Question 57
Given a cyclic quadrilateral $ABCD$. There is a point $P$ on side $BC$ such that $\angle PAB = \angle PDC = 90^\circ$. The medians of vertices $A$ and $D$ in triangles $\triangle PAB$ and $\triangle PDC$ meet at $K$ and the angle bisectors of $\angle PAB$ and $\angle PDC$ meet at $L$. Prove that $KL \perp BC$.