Question 34
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 $PB = QC$.