题目如图,ABABAB为⊙O\odot O⊙O的直径,点DDD在⊙O\odot O⊙O上,连接ADADAD,在ABABAB上截取AC=ADAC=ADAC=AD,连接DCDCDC并延长交⊙O\odot O⊙O于点EEE.若∠A=30∘\angle A=30^{\circ}∠A=30∘,AB=8AB=8AB=8,则DEDEDE的长为______.知识点:切线的判定章节:图形的性质 / 圆