Baltic Way 1998 · Problem 14

Solution:

Since the lines $B D$ and $A C$ are parallel and since $A D$ is the external bisector of $\angle B A C$, we have $\angle B A D = \angle B D A$; denote their common size by $\alpha$ (see Figure 5). Also $\angle C A E = \angle C E A = \alpha$, implying $|A B| = |B D|$ and $|A C| = |C E|$. Let $B'$, $C'$, $F'$ be the feet of the perpendiculars from Figure 5 the points $B$, $C$, $F$ to line $D E$. From $|F C| = |A B|$ we obtain

$$|B' F'| = (|A B| + |A F|) \cos \alpha = |A C| \cos \alpha = |A C'| = |C' E|$$

and

$$|D B'| = |B D| \cos \alpha = |F C| \cos \alpha = |F' C'|,$$

Thus $|D F'| = |F' E|$, whence $|D F| = |F E|$.