题目如图,⊙O\odot O⊙O是△ABC\triangle ABC△ABC的外接圆,∠BAC=60∘\angle BAC=60^{\circ}∠BAC=60∘,AD⊥BCAD\bot BCAD⊥BC于点DDD,延长ADADAD交⊙O\odot O⊙O于点EEE,若BD=3,CD=33BD=\sqrt{3},CD=3\sqrt{3}BD=3,CD=33,则ADADAD的长是______.知识点:解直角三角形章节:图形的变化 / 锐角三角函数