题目如图,ABABAB是⊙O\odot O⊙O的直径,点DDD在直径ABABAB上(D(D(D与AAA,BBB不重合),CD⊥ABCD\bot ABCD⊥AB且CD=ABCD=ABCD=AB,连接CBCBCB,与⊙O\odot O⊙O交于点FFF,在CDCDCD上取一点EEE,使EFEFEF与⊙O\odot O⊙O相切.(1)(1)(1)求证:EF=ECEF=ECEF=EC;(2)(2)(2)若DDD是OAOAOA的中点,AB=4AB=4AB=4,求BFBFBF的长.知识点:垂径定理章节:图形的性质 / 圆