IR-TST 2009 Problem 5

$ABC$ is a triangle and $AA'$ , $BB'$ and $CC'$ are three altitudes of this triangle . Let $P$ be the feet of perpendicular from $C'$ to $A'B'$ , and $Q$ is a point on $A'B'$ such that $QA = QB$ . Prove that : $\angle PBQ = \angle PAQ = \angle PC'C$