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right angles, the sum of the rectangles of the opposite sides is double
the area of the figure.

4. A BCD is a quadrilateral inscribed in a circle, and the diagonal
BD bisects AC : shew that the rectangle AD, AB is equal to the rect-
angle DC, CB.

5. If the vertex A of a triangle ABC be joined to any point in
the base, it will divide the triangle into two triangles such that their
circumscribed circles have radii in the ratio of AB to AC.

6. Construct a triangle, having given the base, the vertical angle,
and the rectangle contained by the sides.

7. Two triangles of equal area are inscribed in the same circle :
shew that the rectangle contained by any two sides of the one is to
the rectangle contained by any two sides of the other as the base of
the second is to the base of the first.

8. A circle is described round an equilateral triangle, and from any
point in the circumference straight lines are drawn to the angular
points of the triangle : shew that one of these straight lines is equal to
the sum of the other two.

9. A BCD is a quadrilateral inscribed in a circle, and BD bisects
the angle ABC : if the points A and C are fixed on the circumference
of the circle and B is variable in position, shew that the sum of AB
and BC has a constant ratio to BD.



THEOREMS AND EXAMPLES ON BOOK VT.



I. ON HARMONIC SECTION.

1. To divide a given straight line internally and externally so that
its segments may be in a given ratio.



H-



'^-.



L M A P\,/'B Q

6

Let AB be the given st. line, and L, M two other st. lines which
determine the given ratio; it is required to divide AB internally and
externally in the ratio L : M ,

Through A and B draw any two par' st. lines AH, BK.
From AH cut off Aa equal to L,
and from BK cut ofi Bb and B6' each equal to M, Bb' being taken in
the same direction as Aa, and B6 in the opposite direction.
Join ab, cutting AB in P;
join ab', and produce it to cut AB externally at Q.
Then AB is divided internally at P and externally at Q,
so that AP : PB = L : M,

and AQ:QB = L:M.

The proof follows at once from Euclid vi. 4.

Obs. The solution is singular; that is, only one internal and one
external point can be found that will divide the given straight line
into segments which have the given ratio.



360 Euclid's elements.



DEFINITION.



A finite straight line is said to be cut harmonically when it
is divided internally and externally into segments which have
the same ratio.



A P B O

Thus AB is divided harmonically at P and Q, if
AP: PBrrAQ : QB.
P and Q are said to be harmonic conjugates of A and B.
If P and Q divide AB internally and externally in the same ratio,
it is easy to shew that A and B divide PQ internally and externally
in the same ratio: hence A and B are harmonic conjugates of P
and Q.

Example. The base of a triangle is divided harmonically by the
internal and external bisectors of the vertical angle :
for in each case the segments of the base are in the ratio of the other
sides of the triangle. [Euclid vi. 3 and A.]

Ohs. We shall use the terms Arithmetic, Geometric, and Harmonic
Means in their ordinary Algebraical sense.

1. If AB is divided internally at P and externally at Q, in the
same ratio, then AB is the harmonic mean between AQ and A P.

For by hypothesis AQ : QB = AP : PB;
.-., alternately, AQ : AP = QB : PB,

that is, AQ : AP = AQ-AB : AB-AP,

which proves the proposition.

2, 1/ AB is divided harmonically at P and Q, and O is the middle
point o/AB;

then shall OP . 0Q = 0A2.



ORB Q



For since AB is divided harmonically at P and Q,
.-. AP : PB = AQ: QB;
.-. AP-PB : AP+PB = AQ-QB : AQ + QB,
or, 20P : 20A = 20A : 20Q;

.-. 0P.0Q = 0A2.

Conversely, if OP . OQ^OA^,

it may be shewn that

AP : PB=:AQ: QB;
that is, that AB is divided harmonically at P and Q.



THEOREMS AND EXAMPLES ON BOOK VI.



361




8. The Arithmetic, Geometric and Harmonic means of tivo straight
lines may be thus represented graphically.

In the adjoining figure, two tan-
gents AH, AK are drawn from any
external point A to the circle PHQK;
HK is the chord of contact, and the
st. line joining A to the centre O cuts
the 0"° at P and Q.

Then (i) AO is the Arithmetic
mean between AP and AQ : for clearly
AO = i(AP + AQ).
(ii) AH is the Geometric mean between AP and AQ:

for AH2=AP.AQ. iii. 36.

(iii) AB is the Harmonic mean between AP and AQ:

for OA.OB = OP2. Ex. 1, p. 238.

.-. AB is cut harmonically at P and Q. Ex. 1, p. 360.
That is, AB is the Harmonic mean between AP and AQ.
And from the similar triangles OAH, HAB,
OA : AH=AH : AB,
.-. A0.AB = AH2; VI. 17.

.-. the Geometric mean hetioeen tioo straight lines is the mean propor-
tional between their Arithmetic and Harmonic means.




4. Given the base of a triangle and the ratio of the other sides, to
find the locus of the vertex.

Let BC be the given base, and let
BAG be any triangle standing upon
it, such that BA : AC = the given
ratio :

it is required to find the locus of A.
Bisect the z BAG internally and
externally by AP, AQ.

Then BC is divided internally at P, and externally at Q,
so that BP : PC= BQ : QC = the given ratio;
.*. P and Q are fixed points.
And since AP, AQ are the internal and external bisectors of the
Z BAC,

.-. the / PAQ is a rt. angle;
.-. the locus of A is a circle described on PQ as diameter.



Exercise. Given three points B, P, C in a straight line: find the
locus of points at which BP and PC subtend equal angles.



362 EUCLID'S ELEMENTS.

DEFINITIONS.

1. A series of points in a straight line is called a range.
If the range consists of four points, of which one pair are har-
monic conjugates with respect to the other pair, it is said to be
a harmonic range.

2. A series of straight lines drawn through a point is called a
pencil.

The point of concurrence is called the vertex of the pencil,
and each of the straight lines is called a ray.

A pencil of four rays drawn from any point to a harmonic
range is said to be a harmonic pencil.

3. A straight line drawn to cut a system of lines is called a
transversal.

4. A system of four straight lines, no three of which are
concurrent, is called a complete quadrilateral.

These straight lines will intersect two and two in six points,
called the vertices of the quadrilateral ; the three straight lines
which join opposite vertices are diagonals.

Theorems on Harmonic Section.

1. If a transversal is drawn parallel to one ray of a harmonic
pencil, the otlier three rays intercept equal parts upon it: and con-
versely.

2. Any transversal is cut harmonically by the rays of a harmonic
pencil.

3. In a harmonic pencil, if one ray bisect the angle between the
other pair of rays, it is perpendicular to its conjugate ray. Conversely
if one pair of rays form a right angle, then they bisect internally arid
externally the angle between the other pair.

4. If A, B, C, D and a, b, c, d are harmonic ranges, one on each
of two given straight lines, and if Aa, Bb, Co, the straight lines which
join three pairs of corresponding points, meet at S; then will Dd also
pass through S.

5. If two straight lines intersect at O, and if O, C, B, D and O, c, b,
d are two harmonic ranges one on each straight line [the points corre-
sponding as indicated by the letters), then Co, Bb, Dd will be con-
current: also Cd, Bb, Do will be concurrent.

6. Use Theorem 5 to prove that in a complete quadrilateral in
which tlie three diagonals are drawn, the straight line joining any pair
of opposite vertices is cut harmonically by the other two diagonals.



THEOREMS AND EXAMPLES ON BOOK VI. 363



II. On centres of similarity and similitude.

1. If any two unequal similar figures are placed so that their
homologous sides are parallel, the lines joining corresponding points in
the two figures meet in a point, lohose distances from any two corre-
sponding points are in the ratio of any pair of homologous sides.




Let ABCD, A'B'C'D'be two similar figures, and let them be placed
so that their homologous sides are parallel; namely, AB, BC, CD,
DA parallel to A'B', B'C, CD', D'A' respectively:
then shall AA', BB', CC, DD' meet in a point, whose distances from
any two corresponding points shall be in the ratio of any pair of
homologous sides.

Let A A' meet BB', produced if necessary, in S.

Then because AB is par^ to A'B'; liyp.

.•. the A^ SAB, SA'B' are equiangular;

.-. SA : SA'=:AB : A'B'; vi. 4.

.-. AA' divides BB', externally or internally, in the ratio of AB to A'B'.
Similarly it may be shewn that CC divides BB' in the ratio of
BC to B'C.

But since the figures are similar,

BC : B'C=AB : A'B';

.-. A A' and CC divide BB' in the same ratio;

that is, AA', BB', CC meet in the same point S.

In like manner it may be proved that DD' meets CC in the

point S.

.-. AA', BB', CC, DD' are concurrent, and each of these lines is
divided at S in the ratio of a pair of homologous sides of the two
figures. Q. E. D.

Cor. If any line is draivn through S meeting any pair of homolo-
gous sides in K and K', the ratio SK : SK' is constant, and equal to the
ratio of any pair of homologous sides.

Note. It will be seen that the lines joining corresponding points
are divided externally or internally at S according as the correspond-
ing sides are drawn in the same or in opposite directions. In either
case the point of concurrence S is called a centre of similarity of the
two figures.



;^t)4 EUCLID'S ELEMENTS.

2. A common tangent STT' to two circles whose centres are C, C,
meets the line of centres in S. If through S anij straight line is
drawn meeting these tico circles in P, Q, and P', Q', respectively,
then the radii CP, CQ shall he respectively parallel to C'P', C'Q'.
Also the rectangles SQ . SP', SP . SQ' shall each he equal to the
rectangle ST . ST'.




Join CT, CP, CQ and C'T', C'P', C'Q'.
Then since each of the z ' CTS, C'T'S is a right angle, in. 18.
.-. CT is pari ^o C'T';
.'. the A* SCT, SC'T' are equiangular;
.-. SC : SC' = CT : C'T'
= CP : C'P';
.-. the A" SCP, SC'P' are similar; vi. 7.

.-. the zSCP = the Z SC'P';

.-. CPis pari to C'P'.
Similarly CQ is par^ to C'Q'.

Again, it easily follows that TP, TQ are par' to T'P', T'Q'
respectively ;

.-. the A" STP, ST'P' are similar.
Now the rect. SP . SQ = the sq. on ST; iii. 37.

.-. SP : ST=:ST : SQ, vi. 10.

and SP
.-. ST
.-. the rect. ST
In the same way it may be proved that

the rect. SP . SQ' = the rect. ST . ST'.

Q.B. D.

Cor. 1. It has been proved that

SC : SC' = CP : C'P';
thus the external common tangents to the two circles meet at a point
S which divides the line of centres externally in the ratio of the radii.

Similarly it may be shewn that the transverse common tangents
meet at a point S' which divides the line of centres internally in the
ratio of the radii.

Cor. 2. CC is divided harmonically at S and S'.

Definition. The points S and S' which divide externally and
internally the line of centres of two circles in the ratio of their radii
are called the external and internal centres of similitude respectively.



ST=:ST


:SQ,


ST-_SP'


:ST'


SQ=SP'


:ST'


ST'=SQ


. SP'.



THEOREMS AND EXAMPLES ON BOOK VI. 365



EXAMPLES.

1. Inscribe a square in a given triangle.

2. In a given triangle inscribe a triangle similar and similarly
situated to a given triangle.

3. Inscribe a square in a given sector of circle, so that two
angular points shall be on the arc of the sector and the other two
on the bounding radii.

4. In the figure on page 278, if Dl meets the inscribed circle in
X, shew tJiat A, X, D^ are collinear. Also if Al^ meets the base in
Y sheiv that llj is divided harmonically at Y and A.

o. With the notation on page 282 shew that O and G are respec-
tively the external and internal centres of similitude of the circum-
scribed and nine-points circle.

6. If a variable circle touches tioo fixed circles, the line joining
their points of contact passes through a centre of similitude. Distinguish
betioeen the different cases.

7. Describe a circle ivhich shall touch two given circles and pass
through a given point.

8. Describe a circle ichich shall touch three given circles.

9. Cj, Cg, C3 are the centres of three given circles; 1^, E^ are the
internal and external centres of similitude of the pair of circles ivhose
centres are C^, C3, and Ig, Eg, I3, Eg, have similar meanings with regard
to the other tivo pairs of circles: sheio that

(i) I^Cj, IgCg, I3C3 are concurrent;

(ii) the six points 1^, Ig, I3, Ej, Eg, E3, lie three and three on four
straight lines.



III. ON POLE AND POLAR.
DEFINITIONS.

(i) If in any straight line drawn from the centre of a circle
two points are taken such that the rectangle contained by their
distances from the centre is equal to the square on the radius,
each point is said to be the inverse of the other.

Thus in the figure given below, if O is the centre of the circle, and
if OP . OGl= (radius)^, then each of the points P and Q is the inverse
of the other.

It is clear that if one of these points is within the circle the other
must be without it.



366



EUCLID'S ELEMENTS.



(ii) The polar of a given point with respect to a given circle
is the straight line drawn through the inverse of the given point
at right angles to the line which joins the given point to the
centre : and with reference to the i)olar the given point is called
the pole.

Thus in the adjoining figure, if OP . OQ = (radius)^, and if through



L


P


M


H ,


/^


~~\ "






^



P and Ql, LM and HK are drawn perp. to OP; then HK is the polar
of the point P, and P is the pole of the st. line HK: also LM is the
polar of the j)oint Q, and Q the pole of LM.

It is clear that the polar of an external point must intersect the
circle, and that the polar of an internal point must fall without it :
also that the polar of a point oii the circumference is the tangent at
that point.



1. Now it has been proved [see Ex. 1,
page 233] that if from an external point P
two tangents PH, PK are drawn to a circle,
of which O is the centre, then OP cuts the
chord of contact H K at right -angles at Q,
so that

0P.0Q=(radius)2,
.-. H K is the polar of P with respect to the
circle. Def. 2.

Hence we conclude that

The Polar of an external point with
reference to a circle is the chord of contact of
tangents drawn from the given point to the circle




The following Theorem is known as the Reciprocal Property of
Pole and Polar.



THEOREMS AXD EXAMPLES ON BOOK VI.



367



2. If A and P are any two points, and if the polar of A with
respect to any circle passes through P, then the polar of P imist pass
through A.

Let BC be the polar of the point A
with respect to a circle whose centre is
O, and let BC pass through P :
then shall the polar of P pass through A.
Join OP; and from A draw AQ perp.
to OP. We shall shew that AQ is the
polar of P.

Now since BC is the polar of A,
.-. the Z ABP is a rt. angle;

Def. 2, page 360.

and the Z AQP is a rt. angle: Gonstr.

.'. the four points A, B, P, Q are concyclic ;

.-. Oa.OP = OA.OB III. 36.

= (radius)2, for CB is the polar of A:

.-. P and Q are inverse points with respect to the given circle.

And since AQ is perp. to OP,

.-. AQ is the polar of P.

That is, the polar of P passes through A.




A similar proof applies to the case when the given point A
without the circle, and the polar BC cuts it.



3. To prove that the locus of the intersection of tangents drawn to
a circle at tlie extremities of all chords which pass throxigh a given point
is the polar of that point.

Let A be the given point within the
circle, of which O is the centre.

Let H K he any chord passing through
A; and let the tangents at H and K
intersect at P :

it is required to prove that the locus of
P is the polar of the point A.

I. To shew that P lies on the polar
of A.

Join OP cutting HK in Q.
Join OA : and in OA produced take the
point B,

so that OA . OB = (radius)^, n. 14.
Then since A is fixed, B is also fixed.
Join PB.




3«)8 EUCLIU'S ELKMKNTS.

Then since H K is the chord of contact of tangents from P,

.-. OP . OQ = (radius)2. Ex. i. p. 233.

But O A . O B = (radius)2 ; Const r.

:. OP.OQ = OA.OB:
.-, the four points A, B, P, Q are concyclic.
.-. the z " at Q and B together = two rt. angles. iii. 22.

But the Z at Gl is a rt. angle; Constr.

.: the Z at B is a rt. angle.
And since the point B is the inverse of A; Constr.

.'. PB is the polar of A ;
that is, the point P lies on the polar of A.

II. To shew that any point on the polar of A satisfies the given
conditions.

Let BC be the polar of A, and let P be any point on it. Draw
tangents PH, PK, and let HK be the chord of contact.

Now from Ex. 1, p. 366, we know that the chord of contact HK
is the polar of P,

and we also know that the polar of P must pass through A ; for P is

on BC, the polar of A: Ex. 2, p. 367.

that is, HK passes through A.

.-. P is the point of intersection of tangents drawn at the ex-
tremities of a chord passing through A.

From I. and II. we conclude that the required locus is the polar
of A.

Note. If A is without the circle, the theorem demonstrated in
Part I. of the above proof still holds good ; but the converse theorem
in Part II. is not true for all points in BC. For if A is without the
circle, the polar BC will intersect it; and no point on that part of
the polar which is within the circle can be the point of intersection of
tangents.



We now see that

(i) The Polar of an external point with respect to a circle is the
chord of contact of tangents drawn from it,

(ii) The Polar of an internal point is the locus of tlie intersections
of tangents drawn at the extremities of all chords which pass through
it.

(iii) The Polar of a point on the circumference is the tangent at
that point.



THEOREMS AND EXAMPLES ON BOOK VI.



369



The following theorem is known as the Harmonic Property oi"
Pole and Polar.

4. Any straight line drawn through a point is cut harmonically
by the point, its polar, and the circumference of the circle.

Let AHB be a circle, P the given
point and HK its polar; let Paqh be any
straight line drawn through P meeting
the polar at q and the o*^" of the circle at
a and b :

then shall P, a, q, b he n harmonic
range.

In the case here considered, P is an
external point.

Join P to the centre O, and let PO
cut the 0*=^ at A and B : let the polar of
P cut the C/« at H and K, and PO at Q.




AHB.



Ex. 1, p. 366.



Then PH is a tangent to the
From the similar triangles OPH, HPQ,
OP : PH = PH : PQ,
.-. PQ. P0=:PH2

= Pa.P&.
.•. the points O, Q, a, b are concyclic:
.-. the ZaQA = the labO

=:the z Oa^ I. o.

=:the Z OQib, in the same segment.
And since QH is perp. to AB,
.-. the ZaQH=the z6QH.
.-. Q.q and QP are the internal and external bisectors of the Z aQ.b :
.: P, a, q, b is a, harmonic range. Ex. 1, p. 360.

The student should investigate for himself the case when P is an
internal point.

Conversely, it may be shewn that if throtigh a fixed point P ang
secant is draum cutting the circumference of a given circle at a, and h,
and if q is the harmonic conjugate at P tvith respect to a,h; then the
locus of q is the polar of P with respect to the given circle.

[For Examples on Pole and Polar, see p. 370.]



DEFINITION.

A triangle so related to a circle, that each side is the polar
of the opposite vertex is said to be self-conjugate with respect
to the circle.

H. E. 24



370 klclid's klkments.



KXAMPLES ON POLE AND POLAR.

1. Tlie straight line which joins any tico points is tlie polar with
respect to a given circle of the point of intersection of their polars.

2. The point of intersection of any two straight lines is the pole of
the straight line which joins their poles.

3. Find the locus of the poles of all straight lines which pas!<
through a given point.

4. Find the locus of the poles, toith respect to a given circle, of tan-
gents draion to a concentric circle.

5. If two circles cut one another orthogonally and PQ he any
diameter of one of them; shew that the polar of P with regard to the
other circle passes through Q.

6. If two circles cut one another orthogonally, the centre of each
circle is the pole of their common chord with respect to the other circle.

7. Any two points subtend at the centre of a circle an angle equal
to one of the angles formed by the polars of the given points.

8. O is the centre of a given circle, and AB a fixed straight line.

P is any point i/i AB; find the locus of the point inverse to P toith
respect to tlie circle.

0. Given a circle, and a fixed point O on its circumference : P /.s
any point on the circle : find the locus of the point inverse to P with
respect to any circle whose centre is O.

10. Given two points A and B, and a circle xchose centre is O;
sJiew that the rectangle contained by OA and the perpendicular from B
on the polar of A is equal to the rectangle contained by OB and the
perpendicular from A on the polar of B.

11. Four points A, B, C, D are taken in order on the circumference
of a circle; DA, CB intersect at P, AC, BD at Q and BA, CD in R :
.<(hew that the triangle PQR is self-conjugate with respect to the circle.

12. Give a linear constructioii for finding the polar of a given
point with respect to a given circle. Hence find a linear construction
for draicing a tangent to a circle from an external point.

13. If a triangle is self-conjugate with respect to a circle, the
centre of the circle is at the orthocentre of the triangle.

14. The polars, with respect to a given circle, of the four points of
a harmonic range form a harmonic pencil : and conversely.



THEOREMS AND EXAMPLES ON BOOK VI.



371



IV. ON THE RADICAL AXIS.

1. To find the locus of points from xchich the tangents drawn to
two given circles are equal.



Fig. 1.



Fig. 2.




Let A and B be the centres of the given circles, whose radii are a
and h ; and let P be any point such that the tangent PQ drawn to the
circle (A) is equal to the tangent PR drawn to the circle (B) :
it is required to find the locus of P.
Join PA, PB, AQ, BR, AB; and from P draw PS perp. to AB.
Then because PQ= PR, .-. PQ2=PR2.
But PQ2=PA2-AQ2; and PR^^PB^-BRS; i. 47.

.-. PA2-AQ2==PB2-BR-^;
that is, PS2 + AS2-a2=PS2 + SB2-Z,2; 1.47.

or, AS2-rt2 = SB2-t2.

Hence AB is divided at S, so that AS^- SB^^a^- fts.

.•. S is B, fixed point.
Hence all points from which equal tangents can be drawn to the
two circles lie on the straight line which cuts AB at rt, angles, so
that the difference of the squares on the segments of AB is equal to the
difference of the squares on. the radii.

Again, by simply retracing these steps, it may be shewn that in
Fig. 1 every point in 8P, and in Fig. 2 every point in SP exterior to
the circles, is such that tangents drawn from it to the two circles are
equal.

Hence we conclude that in Fig. 1 the whole line SP is the required
locus, and in Fig. 2 that part of SP which is without the circles.
In either case SP is said to be the Radical Axis of the two circles.

24-2



372 Euclid's elements.

Corollary. If the circles cut one another as in Fig. 2, it is clear
that the Radical Axis is identical tcith the straight line which passes
through the points of intersection of the circles; for it follows readily
from iiL 36 that tangents drawn to two intersecting circles from any
point in the common chord produced are equal.

2. The Radical Axes of three circles taken in pairs are concurrent.

Zi




Let there be three circles whose centres are A, B, C.
Let OZ be the radical axis of the 0» (A) and (B);
and OY the Badical Axis of the ©• (A) and (C), O being the point of
their intersection :
then shall the radical axis of the ©■ (B) and (C) pass through O.

It will be found that the point O is either without or within all
the circles.

I. When O is without the circles.

From O draw OP, OQ, OR tangents to the o' (A), (B), (C).

Then because O is a point on the radical axis of (A) and (B) ; Hup.

.. OP = OQ.

And because O is a point on the radical axis of (A) and (C), Hyp.

.: OP = OR,

.. OQ = OR;

.-. O is a point on the radical axis of (B) and (G),

i.e. the radical axis of (B) and (C) passes through O.

IL If the circles intersect in such a way that O is within
them all ;

the radical axes are then the common chords of the three circles
taken two and two ; and it is required to prove that these common
chords are concurrent. This may be shewn indirectly by iii. 35.

Definition. The point of intersection of the radical axes of three


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