(1) branch 1: ∠DEC = ∠B+∠BDE

And ∠DEC = ∠DEF+∠FEC.

Where: ∠DEF = ∠B

So: ∠FEC = ∠BDE

AB=AC, so: ∠ B = ∠ C.

BD=CE

So: △ BDE △ CEF (ASA)

∴ ED=EF

△EDF is an isosceles triangle

(2)DEF+∠A = 120

∴∠B+∠A= 120，∠DEF=∠B

∴∠C=60

∴∠B=60

∴∠A=60

∴∠DEF=60

ED=EF

∴∠EDF=∠EFD = 60

△ def is an equilateral triangle.