题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,点DDD是ABABAB中点,连接CDCDCD.点EEE是BDBDBD中点,过点EEE作EF⊥ACEF\bot ACEF⊥AC于点FFF,EFEFEF交CDCDCD于点GGG.若EG=22EG=2\sqrt{2}EG=22,则FG=______.FG= \_\_\_\_\_\_.FG=______.知识点:直角三角形全等的判定章节:图形的性质 / 三角形