Question 126
In triangle $ABC$, perpendicular bisector of the side $AC$ intersects the angle bisector $AK$ in point $P$, $M$ is such point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on different sides from the segment $AC$. Prove that the line $AK$ divides the segment $BM$ in two equal segments.
