Question

Let Γ be a circle with centre O. Let A be the another circle passing through O and intersecting Γ at points A and B. A diameter CD of Γ intersects A at point P different from O. Prove that ∠APC = ∠BPD.

Answer 1

Suppose that A' is a point on A such that ∠A'PC = ∠BPD. Then the segments OA' and OB subtends same angle in the respective minor arcs, so OA' = OB. This shows that A lies on Γ and hence A' =A.

This proves that ∠APC = ∠BPD.