Question 106
Let $P$ be the intersection point of the diagonals $AC$ and $BD$ in a cyclic quadrilateral. A circle through $P$ touches the side $CD$ at the midpoint $M$ of this side and intersects the segments $BD$ and $AC$ at the points $Q$ and $R$, respectively. Let $S$ be a point on the segment $BD$ such that $BS = DQ$. The parallel to $AB$ through $S$ intersects $AC$ at $T$. Prove that $AT = RC$.