题目如图,ABABAB为⊙O\odot O⊙O的直径,DCDCDC与⊙O\odot O⊙O相切于点CCC,过点BBB作BD⊥CDBD\bot CDBD⊥CD于点DDD,连接CBCBCB.(1)(1)(1)求证:BCBCBC平分∠ABD\angle ABD∠ABD;(2)(2)(2)若AC=25AC=2\sqrt{5}AC=25,AB=5AB=5AB=5,求BDBDBD的长.知识点:垂径定理章节:图形的性质 / 圆