题目如图,在△ABC\triangle ABC△ABC中,点EEE为BCBCBC边上一点,以CECECE为直径的半圆OOO交线段ABABAB于点DDD,点FFF,连接CDCDCD,CFCFCF,∠A=∠ACD=∠AFC.∠A=∠ACD=∠AFC.∠A=∠ACD=∠AFC.(1)(1)(1)求证:ACACAC为⊙O⊙O⊙O的切线;(2)(2)(2)若AC=6AC=6AC=6,cosA=35.\cos A=\dfrac{3}{5}.cosA=53.求⊙O⊙O⊙O的半径.知识点:切线的判定与性质章节:图形的性质 / 圆