题目如图,在△ABC\triangle ABC△ABC中,OOO为ACACAC上一点以OOO为圆心,OCOCOC长为半径作圆,与BCBCBC相切于点CCC,过点AAA作AD⊥B0AD⊥B0AD⊥B0交BOBOBO延长线于点DDD,且∠AOD=∠BAD.∠AOD=∠BAD.∠AOD=∠BAD.(1)(1)(1)求证:ABABAB为⊙O⊙O⊙O的切线;(2)(2)(2)若BC=6BC=6BC=6,tan∠ABC=43\tan∠ABC=\dfrac{4}{3}tan∠ABC=34,求ODODOD的长.知识点:切线的判定与性质章节:图形的性质 / 圆