Saudi Arabian IMO Booklet

Let $D$ be the vertex of parallelogram $ABDC$. Then $APDC$ and $AQDB$ are isosceles trapezoids. Therefore the perpendicular bisectors to segments $PD$ and $QD$ coincide with the perpendicular bisectors to $AC$ and $AB$ respectively, the circumcenter $O$ of triangle $ABC$ is also the circumcenter of $DPQ$ and $\angle POQ=2\angle A$. Also since $$\angle XPD=\angle ADP,\angle XQD=\angle ADQ$$ we obtain that $\angle PXQ=2\angle A$. Thus $O,P,Q,X$ are concyclic. $\square$