Problem

Source: Serbian National Olympiad 2012, Problem 1

Tags: geometry, parallelogram, circumcircle, trigonometry, geometry proposed



Let $ABCD$ be a parallelogram and $P$ be a point on diagonal $BD$ such that $\angle PCB=\angle ACD$. Circumcircle of triangle $ABD$ intersects line $AC$ at points $A$ and $E$. Prove that \[\angle AED=\angle PEB.\]