Question 99
Let $\triangle ABC$ be an acute triangle with $|AB| \lt |AC|$ and incenter $I$. Let $D$ be the projection of $I$ onto $BC$. Let $H$ be the orthocenter of $\triangle ABC$. Prove that if $\angle IDH = \angle CBA - \angle ACB$ then $|AH| = 2 \cdot |ID|$.