题目如图,ABABAB是⊙O\odot O⊙O的直径,弦CD⊥ABCD\bot ABCD⊥AB于点EEE,连接ODODOD,若AB=6AB=6AB=6,BE=1BE=1BE=1,则弦CDCDCD的长是( )A.3\sqrt{3}3B.5\sqrt{5}5C.232\sqrt{3}23D.252\sqrt{5}25知识点:垂径定理章节:图形的性质 / 圆