Let ABC be an acute-angled triangle such that AB<AC. Let D be the point of intersection of the perpendicular bisector of the side BC with the side AC. Let P be a point on the shorter arc AC of the circumcircle of the triangle ABC such that DP∥BC. Finally, let M be the midpoint of the side AB. Prove that ∠APD=∠MPB.Proposed by Dominik Burek, Poland geometryperpendicular bisectorcircumcirclememoMEMO 2019sinus