题目如图,ABABAB是⊙O⊙O⊙O的直径,点CCC,DDD在⊙O⊙O⊙O上,CCC是\overparenBD\overparen{BD}\overparenBD的中点,连接BDBDBD交ACACAC于点EEE,延长ACACAC至FFF,使CE=CF.CE=CF.CE=CF.(1)(1)(1)求证:BFBFBF是⊙O⊙O⊙O的切线.(2)(2)(2)若BF=3,sinA=13BF=3,\sin A=\dfrac{1}{3}BF=3,sinA=31,求BDBDBD的长.知识点:切线的判定与性质章节:图形的性质 / 圆