectively the points of
intersection of their internal diagonals.
.*. if a quadrangle be divided into two quadrangles, the
points of intersection of the internal diagonals of the three
quadrangles are collinear.
OF ANHARMONIC PROPERTIES.
269
316", Let A, A', B, B', C, C be six points lying two by two
on two sets of three con
current lines, which meet
at P and O. Then the
points lie upon a third set
of three concurrent lines
meeting at O.
We are to prove that
AG and AA' are in line.
A{0,rC'B} = B'{0:irC'Q}
= B'{CPC>}
= Q{A'P^B}
=A{A';rC'B}.
.'. the pencils A.OjtC'B
and A.A';i'C'B are equi
anharmonic, and have three corresponding rays in common.
Therefore AG and A A' are in line.
Cor. ABC and A'B'C are two As which are in perspective
at both P and Q, and we have shown that they are in per
spective at G also.
As there is an axis of perspective corresponding to
each centre, the joins of the six points, accented letters
being taken together and unaccented together, taken in
every order intersect in three sets of three collinear
points.
Exercises.
If two As have their sides intersecting collinearly, their
corresponding vertices connect concurrently.
The converse of Ex. i.
Three equianharmonic ranges ABCD, A'B'C'D', and
PQRS have their axes concurrent at Y, and AA', BB',
CC, DD' concurrent at X. Then the two groups of
joins AP, BO, CR, DS, and A'P, B'Q, C'R, D'S are
2/0
SYNTHETIC GEOMETRY.
concurrent at two points O and O' which are coUinear
with X.
4. From Ex. 3 show that if a variable A has its sides passing
through three fixed points, and two of its vertices lying
upon fixed lines, its third vertex lies upon a fixed line
concurrent with the other two.
5. If a variable A has its vertices lying on three fixed lines
and two of its sides passing through fixed points, its
third side passes through a fixed point coUinear with
the other two.
The range ABCD is transferred to the circle S by
rectilinear projection through
any point O on the circle.
Then A'B'C'D' is a system of
points on the circle which is
equianharmonic with the range
ABCD.
I. If O' be any other point
on the circle, the
^A'OB' = ^A'0'B',
ABCD ^B'OC' = ^B'0'C', etc.,
and the two pencils O . A'B'C'D' and O'. A'B'C'D' are equi
anharmonic.
.'. four points on a circle subtend equianharmonic pencils
at all fifth points on the circle.
2. Since
sinA'OB' =
A'B'
sin COD':
etc.,
CD'
AB . CD + BC . AD = AC . BD,
A'B'. CD' + B'C. A'D' = A'C'. B'D' ;
which is an extension of Ptolemy's theorem to a concyclic
quadrangle. (205°)
and
(233°)
OF ANHARMONIC PROPERTIES. 2J\
3. If the range ABCD is inverted with O as the centre of
inversion, the axis of the range inverts into a circle S through
O, and A, B, C, D invert into A', B', C, and D' respectively.
Hence, in general, anharmonic relations are unchanged by
inversion, a range becoming an equianharmonic system on a
circle, and under certain conditions vice versa.
4. In the inversion of 3, A and A', B and B', etc., are pairs
of inverse points.
OA.OA' = OB.OB'
= OC.OC'=OD.OD',
and the As OAB and OB'A', OAC and OCA', etc., are
similar in pairs.
And if P be the ± from O to AD, and P^, Pg, P3, and P4 be
the J_s from O to A'B', B'C, CD', and D'A', we have
AB A'B' BC B'C
P
Pi'
P
P2'
CD
P
CD'
P3'
DA
P ■
D'A'
P4'
AB}BC + CD4
DA_
=T^
B'C , CD'
"pT^pr
P
D'A'
„ P4*
But (232°) AB + BC + CD + DA=o,
A'B' ^ B'C , CD' ^ D'A'
And since the same principle applies to a range of any
number of points,
.*. in any concyclic polygon, if each side be divided by the
perpendicular upon it from any fixed point on the circle, the
sum of the quotients is zero.
. In the preceding theorem, as there is no criterion by which
we can distinguish any side as being negative, some of the
perpendiculars must be negative.
Of the perpendiculars one falls externally upon its side of
2/2
SYNTHETIC GEOMETRY.
the polygon and all the others fall internally. Therefore the
theorem may be stated : —
If _Ls be drawn from any point on a circle to the sides of an
inscribed polygon, the ratio of the side, upon which the X
falls externally, to its ± is equal to the sum of the ratios of
the remaining sides to their _Ls.
318°. Theorem. — If two circles be inverted the ratio of the
square on their common tangent to the rectangle on their
diameters is unchanged.
Let S, S' be the
circles and AD be the
common centre line,
and let the circles s
and s' and the circle
Z be their inverses
respectively.
Then Z cuts s and /
orthogonally, and
O{ABCDH0{A'B'C'D'}.
But if abed be the
common centre line of
s and s\ aPJ, bB\ eC,
and dl>' are concurrent
at O. (265', Ex. I)
0{rt^rrt'} = 0{A'B'C'D'}
= 0{ABCD},
{«^^^} = {ABCD},
A.C.BT> _ac.bd
AB.CD ab.ed'
But AC . BD = the square on the common direct tangent to S
and S', and ac . (^^= the square of the corresponding tangent
to s and s'. (i79°j Ex. 2)
And AB . CD and ab.ed diXQ the products of the diameters
respectively.
or
OF ANHARMONIC PROPERTIES. 2/3
And the theorem is proved.
Cor. I. Writing the symbolic expressions {ABCD} and
{adcd} in another form, we have
AD.BC _ad.dc
AB.CD ad.cd'
And AD . BC and ad. be are equal to the squares on the
transverse common tangents respectively.
Cor. 2. If four circles Sj, S2, S3, S4 touch a line at the
points A, B, C, D, and the system be inverted, we have four
circles s^, s^^ s^, ^"4, which touch a circle Z through the centre
of inversion.
Now let </i, </2' ^3» ^4 be the diameters of Sj, S2, etc., and let
^i> ^2> ^35 ^4 be the diameters of j^, s^^ etc., and let t^^ be the
common tangent to s^ and s^, /g^ be that to s^ and J4, etc.
Then AB, etc., are common tangents to S^ and 82, etc.^
and AB.CD + BC.AD + CA.BD=o. (233°)
And
AB2
d,d^
CD2
5354'
AB.CD ^
sld^d^d^^
_ ^12^34
and similar equalities for the remaining terms,
/l2^34 + ^23^1 4 + ^31^4 " ^•
This theorem, which is due to Dr. Casey, is an extension
of Ptolemy's theorem. For, if the circles become point
circles, the points form the vertices of a concyclic quadrangle
and the tangents form its sides and diagonals.
If we take the incircle and the three excircles of a triangle
as the four circles, and the sides of the triangle as tangents,
we obtain by the help of Ex. i, 135°, (^2 _ ^2 + ^2 _ ^2 + ^2 _ ^2
as the equivalent for /12/34 + etc. ; and as this expression is
identically zero, the four circles given can all be touched by
a fifth circle.
S
274
SYNTHETIC GEOMETRY.
319°. Let A, B, C, D, E, F be six points on a circle so con
nected as to form a iiexagram,
i.e.^ such that each point is con
nected with two others.
Let the opposite sides AB,
DE meet in P ; BC, EF in Q ;
and CD, FA in R.
To prove that P, Q, and R are
coUinear.
0{BDER} = Q{CDER}
= F{CDEA}
= B{CDEP}
= Q{BDEP},
.'. QR and QP are in line.
.•. if a hexagram have its
vertices concyclic, the points of
intersection of its opposite sides
in pairs are collinear.
Dcf. — The line of collinearity is called the pascal of the
hexagram, after the famous Pascal who discovered the
theorem, and the theorem itself is known as Pascal's
theorem.
Cor. I. The six points may be connected in 5x4x3x2 or
120 different ways. For, starting at A, we have five choices
for our first connection. It having been fixed upon, we have
four for the next, and so on to the last. But onehalf of the
hexagrams so described will be the other half described by
going around the figure in an opposite direction. Hence, six
points on a circle can be connected so as to form 60 different
hexagrams. Each of these has its own pascal, and there are
thus 60 pascal lines in all.
When the connections are made in consecutive order about
the circle the pascal of the hexagram so formed falls without
the circle ; but if any other order of connection is taken, the
pascal may cut the circle.
OF ANIIARMONIC PROPJiRTIKS. 2/5
Cor 2. In the hexagram in the figure, the pascal is the line
through P, O, R cutting the circle in H and K. Now
C{KFBD} = C(KFQR} ^ —
= F{KCOR} a//\ \c
= F{KCEA}, /W \yf\
{KFBD}={KCEA}, ( \j\)W j
and K is a common point to two equi hV ^/V/^SAP J^
anharmonic systems on the circle. So xTX / ^xf
also is H. B
These points are important in the theory of homographic
systems.
Cor. 3. Let i, 2, 3, 4, 5, 6 denote six points taken consecu
tively upon a circle. Then any particular hexagram is denoted
by writing the order in which the points are connected, as for
example, 2461352.
In the hexagram 246135 the pairs of opposite sides are 24
and 13, 46 and 35, 61 and 52, and the pascal passes through
their intersections.
Now taking the four hexagrams
246135, 245136, 246315, 2453I6,
the pascal of each passes through the intersection of the
connector of 2 and 4 with the connector of i and 3. Hence
the pascals of these four hexagrams have a common point.
It is readily seen that inverting the order of 2 and 4 gives
hexagrams which are only those already written taken in an
inverted order.
.•. the pascals exist in concurrent groups of four, meeting
at fifteen points which are intersections of connectors.
Cor. 4. In the hexagram 1 352461 consider the two triangles
formed by the sides 13, 52, 46 and 35, 24, 61. The sides 13
and 24, 35 and 46, 52 and 61 intersect on the pascal of
1352461, and therefore intersect collinearly.
Hence the vertices of these triangles connect concurrently,
i.e., the line through the intersection of 35 and 61 and the
intersection of 52 and 46, the line through the intersection of
2/6
SYNTHETIC GEOMETRY.
35 and 24 and the intersection of I3 and 46, and the line
through the intersections of 24 and 61 and the intersection
of 13 and 25 are concurrent.
But the first of these Hnes is the pascal of the hexagram
1643521, the second is the pascal of the hexagram 3564213,
and the third is the pascal of the hexagram 4256134.
.*. the pascals exist in concurrent groups of three, meeting
at 20 points distinct from the 1 5 points already mentioned.
Cor. 5. If two vertices of the hexagram coincide, the figure be
comes a pentagram, and the missing side becomes a tangent.
.*, if a pentagram be inscribed in a circle and a tangent at
any vertex meet the opposite side, the point of intersection
and the points where the sides about that vertex meet the
remaining sides are collinear.
Ex. I. The tangents at opposite vertices of a concyclic quad
rangle intersect upon the external diagonal of the quadrangle.
Ex. 2. ABCD is a concyclic quadrangle. AB and CD
meet at E, the tangent at A meets BC at G, and the tangent
at B meets AD at F. Then E, F, G are collinear.
320°. Let six tangents denoted by the numbers i, 2, 3, 4, 5,
and 6 touch a
circle in A, B, C,
D, E, and F.
And let the
points of inter
section of the
tangents be de
noted by 12, 23,
34, etc.
Then the tan
gents form a
hexagram about
the circle.
Now, 12 is
the pole of AB, and 45 is the pole of ED. Therefore the
OF ANHARMONIC PROPERTIES. 277
line 12.45 is the polar of the point of intersection of AB
and ED.
Similarly the line 23 . 56 is the polar of the intersection of
BC and EF, and the line 34.61 is the polar of the intersec
tion of CD and FA.
But since ABCDEF is a hexagram in the circle, these
three intersections are collinear. (319°)
.*. the lines 12 . 45, 23 . 56, and 34. 61 are concurrent at O.
And hence the hexagram formed by any six tangents to a
circle has its opposite vertices connecting concurrently.
Def. — The point of concurrence is the Briaiichon point, and
the theorem is known as Brianchoiis theorei7i.
Cor. I. As the six tangents can be taken in any order to
form the hexagram, there are 60 different hexagrams each
having its own Brianchon point.
Now take, as example, the hexagram formed by the lines
123456 taken in order.
The connectors are 12 . 45, 23 . 56, 34. 61, and these give
the point O.
But the hexagrams 126453, 123546, and T26543 all have
one connector in common with 123456, namely, that which
passes through 12 and 45. Hence the Brianchon points of
these four hexagrams lie upon one connector.
.*. the 60 Brianchon points lie in collinear groups of four
upon 15 connectors of the points of intersection of the
tangents.
Cor. 2. Consider the triangles 12.56.34 and 45.23.61.
These have their vertices connecting concurrently, and there
fore they have their sides intersecting collinearly.
But the point of intersection of the sides 61 . 23 and 56. 34
is the Brianchon point of the hexagram formed by the six
lines 234165 taken in order ; and similar relations apply to
the other points of intersection.
Hence the 60 Brianchon points lie in collinear groups of
three upon axes which are not diagonals of the figure.
2/8 SYNTHETIC GEOMETRY.
Cor. 3. Let two of the tangents become coincident.
Their point of intersection is then their common point of
contact, and the hexagram becomes a pentagram.
.•. in any pentagram circumscribed to a circle the join of
a point of contact with the opposite vertex is concurrent with
the joins of the remaining vertices in pairs.
Ex. I. In any quadrangle circumscribed to a circle, the
diagonals and the chords of contact are concurrent.
Ex. 2. In any quadrangle circumscribed to a circle, the
lines joining any two vertices to the two points of contact
adjacent to a third vertex intersect on the join of the third
and the remaining vertex.
SECTION IV.
OF POLAR RECIPROCALS AND RECIPROCATION.
321°. The relation of pole and polar has already been
explained and somewhat elucidated in Part IV., Section V.
It was there explained that when a figure consists of any
number of points, and their connecting lines, another figure
of the same species may be obtained by taking the poles of
the connectors of the first figure as points, and the polars of
the points in the first figure as connecting lines to form the
second.
And as the first figure may be reobtained from the second
in the same way as the second is obtained from the first, the
figures are said to be polar reciprocals of one another, as
being connected by a kind of reciprocal relation. The word
reciprocal in this connection has not the same meaning as in
184°, Def.
The process by which we pass from a figure to its polar
reciprocal is called polar reciprocation or simply reciprocation.
OF POLAR RECIPROCALS AND RECIPROCATION. 279
322°. Reciprocation is effected with respect to a circle
either expressed or imphed. The radius and centre of this
reciprocating circle are quite arbitrary, and usually no
account need be taken of the radius. Certain problems in
reciprocation, however, have reference to the centre of re
ciprocation, although the position of that centre may gener
ally be assumed at pleasure.
From the nature of reciprocation we obtain at once the
following statements : —
1. A point reciprocates into a line and a line into a point.
And hence a figure consisting of points and lines reciprocates
into one consisting of lines and points.
2. Every rectilinear figure consisting of more than a single
line reciprocates into a rectilinear figure.
3. The centre of reciprocation reciprocates into the line at
00 , and a centreline of the circle of reciprocation reciprocates
into a point at 00 in a direction orthogonal to that of the
centreline.
4. A range of points reciprocates into a pencil of lines, and
the axis of the range into the vertex of the pencil. And
similarly, a pencil of lines reciprocates into a range of points,
and the vertex of the pencil into the axis of the range.
323°. Let O.LMNK be a pencil of four, and C be the
centre of reciprocation. ./
Draw the perpendiculars
C/ on L, C»z' on M, C«'
on N, and Qk' on K.
The poles of L, M, N,
K lie respectively on these
perpendiculars, forming a
range of points as /, m, 11^
k. Then
I. Evidently the
^LOM = ^/C;;/, ^MON = i.wC;/, etc.
.*. the angle between two lines is equal to that subtended at
280 SYNTHETIC GEOMETRY.
the centre of reciprocation by the poles of the lines ; and the
angle subtended at the centre of reciprocation by two points
is equal to the angle between the polars of the points.
2. Any pencil of four is equianharmonic with its polar
reciprocal range. And hence anharmonic or harmonic re
lations are not altered by reciprocation.
Def. — Points are said to be perpendicular to one another
when their joins with the centre of reciprocation are at right
angles. In such a case the polars of the points are perpen
dicular to one another.
324°. In many cases, and especially in rectilinear figures,
the passing from a theorem to its polar reciprocal is quite a
mechanical process, involving nothing more than an intel
ligent and consistent change in certain words in the statement
of the theorem.
In all such cases the truth of either theorem follows from
that of its polar reciprocal as a matter of necessity.
Take as example the theorem of 88°, " The three altitudes
of a A are concurrent,"
To get its polar reciprocal put it in the following form, where
the theorem and its polar reciprocal are given in alternate
lines : —
rp, ^ T^^^^f lines through the vertices) r a ^• ^
The three ^ . ^ , ., ,^of a A Perpendicular
(pomts on the sides )
to the opposite!^' ^^ [are]
( vertices ) (
concurrent
collinear.
To get a point J_ to a vertex we connect the vertex to the
centre of reciprocation, and through this centre draw a line X
to the connector. The point required lies somewhere on this
line. (323°, Def.)
And as the centre may be any point, we may state the
polar reciprocal thus : —
" The lines through any point perpendicular to the joins of
OF POLAR RECIPROCALS AND RECIPROCATION. 28 1
that point with the vertices of a triangle intersect the opposite
sides of the triangle collinearly." (252°, Ex. 8)
325°. Consider any two As These reciprocate into two
As ; vertices giving sides, and sides, vertices.
If the original As are in perspective their vertices connect
concurrently. But in reciprocation the vertices become sides
and the point of concurrence becomes a line of collinearity.
Hence the polar reciprocals of these As have their sides
intersecting collinearly and are in perspective.
.*. As in perspective reciprocate into As in perspective.
But any three concurrent lines through the vertices of a A
intersect the opposite sides in points which form the vertices
of a new A in perspective with the former.
Hence all cases of three concurrent lines passing through
the vertices of a A reciprocate into As in perspective with
the original. Such are the cases of the concurrence of the
three medians, the concurrence of the three altitudes, of the
three bisectors of the angles, etc.
326°. The complete harmonic properties of the tetragram
may be expressed in the two following theorems, which are
given in alternate lines, and are polar reciprocals to one
another : —
Fourj^'"^' determine by their j^^^^^^^^^^^^H six JP^"^^^^
(points) (connectors ) (lines, j
and the~""^«?= iof these by theirl'""='^^"'°"= Ideter
( mtersections ) ( connectors )
mine three newP°™'^l Thel™""^""^ lof any of the
(hnes. ) (mtersections)
three newjP°'"^^lwith the original sixP''^''^^form a har
(hnes ) (lines i
monic jP^"^^^
( range.
Other polar reciprocal theorems, which have been already
282 SYNTHETIC GEOMETRY.
given, are Pascal's and Brianchon's theorems with all their
corollaries, the theorems of Arts. 313" and 314°, of Arts.
315° and 316°, etc.
The circle, when reciprocated with respect to any centre of
reciprocation not coincident with its own centre, gives rise to
a curve of the same species as the circle, ?>., a conic section,
and many properties belonging to the circle, and particularly
those which are unaltered by reciprocation, become properties
of the general curve.
These generalized properties cannot be readily understood
without some preliminary knowledge of the conic sections.
SECTION V.
HOMOGRAPHY AND INVOLUTION.
327°. Let A, B and A', B' be fixed points on two lines, and
• ' ' ' — let P and P' be variable points,
BODE . .
___^ one on each line which so move
B' C D' E' as to preserve the relation
AP_, AT'
BP ' ' B'P"
where >6 is any constant ; and let C, C' ; D, D' ; E, E', etc.,
be simultaneous positions of P and P'.
Then the points A, B, C, D, E, etc., and A', B', C', D', E',
etc., divide homographically the lines upon which they lie.
AC _ ; A'C 1 AD _ ; A'D'
BC^^FC'^^'^BD^'B'D"
A C.BD _A^C.B'D^
AD . BC A'D' . B'C"
or {ABCD} = {A'B'C'D'}.
Similarly, {ABCE} = {A'B'C'E'}, {BCDE} = {B'C'D'E'}, etc.
HOMOGRAPHY AND INVOLUTION. 285
Evidently for each position of P, P' can have only one
position, and conversely, and hence the points of division on
the two axes correspond in unique pairs.
.'. two lines are divided homographically by two sets of
points when to each point on one corresponds one and only
one point on the other, and when any four points on one line
and their four correspondents on the other form equianhar
monic ranges.
Cor. I. If the systems of points be joined to any vertices O
and C, the pencils O.ABCD... and 0'. A'B'C'D'... are
evidently homographic, and cut all transversals in homo
graphic ranges.
Cor. 2. The results of Arts. 304°, Cors. 3 and 4, and of
Arts. 305° and 306° and their corollaries are readily extended
to homographic ranges and pencils.
The following examples of homographic division are given.
Ex. I. A line rotating about a fixed point in it cuts any two
lines homographically.
Ex. 2. A variable point confined to a given line determines
two homographic pencils at any two fixed points.
Ex. 3. A system of ^./.circles determines two homographic
ranges upon any Hne cutting the system.
Consider any two of the circles, let P, Q be the common
points, and let the line L cut one of the circles in A and A'
and the other in B and B'. Then the ^PBB' = Z.PQB', and
_PAB' = ^PQA'. .. £.APB = ^A'OB'.
Hence the segment BA subtends the same angle at P as
the segment at B'A' does at O. And similarly for all the
segments made in the other circles.
Ex. 4. A system of /.^.circles determines two homographic
ranges upon every line cutting the system.
284 SYNTHETIC GEOMETRY.
DOUBLE POINTS OF HOMOGRAPHIC SYSTEMS.
328°. Let ABCD... and A'B'C'D'... be two homographic
^ ^ ^ ^ B c p ranges on a common axis.
° A' B' c D' If any two correspondents
from the two ranges become coincident the point of co
incidence is a double point of the system.
If A and A' were thus coincident we would have the rela
tions {ABCE} = {AB'C'E'}, etc.
Thus a double point is a common constituent of two equi
anharmonic ranges, of which the remaining constituents are
correspondents from two homographic systems upon a
common axis.
ABC and A'B'C being fixed, let D and D' be two variable
correspondents of the doubly homographic system.
Then {ABCD}={A'B'CD'},
, BD.AD __ BC.AC _j»
""^^^"^ AD.B'D'ACTFC"? ^'
Now taking O, an arbitrary point on the axis, let
OD=ji', OV>' = x', OK=a, OA'=a', 0B = ^, 0B'=^'.
ThenBD=x<^, B'D'=x'd', AD=xa, A'D' = x'a',
(xd)(x'a') _p
{xa){x'b')—q'
which reduces to the form
xx' + P;r+ Q;r' + R =0.
When D and D' become coincident x' becomes equal to x
and we have a quadratic from which to determine x^ i.e., the
positions of D and D' when uniting to form a double point.
Hence every doubly homographic system has two double
points which are both real or both imaginary, and of which
both may be finite, or one or both may be at infinity.
Evidently there cannot be more than two double points,
for since such points belong to two systems, three double
points would require the coincidence of three pairs of corre
spondents, and hence of all. (306°)
HOMOGRAPHY AND INVOLUTION. 285
329°. If D be one of the double points of a doubly homo
,. , DB.DA' CB.C'A' P
graphic system, jj^^jj, = ,^^^ = ^, say.
Now DB.DA' and DA.DB' are respectively equal to the
squares on tangents from D to any circles passing through
B, A' and B', A.
But the locus of a point from which tangents to two given
circles are in a constant ratio is a circle coaxal with both.
(275°, Cor. 5>
Hence the following construction for finding the double
points.
Through A, B' and
A', B draw any two
circles so as to intersect
in two points U and
V, and through these
points of intersection pass the circle S", so as to be the locus
of a point from which tangents to the circles S and S' are in
the given ratio VP : VQ.
The circle S" cuts the axis in D, D, which are the required
double points.
Evidently, instead of A, B' and A', B we may take any pairs
of noncorresponding points, as A, C and A', C; or B, C and
B', C. The given ratio ^/P : JQ is different, however, for
each different grouping of the points.
Cor. I. When PQ, i.e., when ^=^^, the circle S"
BC B (_/
takes its limiting form of a line and cuts the axis at one finite
point or at none.
In this case both double points may be at 00 or only one
of them.
Cor. 2. If any disposition of the constituents of the system
causes the circle S" to lie wholly upon one side of the
axis, the double points for that disposition become imagin
ary.
2^6
SYNTHETIC GEOMETRY.
330°. Let L be the axis of a doubly homographic system.
Through any point O on the
circle S transfer the system,
by rectilinear projection, to
the circle. Then ABC...,
A'B'C... form a doubly
homographic system on the
circle.
Now, by connecting any
two pairs of noncorrespond
ents A, B' and A', B ; B, C
and B', C ; C, A' and C, A,
we obtain the pascal line
KH which cuts the circle in
two points such that
{KABC} = {KA'B'C'}.
(319°, Cor. 2)
Hence H and K are double points to the system on the
circle. And by transferring K and H back through the point
O to the axis L, we obtain the double points D, D of the
doubly homographic range.
Cor. I. When the pascal falls without the circle, the double
points are imaginary.
Cor. 2. When one of the joins, KO or HO, is
the double points is at co .
to L, one of
Cor. 3. If the system upon the circle with its double points
H and K be projected rectilinearly through any point on
the circle upon any axis M, it is evident that the projected
system is a doubly homographic one with its double points.
Cor, 4. Cor. 3 suggests a convenient method of finding the
double points of a given axial system.
Instead of employing a circle lying without the axis, employ
the axis as a centreline and pass the circle through any pair
of noncorrespondents.
HOMOGRAPHY AND INVOLUTION. 2Zj
Then from any convenient point on the circle transfer the
remaining points, find the pascal, and proceed as before.
331°. The following are examples of the application of the
double points of doubly homographic systems to the solution
of problems.
Ex. I. Given two nonparallel lines, a point, and a third
line. To place between the nonparallels a segment which
shall subtend a given angle at the given point, and be parallel
to the third line.
Let L and M be the non
parallel lines, and let N be
the third line, and O be the
given point.
We are to place a segment
between L and M, so as to
subtend a given angle at O and be  to N.
On L take any three points A, B, C, and join OA, OB, OC.
Draw A«, B/5, Qc all H to N, and draw Oa\ Ob', Oc' so as to
make the angles AO^', B0<^', COc' each equal to the given
angle.
Now, if with this construction a coincided with <7;', or b with
//, or c with c', the problem would be solved.
But, if we take a fourth point D, we have
OJABCD} = {K^CT^')={abcd) = {a!b'dd'].
.'. abed and a' b'c'd' are two homographic systems upon the
same axis. Hence the double points of the system give the
solutions required.
Ex. 2. Within a given A to inscribe a A whose sides shall
be parallel to three given lines.
Ex. 3. Within a given A to inscribe a A whose sides may
pass through three given points.
Ex. 4. To describe a A such that its sides shall pass
through three given points and its vertices lie upon three
given lines.
288 SYNTHETIC GEOMETRY.
SYSTEMS IN INVOLUTION.
332°. If A, A', B, B' are four points on a common axis,
whereof A and A', as also B and B', are correspondents, a
point O can always be found upon the axis such that
OA.OA' = OB.OB'.
This point O is evidently the centre of the circle to which
A and A', and also B and B', are pairs of inverse points, and
is consequently found by 257°. "Jp^Vb^
Now, let P, P' be a pair of variable conjugate points which
so move as to preserve the relation
OP.OP' = OA.OA' = OB.OB'.
Then P and P' by their varying positions on the axis deter
mine a double system of points C, C, D, D', E, E', etc.,
conjugates in pairs, so that
OA . OA'=OC. OC'OD . OD' = OE . OE' = etc.
Such a system of points is said to be in invohition^ and O
is called the centre of the involution.
When both constituents of any one conjugate pair lie upon
the same side of the centre, the two constituents of every
conjugate pair lie upon the same side of the centre, since
the product must have the same sign in every case.
With such a disposition of the points the circle to which
conjugatesare inverse points is real and cuts the axis in two
_< , , I , , , points F and F'.
able conjugates meet and become coincident
Hence the points F. F' are the double points or foci of the
system.
From Art. 311°, i, FF' is divided harmonically by every
pair of conjugate points, so that
FAF'A', FBF'B', etc., are all harmonic ranges.
When the constituents of any pair of conjugate points lie
upon opposite sides of the centre, the foci are imaginary.
HOMOGRAPHY AND INVOLUTION.
289
333°. Let A, A', B, B', C, C be six points in involution, and
let O be the centre.
Draw any line OPQ through O,
and take P and O so that
OPrOQ=OA.OA',
and join PA, PB, PC, and PC,
and also QA', OB', QC, and QC.
Then, •.• OA. OA' = OP. 0(), o
.. A, P, Q,A' are coney clic. .'. i.OPA = ^OA'Q.
Similarly, B, P, (2, B' are concycUc, and ^OPB = ^OB'0, etc.
z.APB = /.A'OB'.
Similarly, ^BPC = ^B'QC', ^CPC' = z.C'OC, etc
Hence the pencils P(ABCC') and (^(A'B'C'C) are equianhar
monic, or {ABCC} = {A'B'G'C[.
Hence also {ABB'C} = {A'B'BC'J , {AA'BCj = {A'AB'C'J.
And any one of these relations expresses the condition that
the six points symbolized may be in involution.
334°. As involution is only a species of homography, the
relations constantly existing between homographic ranges and
their corresponding pencils, hold also for ranges and pencils
in involution. Hence
1. Every range in involution determines a pencil in involu
tion at every vertex, and conversely.
2. If a range in involution be projected rectilinearly through
any point on a circle it determines a system in involution on
the circle, and conversely.
Ex. The three pairs of opposite connectors of any four
points cut any line in a sixpoint involution.
A, B, C, D are the four points, v. p d c
and P, F the line cut by the six
connectors CD, DA, AC, CB, BD,
and AB. Then
D{PQRR'} = D{CARB}
= B{CARD}
= B{0'P'RR'} = {P'Q'R'R},
T
(302°)
290 SYNTHETIC GKOxMETRY.
{P()RR'} = {P'0'R'R},
and the six points are in involution.
Cor. I. The centre O of the involution is the radical centre
of any three circles through PP', QQ', and RR'; and the
three circles on the three segments PP', OQ', and RR' as
diameters are coaxal.
When the order of PQR is opposite that of P'Q'R' as in the
figure, and the centre O lies outside the points, the coaxal
circles are of the /./.species, and when the two triads of
points have the same order, the coaxal circles are of the
f./.species.
Cor. 2. Considering ABC as a triangle and AD, BD, CD
three lines through its vertices at D, we have —
The three sides of any triangle and three concurrent lines
through the vertices cut any transversal in a sixpoint
involution.
Exercises.
1. A circle and an inscribed quadrangle cut any line through
them in involution.
2. The circles of a coaxal system cut any line through them
in involution.
3. Any three concurrent chords intersect the circle in six
points forming a system in involution.
4. The circles of a coaxal system cut any other circle in
. involution.
5. Any four circles through a common point have their six
radical axes forming a pencil in involution.
INDEX OF DEFINITIONS, TERMS, ETC.
The Numbers refer to the Articles.
Addition Theorem for Sine
and Cosine, .
Altitude, .
Ambiguous Case,
Angle, .
,, Acute, .
,, Adjacent internal
,, Basal, .
,, External,
,, Obtuse,
,, Reentrant, .
„ Right, .
,, Straight,
,, Arms of,
,, Bisectors of, .
,, Complement of,
,, Cosine of,
,, Measure of, . 41
,, Sine of,
,, Supplement of,
,, Tangent of, .
,, Vertex of,
Angles, Adjacent, .
,, Alternate, .
,, Interadjacent,
 ,, Opposite, .
236
87
66
31
40
49
49
49
40
89
36
36
32
43
40
213
207
213
40
213
32
35
11
73
39
Angles, Vertical, . . 39
, , Sum and Difference of, 35
,, of the same Affection, 65
Anharmonic Ratio, .
298
Antihomologous,
289
Apothem,
146
Arc, . . .
lOI
Area,
136
,, of a Triangle,.
1754
Axiom, . . . .
3
Axis of a Range,
230
,, Perspective, .
254
,, Similitude, .
294
,, Symmetry, .
lOI
Basal Angles, .
49
Biliteral Notation, .
22
Bisectors of an Angle,
43
Brianchon's Theorem,
320
Brianchon Point,
320
Centre of a Circle, .
92
,, Inversion,
256
,, Mean Position,
238
,, Perspective,
254
,, Simihtude,
. 281
291
292 SYNTHETIC
GEOMETRY.
Centreline,
• 95
Diameter,
• 95
Centrelocus, .
. 129
Difference of Segments, . 29
Centroid,
• 85
Dimension,
• 27
Circle,
. 92
Double Point, .
109, 328
Circle of Antisimilitude,
. 290
,, Inversion, .
. 256
Eidograph,
. 211^
,, Similitude,
. 288
Endpoints,
22
Circumangle, .
. 36
Envelope,
. 223
Circumcentre, .
86, 97
Equal, .
27, 136
Circumcircle, .
. 97
Equilateral Triangle
• 53
Circumference,
. 92
Excircle, .
. 131
Circumradius, .
• 97
External Angle,
. 49
Circumscribed Figure,
• 97
Extreme and Mean ]
Ratio, 183
Chord, .
95
Extremes,
• 193
Chord of Contact, .
114
Coaxal Circles,
273
Finite Line,
21
Collinear Points, . 13
I, 247
Finite Point, .
21
Commensurable,
. 150
Complement of an Angle,
40
Generating Point,
. 69
Concentric Circles, .
• 93
Geometric Mean,
. 169
Conclusion,
4
Given Point and Lin
e, . 20
Concurrent Lines, . S
5, 247
Congruent,
51
Harmonic Division,
. 208
Concyclic,
97
Harmonic Ratio,
• 299
Conjugate Points, etc..
267
Harmonic Systems, .
313, 314
Contact of Circles, .
291
Homogeneity, .
. 160
Continuity, Principle of,
104
nomographic Syster
IS, . 327
Corollary,
8
Homologous Sides, .
. 196
C.P. Circles, .
274
Homologous Lines
and
Cosine of an Angle,
213
Points,
. 196
Constructive Geometry,
117
Hypothenuse, .
88, 168
Curve, .
15
Hypothesis,
4
Datum Line, .
20
Incircle of a Triangle, . 131
Degree, .
41
Incommensurable, .
. 150
Desargue's Theorem,
253
Infinity, .
21, 220
Diagonal, . . 8
0, 247
Initial Line, .
20
Diagonal Scale,
211^
Inscribed Figure,
• 97
INDEX OF DEFINITIONS, TERMS, ETC. 293
Interadjacent Angles,
Inverse Points, . i79>
Inverse Figures,
Involution,
Isosceles Triangle, .
Join,
Limit,
Limiting Points,
Line,
Line in Opposite Senses,
Linesegment, .
Locus,
L.P. Circles, .
Magnitude,
Major and Minor, .
Maximum,
Mean Centre, .
Mean Proportional, .
Means, .
Measure,
Median, .
Median Section,
Metrical Geometry,
Minimum,
73
256
260
332 i
53 I
167
148
274
12
156
21
69
274
190
102
175
238
169
193
150
55
183
150
175
Orthogonal Projection, 167, 229
ril6, Ex. 6
Ninepoints Circle, I265, Ex. 2
Normal Quadrangle, . 89
Obtuse Angle, .
Opposite Angles,
Opposite Internal Angles,
Origin, . . . .
Orthocentre, .
Orthogonally, .
40
39
49
20
88
115
Pantagraph, .
Parallel Lines,
Parallelogram,
Pascal's Hexagram
Pascal Line, .
Peaucellier's Cell,
Pencil,
Perigon, .
Perimeter,
Perspective,
Perspective, Axis of,
,, Centre of.
Perpendicular, .
Physical line, .
Plane, . • • 10,
Plane Geometric Figure
Plane Geometry,
Point,
Point of Bisection,
,, Contact,
Point, Double, . 109
Pole,
Polar,
Polar Reciprocal
Polar Circle, Centre, etc.
Polygon, .
Prime Vector, .
Projection,
Proportion,
Proportional Compasses
Protractor,
70
80
319
319
211
203
36
146
254
254
254
40
12
7
10
II
13
30
109
328
266
266
268
266
132
20
167
192
2Ili
123
—r
Quadrangle, ^
Quadrilateral, j
Quadrilateral, complete,
80, 89
. 247
!94
SYNTHETIC GEOMETRY.
Uadian, .
Radical Axis, .
Radical Centre.
Radius, .
Radius Vector,
Range, .
Ratio,
Reciprocation,
Reciprocal Segments
Rectangle,
Rectilinear Figure,
Rediidio ad abstudtuii ,
Reentrant Angle,
Rhombus,
Right Angle, .
Right Bisector,
Rotation of a Line,
Rule,
Rule of Identity,
Scalene Triangle
Secant, .
Sector, .
Segment,
Semicircle,
Selfconjugate
Selfreciprocal
Sense of a Line,
Sense of a RectangI
Similar Figures,
Similar Triangles,
Sine of an Angle,
Spatial Figure,
"'1
alj
Square,
78,
207
273
276
92
32
230
188
321
184
82
14
54
89
82
36
42
2, 222
16
7
57
95
211^
21
102
268
156
161
206
n. 196
213
19
82, 121
Straight Angle, . . 36
„ Line, . . .14
Edge, . . 16
Sum of Linesegments, . 28
Sum of Angles, . . 35
Superposition, ... 26
Supplement of an Angle, 40
Surface, .... 9
Tangent, . . . 109
Tangent of an Angle, . 213
Tensor, . . . .189
Tetragram, — . . . 247
Theorem, ... 2
Trammel, . . .107
'Transversal, • • • 73
Trapezoid, ... 84
Triangle, ... 48
,, Acuteangled, etc., 77
,, Base of, . . 49
,, Equilateral, . 53
,, Isosceles,. . 53
,, Obtuseangled, TJ
Scalene, . . 57
,, Right Angled, . 77
Uniliteral Notation, . li
Unitarea, \ ^
Unitlength, /
Versed Sine of an Arc, . 176
Vertex of an Angle, . . 32
Vertex of a Triangle, . 49
Vertical Angles, . . 39
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