In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$. Furthermore $AB=DE$, $BC=EF$, and $CD=FA$. Prove that diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent.