Let $ABC$ be an isosceles triangle with $\angle B = 120^o$ . Points $P$ and $Q$ are chosen on the prolongations of segments $AB$ and $CB$ beyond point $B$ so that the rays $AQ$ and $CP$ intersect and are perpendicular to each other. Prove that $\angle PQB = 2\angle PCQ$. (A.Akopyan, D.Shvetsov)