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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.
Solution
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.
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