Question 51
Let $ABC$ be a triangle with $AB \neq AC$. The angle bisector of $\angle BAC$ intersects $BC$ in $D$. The circle with diameter $AD$ intersects $AC$ again in $P$, and $BC$ again in $Q$. The point $R \neq Q$ lies on the line parallel to $AD$ through $Q$. Suppose that $AQ = AR$. Prove that the points $B, P, Q$, and $R$ lie on a common circle.