(1)证明:∵△ABC为等边三角形,∴∠BAC=∠C=60°,AB=CA.在△ABE和△CAD中,AB=CA,∠BAE=∠C,AE=CD∴△ABE≌△CAD.(2)解:∵∠BFD=∠ABE+∠BAD,又∵△ABE≌△CAD,∴∠ABE=∠CAD.∴∠BFD=∠CAD+∠BAD=∠BAC=60 °.