OMM 2001 Problem 3

$ABCD$ is a cyclic quadrilateral. $M$ is the midpoint of $CD$ . The diagonals meet at $P$ . The circle through $P$ which touches $CD$ at $M$ meets $AC$ again at $R$ and $BD$ again at $Q$ . The point $S$ on $BD$ is such that $BS = DQ$ . The line through $S$ parallel to $AB$ meets $AC$ at $T$ . Show that $AT = RC$ .