Irska

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 midpoint of $AD$ and $X$ be the midpoint of $AH$. Let $Y$ be the point on $EB$ such that $XY$ is perpendicular to $BE$. Prove that $|XY|=|XH|$.