KMO 2002 Problem 5

Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the perpendicular line from $A$ to $BC$ meet $\omega$ at $D$ . Let $P$ be a point on $\omega$ , and let $Q$ be the foot of the perpendicular line from $P$ to the line $AB$ . Prove that if $Q$ is on the outside of $\omega$ and $2\angle QPB = \angle PBC$ , then $D,P,Q$ are collinear.