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∠CXA=∠CMA. Let Y be a point on the ray AC such that 2∠AYB=∠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 geometrypower of a pointcircumcircleMEMO 2021memo