Not difficult. It will be done in 5 minutes, which is the level of training questions. It's not a contest topic.

Extend DF to g, make FG=FD, and connect AG and eg.

It is easy to get △ BFD △ AFG, and get AG=BD, ∠GAB=∠ABC.

Due to △AOE∽△BOD, AE/BD=OE/OD and AE/AG=OE/OD are obtained.

In the quadrilateral ODCE, ∠ DOE =180-∠ C = ∠ BAC+∠ ABC = ∠ BAC+∠ GAE.

So △GAE∽△DOE, we get ∠AEG=∠OED.

∠ GED =∠ GEO +∠ OED =∠ GEO +∠ AEG = 90，FG=FD。

That is, EF is the center line on the hypotenuse of a right triangle, and EF=FG=FD is obtained.