因为三角形BDE和三角形GDF全等,则FG=BE,
又因为BE=AF,则,FG=AF=BE
则AF/AE=FG/AE=CF/AC
又因为CF/AC=CF/(AF+CF),
而且由题目已知,CF=AE,即是CF/(AF+CF)=AE/(AF+AE)
则得结果,AF/AE=FG/AE=CF/AC=CF/(AF+CF)=AE/(AF+AE)
即AF/AE=AE/(AF+AE)
因为三角形BDE和三角形GDF全等,则FG=BE,
又因为BE=AF,则,FG=AF=BE
则AF/AE=FG/AE=CF/AC
又因为CF/AC=CF/(AF+CF),
而且由题目已知,CF=AE,即是CF/(AF+CF)=AE/(AF+AE)
则得结果,AF/AE=FG/AE=CF/AC=CF/(AF+CF)=AE/(AF+AE)
即AF/AE=AE/(AF+AE)