题目如图,CDCDCD是⊙O\odot O⊙O的直径,ABABAB是⊙O\odot O⊙O的弦,AB⊥CDAB\bot CDAB⊥CD,垂足为MMM,EEE为AD^\widehat {AD}AD上一点,且AE^=AC^\widehat {AE}=\widehat {AC}AE=AC,连接ECECEC交ABABAB于点FFF,连接ACACAC.(1)(1)(1)求证:∠BAC=∠ECA\angle BAC=\angle ECA∠BAC=∠ECA;(2)(2)(2)若OM=3OM=3OM=3,OC=5OC=5OC=5,求ABABAB的长.知识点:垂径定理章节:图形的性质 / 圆