Let $ ABCD$ be a parallelogram with $ \angle BAD = 60$ and denote by $ E$ the intersection of its diagonals. The circumcircle of triangle $ ACD$ meets the line $ BA$ at $ K \ne A$, the line $ BD$ at $ P \ne D$ and the line $ BC$ at $ L\ne C$. The line $ EP$ intersects the circumcircle of triangle $ CEL$ at points $ E$ and $ M$. Prove that triangles $ KLM$ and $ CAP$ are congruent.
Problem
Source: MEMO 2009, problem 5, team competition
Tags: geometry, parallelogram, circumcircle, geometry proposed