IrMO 2009 Problem 2

Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $|AB|\cdot |BH|=|AH|^2$ . Let $E$ be the midpoints of $AD$ and $X$ be the midpoint of $AH$ . Let $Y$ be a point on $EB$ such that $XY$ is perpendicular to $BE$ . Prove that $|XY|=|XH|$ .