解:∵∠ABC+∠DCB=180°
∴CG∥AB,
∴∠1=∠FEA,( )
∵∠1=∠2,∴∠2=∠FEA,( )
∴EG∥ ▲ , ( )
∴ ▲ +∠FDB=180°,
∵∠GFA=∠DFE,( )
∴∠GFA+∠FDB=180°.