NOTE. There is a very close analogy between Ths. 58 (i) and 58 (ii); notice that the proofs are nearly identical. Ex. 6. State and prove the converse of this theorem. THEOREM 59 (i). If two chords of a circle intersect inside the circle, the rectangle contained by the segments of one is equal to the rectangle contained by the segments of the other*. B Fig. 125. Data AB and CD are two chords of a circle meeting in P. LAPD = < CPB (vert. opp.), and D = LB (in the same segment), .. the third angles are equal, and the As are equiangular, COR. The rectangles contained by the segments of all chords of a circle through the same point are equal. E To calculate the value of PA. PB. A Suppose that EPF is the chord which is bisected B at P, and O the centre of the circle. Then PA. PB PE. PF = PE2 = OE2 - OP2. (Why?) = *See note on p. 61. Fig. 126. THEOREM 59 (ii). If two chords of a circle intersect outside the circle, the rectangle contained by the segments of one is equal to the rectangle contained by the segments of the other. ▲ P is common, and ▲ D = ▲ B (in the same segment), .. the third angles are equal, and the As are equiangular, COR. (A PARTICULAR CASE.) If PT is a tangent to a circle and AB a chord of the circle passing through P, then PT2 = PA. PB (see fig. 128). NOTE. Theorems 59 (i) and 59 (ii) are really two different cases of the same theorem; notice that the proofs are nearly identical. B P Fig. 128. The Corollary is so important that an independent proof of it is given below. THEOREM 59 (COR.). If a chord of a circle is produced to meet a tangent, the square on the tangent from the point of intersection is equal to the rectangle contained by the segments of the chord. B Fig. 129. Data BA is a chord produced to meet the tangent at T in P. .. the third angles are equal, and the triangles are equiangular. COR. For all chords of a circle passing through P the rectangle PA. PB is constant and equal to OP2 - OT2 where O is the centre of the circle and PT is the tangent from P. TEX. 7. To construct a circle to pass through two given points A, B and to touch a given straight line XY. Show that P, the point of contact, may be found by the following construction. Draw any circle through A, B to cut XY in C, D. From O, the point where AB produced cuts XY, draw a tangent OT to this circle; with centre O and radius OT, draw a circle cutting XY in P, P'. Then the required circle touches XY at P or P'. How is the construction to be completed? What happens if A, B are on opposite sides of X, Y? What arises from the fact that either P or P' may be used? Discuss the case in which AB is parallel to XY. THEOREM 60. If a perpendicular is drawn from the right angle of a right-angled triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to one another. B Fig. 130. Data ABC is a triangle right-angled at A. AD is the perpendicular from A to BC. To prove that As ABC, DBA, DAC are similar to one another. ▲ B is common and ▲ BAC = ▲ BDA (rt. ▲ s). .. the third angles are equal. .. As ABC, DBA are equiangular with one another. Similarly As ABC, DAC,, .. As ABC, DBA, DAC are similar to one another. Q. E. D. COR. 1. AD2 = BD. DC [since DBA and DAC are similar]. CA2 CD. CB = [since DBA and ABC are similar]. [since DAC and ABC are similar]. COR. 3. BA2 + CA2= BC2 [by adding results in Cor. 2]. Ex. 8. ABC is a triangle right-angled at A; AD is an altitude of the triangle. Theorem 60 proves that As ABD, CAD are similar. Write down the three equal ratios; and, by taking them in pairs, deduce the corresponding rectangle properties. Ex. 9. Prove Cor. 1 by means of Th. 59 (i). CONSTRUCTIONS. To divide a line internally, or externally, in a given ratio. Ρ AB is the given line. Let the given ratio be the ratio of the lines and 9. Construction Draw a line from A at any angle to AB. Join YB and draw XC || to YB. Then C divides AB in the given ratio. |