Question 197
Let $\triangle ABC$ be an acute triangle, $D$ the foot of the perpendicular from $A$ to $BC$. With $AD$ as a diameter, draw a circle intersecting $AB$ and $AC$ at $E$ and $F$ respectively. Suppose $AD$ and $EF$ intersect at $G$, and $AD$ is extended to meet the circumcircle of $\triangle ABC$ at $H$. Show that $AD^2 = AG \cdot AH$.