Problem

Source: own, Iran MO 3rd round 2019 mid-terms - Geometry P1

Tags: geometry, cyclic quadrilateral



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