Question 68
In convex quadrilateral $A B C D$, $\angle C A B = \angle B C D$. $P$ lies on line $B C$ such that $A P = P C$, $Q$ lies on line $A P$ such that $A C$ and $D Q$ are parallel, $R$ is the point of intersection of lines $A B$ and $C D$, and $S$ is the point of intersection of lines $A C$ and $Q R$. Line $A D$ meets the circumcircle of $A Q S$ again at $T$. Prove that $A B$ and $Q T$ are parallel.