Given a triangle $ABC$, the perpendicular bisector of the side $AC$ intersects the angle bisector of the triangle $AK$ at the point $P$, $M$ - such a point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on opposite sides of the line $AC$. Prove that the line $AK$ bisects the segment $BM$. (Anton Trygub)