数学题急用！～如图,ABCD是正方形.点G是BC边上的任意一点,
如图,AB是正方形.点G是BC边上的任意一点,DE⊥AG于E,BF∥DE,交AG于F.求证AF-BF=EF
∵ABCD是正方形.∴AB=AD,∴∠BAD=90,∴∠BAF+∠EAD=90.∵DE⊥AG,∴∠AED=90,∴∠EAD+∠ADE=90,∴∠BAF=∠ADE.∵BF∥DE,∴BF⊥AG,∴∠BFA=90,∴∠BFA=∠AED.∴△ABF≌△DAE,∴AE=BF.∵AF-AE=EF,∴AF=BF=EF.