Question 82
Let $ABC$ be a given acute-angled triangle. $D$ is the foot of the altitude from $A$ to $BC$. Let $E$ be a point on $AD$ such that $\frac{AE}{ED} = \frac{CD}{DB}$ and let $F$ be the foot of the altitude from $D$ to $BE$. Prove that $\angle AFC = 90^\circ$.