题目如图,ABABAB是⊙O\odot O⊙O直径,弦CD⊥ABCD\bot ABCD⊥AB于点EEE,过点CCC作DBDBDB的垂线,交ABABAB的延长线于点GGG,垂足为点FFF,连结ACACAC.(1)(1)(1)求证:AC=CGAC=CGAC=CG;(2)(2)(2)若CD=EG=8CD=EG=8CD=EG=8,求⊙O\odot O⊙O的半径.知识点:切线的判定章节:图形的性质 / 圆