Question 69
Let $ABC$ be a triangle with $AB \neq AC$ and let $M$ be the midpoint of $BC$. The bisector of $\angle BAC$ meets the line $BC$ at $Q$. Let $H$ be the foot of the altitude from $A$ onto $BC$. The perpendicular to $AQ$ through $A$ meets the line $BC$ at $S$. Show that $MH \cdot QS = AB \cdot AC$.