Let $ABC$ be a triangle with $AB \neq AC$ and let $M$ be the middle of $BC$. The bisector of $\angle BAC$ intersects the line $BC$ in $Q$. Let $H$ be the foot of $A$ on $BC$. The perpendicular to $AQ$ passing through $A$ intersects the line $BC$ in $S$. Show that $MH \times QS=AB \times AC$.