Berkeley Math Circle

Solution:

We have $\angle D F C = \pi / 2 = \angle D E B$. Also, $\angle C D F = \angle B D E$ since they are vertical angles. (It seems possible that $E$ and $F$ could lie on the same side of $D$, so that $\angle C D F$ and $\angle B D E$ would be supplementary rather than equal; however, if they were supplementary and unequal, then one of them, say $\angle B D E$, would be $> \pi / 2$, so the sum of the angles of $\triangle B D E$ would be $> \pi$, a contradiction.) These equal angles imply $\triangle D F C \sim \triangle D E B$. But $C D = B C / 2 = B D$, so in fact $\triangle D F C \cong \triangle D E B$, giving $C F = B E$.

Solution:

Let $H$ be the foot of the perpendicular from $A$ to $B C$. Then, using the $b h / 2$ formula and the fact that $D$ is the midpoint of $B C$, we get
$$\frac{A D \cdot B E}{2} = \operatorname{Area}(\triangle A B D) = \frac{B D \cdot A H}{2} = \frac{C D \cdot A H}{2} = \operatorname{Area}(\triangle A C D) = \frac{A D \cdot C F}{2} .$$
Multiplying by $2 / A D$ now gives $B E = C F$.