(1) As shown in figure 1, △ABC is an equilateral triangle, point D is a point on the side of BC, and DE∑AC intersects with AB at point E, indicating that △BDE is also an equilateral triangle. (2)

(1) proves that ∵△ABC is an equilateral triangle,

∴∠A=∠B=∠C=60，

∫DE∑AC，

∴∠BED=∠A=60，∠BDE=∠C=60 ,∴∠B=∠BED=∠NDE=60，

∴△BDE is also an equilateral triangle;

(2) Prove that DK∨AC and AB are in K,

∫△ABC is an equilateral triangle,

∴ with (1), △BDK is an equilateral triangle, ∠EKD=∠EAC,

∴DK=BD，

ED = EC，

∴∠EDC=∠ECD，

∴∠B+∠KED=∠EDC，

∠∠ECA+∠ACB =∠ECD，

∴∠B+∠KED=∠ECA+∠ACB，

∠∠B =∠ACB = 60，

∴∠KED=∠ECA，

In △DKE and △ East Africa,

∠EKD=∠EAC∠KED=∠ECAED=EC，

∴△DKE≌△EAC(AAS)，

∴AE=DK，

∴BD=AE.