(2) cross e as EF∨BC and cross AC as f,

Equilateral triangle ABC,

∴∠ABC=∠ACB=∠A=60，AB=AC=BC，

∴∠AEF=∠ABC=60，∠AFE=∠ACB=60，

That is ∠ AEF =∠ AFE =∠ A = 60,

△ AEF is an equilateral triangle,

∴AE=EF=AF，

∠∠ABC =∠ACB =∠AFE = 60，

∴∠DBE=∠EFC= 120，∠D+∠BED=∠FCE+∠ECD=60，

DE = EC，

∴∠D=∠ECD，

∴∠BED=∠ECF，

In Debbie and △ECF.

∴△DEB≌△ECF，

∴BD=EF=AE, that is, AE=BD.

So the answer is: =;

(3)CD= 1 or 3.