Because △ABC is an equilateral triangle, AB⊥AD. Then cross E to make eh ⊥ AD on AD, and cross AD to H, assuming AB=2.
In this way, through calculation, it can be concluded that EH=2, through BAF = EHF = 90? , ∠ AFB = ∠ EFH, AB = Eh, the principle that all two angles and a line are equal is an equal triangle. You can get BF=EF.
(2) Connect BH and AC to G because ∠CAD=2∠DAE, ∠CAD=60? ,∠DAE=30? Through calculation, it is found that AH=2, AB=AH, ∠BAC=∠CAD, ***AG are on the same side, so ∠ABC=∠ACB=30?
∠ADC=30? , so ∠ADC=∠ACB, that is, BH∪CD∪AE, and then ∠AFE=∠BFH, because it is diagonal.
So △AFE is similar to △BFH, so BF/EF=BH/AE=3/2.