98. Scholium. Let AB be a given segment and M its middle point, and let C move along the line AB. The conjugate point D will then also move along the same line. When C is at B, D is at the same point. As C moves toward M, the motion of D from B is at first very little greater than that of C; but as C approaches M, D moves more rapidly, and when C is indefinitely near to M, the distance of D is infinite. When C is to the left of M, D is to the left of A, and the two points approach A simultaneously. If C is in AB produced, D is on the segment AB.

.

99. Def. A system of straight lines diverging from a point is called a pencil; each diverging line is called a ray, and the point from which they diverge is called the vertex of the pencil.

If from the point V straight lines be drawn to four harmonic points, A, C, B, D, the four lines, VA, VC, VB, VD, form an harmonic pencil.

It is evident that if AB is divided harmonically by the pencil, any straight line, acbd, drawn parallel to AD will also be divided harmonically by the pencil.

PROPOSITION III.

100. Every straight line which cuts an harmonic pencil is divided harmonically by the pencil.

Let the pencil V.ACBD divide the transversal ACBD harmonically, so that AC.BD=AD.CB; then will it divide any other transversal afgh harmonically.

Through the point a draw ad parallel to AD. Since the triangle agb is cut by the transversal VC, we have (Art. 84)

And since the same triangle is cut by the transversal Vd, we have

Multiplying together (1) and (2), and omitting common factors, we have af.gh.bc.ad ah.fg.ac.bd.

But since ad is parallel to AD, we have

that is, ag is divided harmonically.

101. Hence the rays VA, VB are called conjugate rays with respect to

the rays VC, VD, and are said to divide the angle AVB harmonically; also the rays VC, VD are conjugate rays with respect to the rays VA, VB, and divide the angle AVB harmonically.

102. Scholium. The preceding demonstration applies when the transversal cuts one or more of the rays on opposite sides of the vertex, as in the an- a nexed figure.

PROPOSITION IV.

103. Three straight lines of an harmonic pencil being given, it is required

to find the fourth ray of the pencil.

Let VA, VB, VC be three rays of an harmonic pencil;. it is required to find the fourth ray VD, conjugate to VC.

On VC take any point C, and draw CE parallel to VA. Take EF equal to VE; join CF, and produce it to G; and through V draw VD parallel to GF.

The line VD is conjugate to VC.

Since the triangles AGC, AVD are similar, we have

GC:VD::AC: AD.

Also, since the triangles CFB, VBD are similar, we have

VD: CF::BD: CB.

B

Multiplying together these proportions, and observing that GC=CF, AC.BD AD.CB.

since VE EF, we have

Hence AD is divided harmonically in C and B, and the ray VD is conjugate to the ray VC.

.104. Def. If the opposite sides of any quadrilateral be produced to meet, the figure thus formed is called a complete quadrilateral.

Let ABCD be any quadrilateral, and let its opposite sides be produced to meet in E and F; the line EF is called the third diagonal of the quadrilateral, and the whole figure formed by the four lines meeting in six points forms a complete quadrilateral. The complete quadrilateral has three diagonals: viz., two interior, AC, BD, and one exterior, EF.

PROPOSITION V.

105. Each diagonal of a complete quadrilateral is harmonically divided by the two other diagonals.

Let ABCDEF be a complete quadrilateral, and let AC, BD, EF be its diagonals; any one of them, as EF, is divided harmonically by the two others in the points G and H.

The triangle AEF cut by the transversal BDH (Art. 84)*gives

E

B

G

F

H

In the same triangle the straight lines AG, ED, BF, drawn from the vertices and intersecting in C (Art. 86), give the equation FG.AD.EB EĢ.FD.AB. (2)

Multiplying together (1) and (2), and omitting the common factors, we have EH.FG FH.EG.

In a similar manner it may be proved that each of the diagonals AC, BD is harmonically divided by the two other diagonals.

PROPOSITION VI. ·

106. The middle points of the three diagonals of a complete quadrilateral are in a straight line.

K, the middle point of AC, and NO passes through L, the middle point of BD.

If we regard ADF as a transversal of the triangle BCE, we obtain (Art. 84) AB.DE.FC-AE.DC.FB.

Dividing each factor by 2, and observing that AB KN, DE=LO, FC MP, etc., we have KN.LO.MP.=KP.LN.MO.

Therefore the points K, L, M, lying in the sides of the triangle NOP, satisfy Art. 85, and are in the same straight line.

POLES AND POLARS WITH RESPECT TO AN ANGLE. 107. Def. If through a fixed point, P, a line, PAB, be drawn cutting

P

B

the two sides of an angle, XVY, and if on PB a point Q be taken, which is the harmonic conjugate of P with respect to the points A and B, the line VQ is called the polar of the point P with respect to the sides of the angle XVY, and P is called the pole of VQ.

PROPOSITION I.

108. Any straight line drawn through a pole to meet the sides of an angle is divided harmonically by these lines and the polar.

Let P be the pole and VQ the polar with respect to the lines VX, VY, and let A and B be the points in which a straight line drawn through P meets the two lines. Since VQ is the polar of the point P, the points P, A, Q, and B form an harmonic system; and if we draw the line VP, the lines VP, VA, VQ, VB form an harmonic pencil, and any straight line Pb drawn through P will be divided harmonically by the pencil (Prop. 100).

PROPOSITION II.

109. A point P and an angle XVY being given, if from P we draw two lines cutting the sides of the angle in A, B, a, b, and if we draw the diagonals Ab, Ba intersecting in Q, the line VQ will be the polar of the point P with respect to the lines VX, VY.

If we consider the complete quadrilateral VaQb, AB, we find the diagonal AB to be divided harmonically in P and C by the two other diagonals ab, VQ (Art. 105); hence the four lines VP, VX, VQ, VY form an harmonic pencil, and the straight line VQ is the polar of the point P with respect to the two lines VX, VY.

P

B

C

X

110. Scholium. The polar of any given point with respect to an angle may also be found by the method of Art. 103.

POLES AND POLARS WITH RESPECT TO A CIRCLE.

111. Def. If from a fixed point P a line be drawn to C, the centre of a circle, and on CP a point p be taken such that CP.Cp is equal to the square of the radius of the circle, the straight line Dp perpendicular to the line CP at the point p is called the polar of P with respect to the circle, and the point P is called the pole of Dp with respect to the circle.

PROPOSITION I.

112. Any straight line drawn through the pole to meet a circle is divided harmonically by the circle and the polar.

Let P be the pole and pD be the polar, and let A and B be the points in which a straight line drawn through P meets the circumference. Join CA, CB, pA, pB, and produce Bp to meet the circumference in F.

t

E

G

C

F

T

B

H

Then, since CP.Cp=CB', we have
CP:CB::CB:Cp;

and therefore the angle CBP is equal
to the angle CpB (B. IV., Pr. 21).
Also, since CP.Cp-CA2, we have
CP:CA::CA:Cp;
and therefore the angle CAP is equal
to the angle CpA. But CBP is equal
to CAB, which is the supplement of
CAP; therefore CpB is the supple-

ment of CpA. Hence CpB, which is equal to PpF, is equal to PpA. Also ApD and BpD are equal, being the complements of ApP and BpC. Therefore Dp bisects the vertical angle of the triangle ApB, and Pp bisects the exterior angle of the same triangle; hence the line AB is divided harmonically (Art. 95) in P and E.

113. Cor. 1. To find the pole of a given straight line DT, draw a diameter GH perpendicular to the given line intersecting it in p, and on this diameter produced, take P such that CP.Cp=CG2; the point P is the pole of DT with respect to the circle.

114. Cor. 2. When the point P is without the circle, its polar is the line Tt joining the points of contact of the tangents from P. For let T and t be the points where Dp intersects the circumference, then, since CP.Cp=CT2, we have CP:CT::CT: Cp.

Hence the two triangles CPT, CpT are similar; therefore CTP is a right angle, and PT is a tangent to the circle at the point T. For the same reason a line drawn from P to the point t is a tangent to the circle.

115. Cor. 3. If the point P approaches the circumference of the circle, the point p will also approach it, and when PC becomes equal to the radius of the circle, Cp will also be equal to the radius of the circle; therefore the polar of any point on the circumference is the tangent at that point.

PROPOSITION II.

116. The polars of all the points of a straight line pass through the pole of that line.

Let AB be any straight line, and P any point in it. Draw CP, and take p such that CP.Cp=CD2, the straight line pE perpendicular to CP will contain the pole of the straight line AB.

Draw CF perpendicular to AB cutting Ep in Then, since the triangles CpE, CFP are similar, we have

E.

CF.CE CP.Cp.
CP.Cp=CD*.