题目如图111,ABABAB为⊙O\odot O⊙O直径,CBCBCB与⊙O\odot O⊙O相切于点BBB,DDD为⊙O\odot O⊙O上一点,连接ADADAD、OCOCOC,若ADADAD∥OC.OC.OC.(1)(1)(1)求证:CDCDCD为⊙O\odot O⊙O的切线；(2)(2)(2)如图222,过点AAA作AE⊥ABAE\bot ABAE⊥AB交CDCDCD延长线于点EEE,连接BDBDBD交OCOCOC于点FFF,若AB=3AE=12AB=3AE=12AB=3AE=12,求BFBFBF的长.知识点：圆周角定理I章节：图形的性质 / 圆
