Question 59
Let $ABC$ be a triangle with $\angle A \lt 90^\circ$ and $AB \neq AC$. Denote by $H$ the ortocenter of triangle $ABC$, by $N$ the midpoint of $[AH]$, by $M$ the midpoint of the side $[BC]$ and by $D$ the intersection point of the angle bisector of $\angle BAC$ with $[MN]$. Prove that $\angle ADH = 90^\circ$.