Question

Two circles intersect at A and B. From a point, P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.

Answer 1

Proof:

A and B are the points intersecting the circles. Join AB.

∠P’PB = ∠PAB   ...(Alternate segment theorem)

∠PAB + ∠BAC = 180°   ...(1)  ...(PAC is a straight line)

∠BAC + ∠BDC = 180°  ...(2)

ABDC is a cyclic quadrilateral.

From (1) and (2) we get

∠P’PB = ∠PAB = ∠BDC

P’P and DC are straight lines.

PD is a transversal alternate angles are equal.

∴ P’P || DC.