Question

Let $\Gamma$ be a circle with centre $O.$ Let $\Lambda$ be another circle passing through $O$ and intersecting $\Gamma$ at points $A$ and $B.$ A diameter CD of at points A and B. A diameter $CD$ of $\Gamma$ intersects $\Lambda$ at a point $P$ different from $O.$ Prove intersects $\Lambda$ at a point P different from O. Prove that
$\angle$APC = $\angle$BPD .

Answer 1

Suppose that A' is a point on $\Lambda$ such that $\angle$A'PC = $\angle$BPD
Segments OA' and OB subtends same angle in the respective minor arcs, so OA' = OB
This shows that A lies on $\Gamma$ and hence A' = A
$\therefore \angle$APC = $\angle$BPD .