The diagonals of a cyclic convex quadrilateral $ABCD$ intersect at $P$. A circle through $P$ touches the side $CD$ at its midpoint $M$ and intersects the segments $BD$ and $AC$ again at the points $Q$ and $R$ respectively. Let $S$ be the point on segment $BD$ such that $BS = DQ$. The line through $S$ parallel to $AB$ intersects $AC$ at $T$. Prove that $AT = RC$.