题目如图,在矩形ABCDABCDABCD中,BC=8BC=8BC=8,以ABABAB为直径作⊙O\odot O⊙O,将矩形ABCDABCDABCD绕点BBB旋转,使所得矩形A\’BC\’D\’{A\’}BC\’{D\’}A\’BC\’D\’的边C\’D\’{C\’}{D\’}C\’D\’与⊙O\odot O⊙O相切,切点为EEE,边A\’B{A\’}BA\’B与⊙O\odot O⊙O相交于点FFF.若BF=8BF=8BF=8,则ABABAB长为( )A.999B.101010C.838\sqrt{3}83D.121212知识点:切线的判定章节:图形的性质 / 圆