题目如图,ABABAB是⊙O\odot O⊙O的直径,弦CD⊥ABCD\bot ABCD⊥AB于EEE,若∠ABC=30∘\angle ABC=30^{\circ}∠ABC=30∘,OE=1OE=1OE=1,则ODODOD长为( )A.333B.6\sqrt{6}6C.232\sqrt{3}23D.222知识点:切线的性质章节:图形的性质 / 圆