Question 28
Let $ABC$ be a triangle with $AB \lt AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen in such a way that $BP = CP$. Let $BP$ and the angle bisector of $\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$.