题目如图,ABABAB为⊙O\odot O⊙O的直径,点CCC,DDD为直径ABABAB同侧圆上的点,且点DDD为AC^\widehat {AC}AC的中点,过点DDD作DE⊥ABDE\bot ABDE⊥AB于点EEE,延长DEDEDE,交⊙O\odot O⊙O于点FFF,ACACAC与DFDFDF交于点GGG.(Ⅰ)(Ⅰ)(Ⅰ)如图①,若点CCC为DB^\widehat {DB}DB的中点,求∠AGF\angle AGF∠AGF的度数；(Ⅱ)(Ⅱ)(Ⅱ)如图②,若AC=12AC=12AC=12,AE=3AE=3AE=3,求⊙O\odot O⊙O的半径.知识点：切线的判定章节：图形的性质 / 圆
