题目如图,在△ABC\triangle ABC△ABC中,BE⊥ACBE\bot ACBE⊥AC于点EEE,BCBCBC的垂直平分线分别交ABABAB、BEBEBE于点DDD、GGG,垂足为HHH,CD⊥ABCD\bot ABCD⊥AB,CDCDCD交BEBEBE于点FFF(1)求证:△BDF\triangle BDF△BDF≌△CDA\triangle CDA△CDA(2)若DF=DGDF=DGDF=DG,求证:①BEBEBE平分∠ABC\angle ABC∠ABC②BF=2CEBF=2CEBF=2CE.知识点:线段垂直平分线的性质;全等三角形的判定与性质章节:未标注