题目如图,⊙O⊙O⊙O是△ABC\triangle ABC△ABC的外接圆,\overparenAC=\overparenBC\overparen{AC}=\overparen{BC}\overparenAC=\overparenBC,过点AAA作AD/ /BCAD/\!/BCAD//BC交⊙O⊙O⊙O于点DDD,连接CDCDCD,延长DADADA到点EEE,连接CECECE,∠D=∠E.∠D=∠E.∠D=∠E.(1)(1)(1)求证:CECECE是⊙O⊙O⊙O的切线;(2)(2)(2)若CE=8CE=8CE=8,AE=5AE=5AE=5,求⊙O⊙O⊙O半径的长.知识点:切线的判定与性质章节:图形的性质 / 圆