SHARYGIN 2013 Problem 6

Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at point $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.