RS-SMO 2010 Problem 4

Let $O$ be the circumcenter of triangle $ABC$ . A line through $O$ intersects the sides $CA$ and $CB$ at points $D$ and $E$ respectively, and meets the circumcircle of $ABO$ again at point $P \neq O$ inside the triangle. A point $Q$ on side $AB$ is such that $\frac{AQ}{QB}=\frac{DP}{PE}$ . Prove that $\angle APQ = 2\angle CAP$ .

Proposed by Dusan Djukic