Question 55
Let $\triangle ABC$ ($BC \gt AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD$, $OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q$, $P$ and $K$, respectively, such that $P \in (AK)$ and $K \in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.