Dividing one proportion by the other, cB aB dB Similarly, [A'B'C'D'] = B' cB dB But ; aB 6B hence, [ABCD] =[A'B'C'D'). Scholium.--The anharmonic ratio of a pencil of four rays is the anharmonic ratio of four points cut from the rays by any transversal. The angles of a pencil are the six angles formed by the four rays with each other, taken two and two. Hence, the values of the anharmonic ratios of two pencils are equal, when the rays make equal angles with each other, each to each. The anharmonic ratio of a pencil is represented by [O- ABCD]. Proposition 16. Theorem.-If four points be taken in the circumference of a circle, and lines be drawn to any other point in the circumference, the anharmonic ratio of the pencil is the same, whatever the position of the fifth point. For the angles at O and Oʻ are equal; hence (Prop. 15, Sch.), [O-ABCD] =[O-ABCD]. B Preposition 17. Theorem.-If the centre of a circle be joined with the points of intersection of four fixed tangents with a fifth tangent, the anharmonic ratio of the pencil formed is the same, whatever be the position of the fifth tangent. Let four tangents touch the circle in A, B, C, D, and intersect a fifth tangent in E, F, G, H; then will [O-EFGH] be the same, whatever the position of EFGH. For the angle EOK is one half AOK, and FOK is one half BOK; therefore EOF is one half the fixed angle A OB, and is therefore constant, whatever the position of EFGH. The same may be proved of FOG and GOH. Hence the anharmonic ratio of the pencil is the same, whatever the position of EFGH. K M N B A D Proposition 18. Problem.-To divide a straight line harmonically in a given ratio. Let AB be the line and M:N the given ratio ; it is required to divide AB har c monically in the ratio M:N. Lay off in any direction AC: BC :: AD : BD :: M : N. E proven that F Scholium.-- If any point not in AB be taken, and a pencil of four rays drawn through A, C, B, D, it is said to be a harmonic pencil. Proposition 19. Problem.—To find the locus of all points whose distances from two given points are in a given ratio. A or, Let A and B be the points, and let M : N be the given ratio. Divide (Prop. 18) AB harmonically in C and D; on CD describe a circle. The circumference of this A circle will be the required locus. Take any point, P, and join PA, PC, PB, PE, and PD. Because AB is divided harmonically in C and D, AD : AC :: BD : BC, AE+EC : AE-EC :: CE+EB : CE-EB; hence, (IV. 11), AE : EC :: EC : EB, or, AE: EP :: EP: EB. Therefore the sides about the common angle E of the triangles AEP, BEP are proportional; hence the triangles are similar, and the angle BPE is equal to A. Also ECP= CAP+CPA, and its equal EPC= EPB+BPC. But EPB = CAP; .. APC=BPC, and (V.3), AP : PB :: AC : CB :: M : N. But P is any point of the circumference; hence the circumference is the required locus. Corollary 1.-Because CPD is a right angle, and PC bisects the angle APB, PD must also bisect the angle BPF. Corollary 2.-If the angles APB, BPF be bisected, then AB is divided harmonically in C and D; and, conversely, if AB be divided harmonically in C and D, the angles APB, BPF will be bisected by PC and PD. SECTION VI. POLE AND POLAR IN THE CIRCLE. DEFINITIONS. B IF from a point, 0, without the circle ABD, the secant OAB be drawn, and C, the harmonic conjugate of O with respect to A and B, be taken, then the locus of C, as OAB revolves about 0, is called the polar of the point 0; and O is the pole of this locus. Proposition 20. Theorem.—The polar of a point is a straight line at right angles to the diameter through the point. Let 0 and C be harmonic conjugates with respect to A and B on the circumference ABD. Draw CE at right angles to the diameter DE; and on OC describe a semicircle passing B A K through F. Join BF, and let it cut the circle in G; join AF, BD, BE. Because AB is divided harmonically in 0 and C, and F is a point of the circumference OFC, B OF bisects (Prop. 19, Cor. 2) AFG; and DE is a diameter; hence the arcs AD, DG are equal, and consequently the H angles OBD, DBF; hence, also BE, drawn from B in the circumference DBE, bisects FBH, and therefore OF is divided (Prop. 19, Cor. 2) harmonically in D and E, and F is the harmonic conjugate of 0. Hence F is a fixed point, and C is always in the perpendicular to OE through F; therefore CF is the polar of the G K 0 с pole O. Corollary 1.—The radius of the circle is a mean proportional between the distances of the centre from the pole and the polar. For, because OE : OD :: FE : FD, as in last Prop., OK : DK :: DK: FK, or, OK. FK = DK? = RP. Corollary 2.—The pole and polar are on the same side of the centre; if the pole be exterior, the polar is interior to the circle, and vice versâ. Corollary 3.-The polar passes through the point of contact of the tangent through the pole, |