Prove:
(1) If there are two sides in AB, BC and AC, AB=AC.
Then AF=BD.
BE = AD，FD=DE
∴△AFD≌△BDE
∴∠A=∠B.
AB = AC
∴∠A=∠B=∠C.
That is, △ABC is a regular triangle.
(2) reduction to absurdity: pure geometric proof is used as an auxiliary line, as shown in the following figure.
If ∠ A ∠ B ∠ C
Then at least one of these three angles is greater than 60 and the other is less than 60.
Let ∠A > 60∠B < 60, and take points G and H on BA and its extension line respectively, so that ∠ DGE = 60 and ∠ AHF = 60.
∠∠ADF+∠FDE+∠EDG = 180, and the sum of internal angles of △DGE is 180. Note that △ dge = ∠ FDE = 60.
∴∠DEG=∠ADF
At △HDF and △GED,
∠DHF=∠EGD
∠HDF =∠ degrees
DF=DE
∴△HDF≌△GED
∴DH=GE
∫∠BGE is an obtuse angle.
∴BE>GE
That is, AD=BE is greater than DH.
Obviously not established;
So △ABC must be a regular triangle.