In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.