题目如图,在△ABC\triangle ABC△ABC中,以ACACAC为直径的⊙O⊙O⊙O交BCBCBC边于点DDD,在BCBCBC边上取一点EEE,使得CA=CECA=CECA=CE,连结AEAEAE,交⊙O⊙O⊙O于点FFF,且∠BAE=12∠ACE.∠BAE=\dfrac{1}{2}∠ACE.∠BAE=21∠ACE.(1)(1)(1)求证:BABABA是⊙O⊙O⊙O的切线;(2)(2)(2)若⊙O⊙O⊙O的直径为444,BE=2CDBE=2CDBE=2CD,求\overparenCD\overparen{CD}\overparenCD的长.知识点:切线的判定与性质章节:图形的性质 / 圆