题目如图,以点OOO为圆心,ABABAB长为直径作圆,在⊙O⊙O⊙O上取一点CCC,延长ABABAB至点DDD,连接DCDCDC,∠DCB=∠DAC∠DCB=∠DAC∠DCB=∠DAC,过点AAA作AE⊥ADAE⊥ADAE⊥AD交DCDCDC的延长线于点E.E.E.(1)(1)(1)求证:CDCDCD是⊙O⊙O⊙O的切线;(2)(2)(2)若CD=4CD=4CD=4,DB=2DB=2DB=2,求AEAEAE的长.知识点:切线的判定与性质章节:图形的性质 / 圆