∴∠A=60。 ∠ABC+∠ACB= 120

∠BFE =∠FBC+∠FCB = 1/2∠ABC+ 1/2∠ACB = 60

cos∠BFE= 1/2

∴ ① Correct.

∠BFE=60

∴∠DFE= 120

∴∠DFE+∠A= 180

∴∠FDA+∠FEA= 180

∴∠FDC=∠FEA

F is FM⊥CD in M, FN⊥AE in N.

∫F is the heart of △ABC.

∴FM=FN

∴△FDM≌△FEN

∴FD=FE,③ Correct.

If ② is correct, ∠BCD=∠FDC=∠FEA.

∠∠A+∠BCD+∠ABC = 180

∠A+∠FEA+∠FCA= 180

∴∠FCA=∠ABC

∠BCA=2∠FCA

Then < BCA = 2 < ABC

∫∠A = 60，

∴∠BCA+∠ABC= 120

Therefore, ∠ BCA = 80, ∠ ABC = 40.

The title does not have this condition, so it is wrong.

If ④ is correct. FD = FE

∴BF=2FE

And < bfe = 60.

∴△BFE is a right triangle (three angles are 30 degrees, 60 degrees and 90 degrees respectively).

∠FBE=30

∴∠ABC=2∠FBE=60。

It can be concluded that △ABC is a regular triangle. There is no such condition in the question. Therefore, 4 errors.