TUYMAADA 2009 J6

$M$ is the midpoint of base $BC$ in a trapezoid $ABCD$ . A point $P$ is chosen on the base $AD$ . The line $PM$ meets the line $CD$ at a point $Q$ such that $C$ lies between $Q$ and $D$ . The perpendicular to the bases drawn through $P$ meets the line $BQ$ at $K$ . Prove that $\angle QBC = \angle KDA$ .

Proposed by S. Berlov