题目如图,ABABAB是⊙O\odot O⊙O的直径,点CCC,DDD在⊙O\odot O⊙O上,若AD^=CD^\widehat {AD}=\widehat {CD}AD=CD,求证:ODODOD∥BC.BC.BC.知识点:切线的判定章节:图形的性质 / 圆