Question 148
A circle $\Omega$ centered at $O$ is circumscribed about an acute-angled triangle $ABC$ with $AB \lt BC$. Let $H$ be the orthocenter of $ABC$. A point $D$ is chosen on the extension of $BO$ beyond $O$ so that $\angle ADC = \angle ABC$. A line through $H$ parallel to $BO$ meets the smaller arc $AC$ of $\Omega$ at $E$. Prove that $BH = DE$. (A. Kuznetsov)