题目已知:如图,ABABAB是⊙O⊙O⊙O的直径,点EEE为⊙O⊙O⊙O上一点,点DDD是\overparenAE\overparen{AE}\overparenAE上一点,连接AEAEAE并延长至点CCC,使∠CBE=∠BDE∠CBE=∠BDE∠CBE=∠BDE,BDBDBD与AEAEAE交于点F.F.F.(1)(1)(1)求证:BCBCBC是⊙O⊙O⊙O的切线;(2)(2)(2)若BDBDBD平分∠ABE∠ABE∠ABE,DF=1DF=1DF=1,BF=5BF=5BF=5,求ADADAD的长.知识点:切线的判定与性质章节:图形的性质 / 圆