Hint: Can you prove that all interior angles are 60 degrees?
Over the sides of an arbitrary triangle ABC from the outside are constructed isosceles triangles with an angle of at the top.
Prove that the three vertices of this triangles form an equilateral triangle (triangle DFE on picture).
Chiro has given a very good hint.
angle FCD = 120 given. triangle FCD is an isosceles triangle thus angle CFD = angle CFD = 30 degree. similarly from triangle EAF we will get
angle AEF = angle AFE = 30 degree
We have angle AFC = 120
I am sure now you can establish that angle EFD = 60 degree and similarly other angles.