Question 53
Let $ABCD$ be a cyclic quadrilateral such that $AB = AD + BC$ and $CD \lt AB$. The diagonals $AC$ and $BD$ intersect at $P$, while the lines $AD$ and $BC$ intersect at $Q$. The angle bisector of $\angle APB$ meets $AB$ at $T$. Show that the circumcenter of $\triangle CTD$ lies on the circumcircle of $\triangle CQD$.