Let $ABCD$ be a parallelogram with $\angle BAD<90^o$ and $AB> BC$ . The angle bisector of $BAD$ intersects line $CD$ at point $P$ and line $BC$ at point $Q$. Prove that the center of the circle circumscirbed around the triangle $CPQ$ is equidistant from points $B$ and $D$.
Problem
Source: Czech-Polish-Slovak Junior Match 2014, Team p2 CPSJ
Tags: geometry, circumcircle, parallelogram, equal segments