THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Triangles $ABD$ and $FBD$ are congruent (S.A.S.), so that $m(\angle FDB)=m(\angle ADB)=60^{\circ }$, and $m(\angle FDC)=60^{\circ }$.

Triangle $EBC$ is isosceles ($BE=BC$), with $m(\angle EBC)=20^{\circ }$, hence $m(\angle BCE)=80^{\circ }$, and from the hypothesis we have $m(\angle ACB)=40^{\circ }$, so $\angle FCD=\angle DCE$.

Thus, triangles $FCD$ and $ECD$ are congruent (A.S.A.), hence triangle $FCE$ is isosceles. $CD$ is an internal angle bisector, hence also a height. We conclude that $AC\perp FE$.