Problem

Source: Polish Junior Math Olympiad 2nd Round 2020

Tags: geometry, parallelogram, perpendicular bisector



Let $ABCD$ be the parallelogram, such that angle at vertex $A$ is acute. Perpendicular bisector of the segment $AB$ intersects the segment $CD$ in the point $X$. Let $E$ be the intersection point of the diagonals of the parallelogram $ABCD$. Prove that $XE = \frac{1}{2}AD$.