Question 45
Given is a right triangle $ABC$ with $\angle A = 90^\circ$. Let $M$ be the midpoint of $BC$ and $P$ be an arbitrary point on $AM$. The reflection of $BP$ with respect to $AB$ intersects the lines $AC$ and $AM$ at $T$ and $Q$, respectively. Let the circumcircles of $BPQ$ and $ABC$ meet for the second time at $F$. Prove that the circumcenter of $CFT$ lies on $BQ$.