Problem

Source: 2019 Balkan MO Shortlist G5 - BMO

Tags: geometry, circumcircle, equal segments



Let $ABC$ ($BC > AC$) 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 $TB = TQ$.