如图,已知△ABC和△BDE都为等腰直角三角形,点E在AB上,点F为CD的中点,连接BF.

1、∵△ABC和△BDE是等腰直角三角形(AB=AC,BE=BD)∴∠ABC=∠DBE=45°∴∠DBC=∠DBE+∠ABC=90°∵F是CD中点∴BF=1/2CD=CF=DF∴∠BCF=∠CBF2、∵BF=DF,EF=EF,DE=BD∴△BEF≌△DEF(SSS)∴∠DFE=∠BFE延长EF,交AC于G∵∠DFE=∠CFG∴∠BFE=∠CFG∵∠ABC=∠ACB,∠BCF=∠CBF∴∠ABC-∠CBF=∠ACB-∠BCF,即∠ABF=∠ACF∴∠EBF=∠GCF∵BF=CF∴△BEF≌△CGF(ASA)∴EF=FG,BE=CG∵AB-BE=AC-CG∴AE=AG∵AF=AF,EF=FG,AE=AG∴△AEF≌△AGF(SSS)∴∠AFG=∠AFE∵∠AFE+∠AFG=180°∴∠AFE=90°即AF⊥EF