题目如图,在△ABC\triangle ABC△ABC中,∠BAC=90∘\angle BAC=90^{\circ}∠BAC=90∘,点EEE是BCBCBC延长线上一点,DDD为BCBCBC下方一点,连接DEDEDE,DE=ABDE=ABDE=AB,过点DDD作DF⊥DEDF\bot DEDF⊥DE交BEBEBE于点FFF,且FB=CEFB=CEFB=CE.(1)(1)(1)求证:∠ACB=∠DFE\angle ACB=\angle DFE∠ACB=∠DFE;(2)(2)(2)连接ADADAD交BCBCBC于点GGG,若AG=5AG=5AG=5,求ADADAD的长.知识点:切线的判定章节:图形的性质 / 圆