MathNet 2009-part-02 — 175 problems (EN–VI) Question 124

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 the intersection, other than $C$, of the circumcircles of triangles $AMC$ and $BDC$. Prove that $CP$ is the bisector of angle $BCA$.

(Iran, 2008)