(1) Make a circle through a, e, c and f, as shown in the figure. As can be seen from the topic AB=BC=AC, the triangle ABC is an equilateral triangle with AD//BC, ∠ BAC = ∠ D, ∠ AE=AF = ∠ ACD.

(2) The conclusion remains unchanged, AE=EF? . Connecting Ge and GC, from AD//BC and AB=BC, we get ∠BAC=∠BCA=∠CAD, from arc theorem ∠ DAC = ∠ GE, GC = ∠ ACB = ∠ AE=EF, and ∠.

(3) unchanged, AE=AF.