BALTIC-WAY 1998 Problem 15

Given an acute triangle $A B C$. Point $D$ is the foot of the perpendicular from $A$ to $B C$. Point $E$ lies on the segment $A D$ and satisfies the equation

$$\frac{|A E|}{|E D|}=\frac{|C D|}{|D B|}$$

Point $F$ is the foot of the perpendicular from $D$ to $B E$. Prove that $\angle A F C=90^{\circ}$.