the three remaining pairs of rays are in the same straight line, AD (7); that is, the diagonals AD, BE, and CF, intersect in the same point, O.

SCHOLIUM. Four corollaries follow from this theorem, similar to those in the preceding article. The number of sides is diminished by increasing an angle A, until the sides BA and AF are in the same straight line, and A is a point of contact. (Give the corollaries, and construct the figures.)

particular value, -1, the points A, B, C, and D, are called harmonic points, and the line AB is said to be divided harmonically in the points C and D.

agrees with Definition, p. 82, when the direction of the lines is considered.

DEF. 2. A harmonic pencil is a pencil of four rays, whose anharmonic ratio is equal to — 1.

COR. Any transversal cutting a harmonic pencil is divided harmonically (5).

XV.

Theorem. If a straight line, AB, is divided harmonically in the points C and D, and O is the centre of AB, then

A

OB2 = OC X OD;

C B D for, since AC: CB=AD : DB,

we have, by composition and division,

or,

hence

AC - CB: AC+CB=AD—DB: AD+DB;

20C: 20B = 20B : 20D:

OB2= OC X OD.

COR. Conversely, if O is the centre of AB, and OB2=

OC X OD, the line AB is divided harmonically in C and D. (Show this.)

XVI.

Theorem. If two circles cut each other orthogonally (that is, so that the tangents at a point of intersection are perpendicular to each other), any line passing through the centre of one circle, and cutting the other, is divided harmonically.

The tangent, OT, to one circle, passes through the centre, O, of the other (II., 11, Cor. 8); then OT2 = OB2 = OC × OD (III., 37, Cor. 2) hence AB is divided harmonically in C and D (15 Cor.).

COR. Conversely, if AB is cut harmonically in C and D, any circle passing through the conjugate points; C and D, is cut orthogonally by the circle whose diameter is AB.

T

B

XVII.

Theorem. In a complete quadrilateral, each diagonal is divided harmonically by the other two (I., 36).

In the complete quadrilateral, ABCDEF, the diagonals,

AC, DB, and EF, cut each other harmonically in the points L, M, N.

PROOF. In the triangle FDE, the lines FC, DL, and EA, pass through a common point, B: hence

AF X DCX EL=AD × CE × LF (III., 35).

Also, since the transversal CAM cuts

the triangle DEF, we have,

D

M

L

B

C

E

AD X CE X MF = AFX DC X EM (III., 33).

Multiplying these equations, member by member, we have, EL X MF=EM × LF :

hence

EL: LF EM: MF.

The lines DE, DL, DF, and DM, evidently form a harmonic pencil (14), and the line MC is cut harmonically (14, Cor.); that is,

CN: NA = CM: MA.

Also FM, FA, FN, and FC form a harmonic pencil: hence the line DL, which cuts them (produced), is divided harmonically; that is,

DN: NB = DL : LB.

POLES AND POLARS WITH RESPECT TO A

CIRCLE.

XVIII.

DEF. The perpendicular, P, drawn through one of two conjugate harmonic points, C and D, on the diameter of a circle, is called the polar of the other point; conversely, this point is the pole of the line P (Fig. p. 199).

From III., 19, Cor. 2, it follows, 1st, If the pole is within the circle, the polar is without, and conversely; 2d, As the one approaches the centre, the other recedes to an infinite distance from the circle; 3d, The pole and polar approach the circumference at the same time, and the pole of a tangent is its point of contact to the circle.

T

XIX.

Theorem. If from a point, D, without a circle, two tạngents be drawn, the line TT, passing through the points of contact, is the polar of the given point.

C B

A

T

For TT'L AD (II., 18, Cor. and 14); also ▲ OTC

~

OTD

(III., 22 and II., 11, Cor. 6) :

hence

or,

OD: OTOT: OC;

OTOB2 OCX OD,

and AB is divided harmonically, in the points C and D (15, Cor.): hence TT' is the polar of the point D.

This theorem furnishes an easy solution to the following problems.

XX.

PROBLEM. Given a point to construct a polar with respect to a circle, and, conversely, given a straight line to find its pole with respect to a circle. (Give solutions of the four cases.)

XXI.

PROBLEM. Given three points in a straight line to find the fourth harmonic point.

XXIL

Theorem. If, about a point in the plane of a circle, a secant be turned, the polar of the fixed point will be the locus of the fourth harmonic point.

HYPOTH. DP is the polar

of the fixed point C, and MNP is any position of the secant turned about C.

TO BE PROVED. P is the A fourth harmonic point of M, C, and N.

PROOF. Draw ON, OM, DN, and DM.

Then

hence

and

hence

C B

▲ OCN ~ OND (III., 23):
ONCODN.

In the same manner it may be shown that

Therefore

and the complements of these angles are equal. DA and DP bisect the angles made by MD and ND, and hence M, C, N, and P are four harmonic points (III., 19).

If the fixed point C is without the circle, and its polar within, the demonstration is the same, except that the angles ONC and OMC are supplements of each other.

COR. 1. If any number of points on a straight line be taken as poles, their polars with respect to a circle pass through a common point, which is the pole of the given line, and conversely. For any point P on a line DP is the conjugate harmonic point of C, the pole of DP.

COR. 2. If, from any number of points on a straight line, pairs of tangents to a circle be drawn, the chords joining their point of contact pass through one common point, which is the pole of the given straight line, and conversely (19).

Hence, as the pole moves on a straight line, DP, its polar, revolves about a fixed point, C, and conversely.

EXERCISE. If the pole moves on a line passing through the centre of the circle, where is the fixed point about which the polar revolves? If DP is a tangent, where are the points M, N, and C? If DP passes through the centre of the circle, construct those points.

The following is another demonstration of Cor. 1.

XXIII.

Theorem. The polars of all the points of a straight line pass through a common point, which is a pole of that line.