Question 32
Let $ABC$ ($AC \neq BC$) be an acute triangle with orthocenter $H$ and incenter $I$. The lines $CH$ and $CI$ meet the circumcircle of $\triangle ABC$ at points $D$ and $L$, respectively. Prove that $\angle CIH = 90^\circ$ if and only if $\angle IDL = 90^\circ$.