SHARYGIN 2013 Problem 14

Let $M$ , $N$ be the midpoints of diagonals $AC$ , $BD$ of a right-angled trapezoid $ABCD$ ( $\measuredangle A=\measuredangle D = 90^\circ$ ). The circumcircles of triangles $ABN$ , $CDM$ meet the line $BC$ in the points $Q$ , $R$ . Prove that the distances from $Q$ , $R$ to the midpoint of $MN$ are equal.