解:当AE=2EF时,FG=3EF.
证明:∵四边形ABCD是正方形,
∴AB∥CD,AD∥BC,
∴△ABE∽△FDE,△ADE∽△GBE,
∵AE=2EF,
∴BE:DE=AE:EF=2,
∴BC:AD=BE:DE=2,
即BG=2AD,
∵BC=AD,
∴CG=AD,
∵△ADF∽△GCF,
∴FG:AF=CG:AD,
即FG=AF=AE+EF=3EF.
FG= 3 EF
△DEF∽ △BEA
DF/AB = EF / AE = 1/2
F是CD中点
△GFC≌ △AFD
FG=AF
FG=3EF