证明：取AC的中点H,连接HE
∵CH=AH,CE=BE
∴HE//AB,HE=AB/2,
∴∠G=∠HEF
∵AF=DF
∴HF=AH-FA=CH-DF=CD-DH-DF=AB-(DH+DF)=AB-HF
∴HF=AB/2=HE
∴∠HEF=∠HFE=∠GFA
∴∠GFA=∠G
∴AF=AG