ELMO SL 2012 G7

Let $\triangle ABC$ be an acute triangle with circumcenter $O$ such that $AB<AC$ , let $Q$ be the intersection of the external bisector of $\angle A$ with $BC$ , and let $P$ be a point in the interior of $\triangle ABC$ such that $\triangle BPA$ is similar to $\triangle APC$ . Show that $\angle QPA + \angle OQB = 90^{\circ}$ .

Alex Zhu.