证明：EF‖CD,∠AEG=∠AGE=60°∴△AEG是等边三角形
AG=GE=AE=BD
EF=CD,GF=EF+GE,BC=CD+BD
∴GF=BC=AC
GF=AC,∠FGA=∠CAE,AG=AE
∴△AGF≌△EAC