Question 12
Let $O$ be the center of the circumscribed circle $\triangle ABC$. The line through $O$ intersects the sides $CA$ and $CB$ at points $D$ and $E$, respectively, and the inscribed circle $\triangle ABO$ at point $P$ inside the triangle (different from $O$). Point $Q$ on page $AB$ is such that $\frac{AQ}{QB} = \frac{DP}{PE}$. Prove that $\angle APQ = 2 \cdot \angle CAP$. (Dusan Djukic)