MEMO 2021 Problem 6

Let $ABC$ be a triangle and let $M$ be the midpoint of the segment $BC$ . Let $X$ be a point on the ray $AB$ such that $2 \angle CXA=\angle CMA$ . Let $Y$ be a point on the ray $AC$ such that $2 \angle AYB=\angle AMB$ . The line $BC$ intersects the circumcircle of the triangle $AXY$ at $P$ and $Q$ , such that the points $P, B, C$ , and $Q$ lie in this order on the line $BC$ . Prove that $PB=QC$ .

Proposed by Dominik Burek, Poland