Let $ABC$ be an acute non isosceles triangle. The angle bisector of angle $A$ meets again the circumcircle of the triangle $ABC$ in $D$. Let $O$ be the circumcenter of the triangle $ABC$. The angle bisectors of $\angle AOB$, and $\angle AOC$ meet the circle $\gamma$ of diameter $AD$ in $P$ and $Q$ respectively. The line $PQ$ meets the perpendicular bisector of $AD$ in $R$. Prove that $AR // BC$.
Problem
Source: Danube 2018 P3
Tags: geometry, parallel, angle bisector, circumcircle, perpendicular bisector