USAJMO

Instead of trying to find a synthetic way to describe $EF$ being tangent to the $A$ -excircle (very hard), we instead consider the foot of the perpendicular from the $A$ -excircle to $EF$ , hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe $EF$ , something more closely related to the $A$ -excircle; as we are considering perpendicularity, if we could generate a line parallel to $EF$ , that would be good.
So we recall that it is well known that triangle $AEF$ is similar to $ABC$ . This motivates reflecting $BC$ over the angle bisector at $A$ to obtain $B'C'$ , which is parallel to $EF$ for obvious reasons.
Furthermore, as reflection preserves intersection, $B'C'$ is tangent to the reflection of the $A$ -excircle over the $A$ -angle bisector. But it is well-known that the $A$ -excenter lies on the $A$ -angle bisector, so the $A$ -excircle must be preserved under reflection over the $A$ -excircle. Thus $B'C'$ is tangent to the $A$ -excircle.Yet for all lines parallel to $EF$ , there are only two lines tangent to the $A$ -excircle, and only one possibility for $EF$ , so $EF = B'C'$ .
Thus as $ABB'$ is isoceles, $$[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,$$ contradiction. -alifenix-