过点C作CG‖AB,交ED的延长线于点G,连接FG；则有：∠DBE = ∠DCG .在△BDE和△CDG中,∠DBE = ∠DCG ,BD = CD ,∠BDE = ∠CDG ,所以,△BDE ≌ △CDG ,可得：BE = CG ,DE = DG .在△DEF和△DGF中,DE = DG ,∠EDF = 90°...