题目如图,△ABC\triangle ABC△ABC内接于⊙O\odot O⊙O,ABABAB是⊙O\odot O⊙O的直径,CECECE平分∠ACB\angle ACB∠ACB交⊙O\odot O⊙O于点EEE,过点EEE作EFEFEF∥ABABAB,交CACACA的延长线于点FFF.(1)(1)(1)求证:EFEFEF与⊙O\odot O⊙O相切;(2)(2)(2)若∠CAB=30∘\angle CAB=30^{\circ}∠CAB=30∘,AB=8AB=8AB=8,过点EEE作EG⊥ACEG\bot ACEG⊥AC于点MMM,交⊙O\odot O⊙O于点GGG,交ABABAB于点NNN,求AG^\widehat {AG}AG的长.知识点:垂径定理章节:图形的性质 / 圆