45th Mongolian Mathematical Olympiad

In the right triangle $ABC$, $\angle A=90^{\circ }$. The midpoint of side $BC$ is denoted by $M$ and the point $D$ lies on $AC$ such that $AD=AM$. Let $P$ be a different from $C$ and intersection of the circumcircles of triangles $AMC$ and $BDC$. Prove that $CP$ is the bisector of angle $BCA$.

(Iran, 2008)