延长BA至H,使AH＝BD.
∵BD＝AH、DF＝FA,∴BD＋DF＝FA＋AH,∴BF＝FH,又BE＝EC,
∴EF是△BCH中过BC、BH的中位线,∴EF∥CH,∴∠BFE＝∠AHC,而∠BFE＝∠AFG,
∴∠AFG＝∠AHC.······①
∵EF∥CH,∴EG∥CH,∴∠ACH＝∠AGF.······②
∵AH＝BD、BD＝AC,∴AH＝AC,∴∠ACH＝∠AHC.······③
由①、②、③,得：∠AFG＝∠AGF,∴△AFG是以FG为底边的等腰三角形.