如图,角ABC=90度,AB=BC,D为AC上一点,分别过AC作BD的垂线,垂足分别为EF 求证：EF=CF—AE

∵CF⊥BE AE⊥BE
∴∠CFE=∠AEF=90
∵AB=BC
∴ ∠BAC= ∠BCA
∴∠BAC= ∠BCA=(180- ∠ABC)/2=90/2=45
∵ ∠BAE+ ∠ABE=90 ∠FBC=FCB=90 ∠ABE+∠FBC=90
∴ ∠BAE=∠FCB
在△ABE与△BFC中
∠CFE=∠AEF
∠BAE=∠FCB
AB=AC
△ABE≌△BFC
∴AE=BF BE=CF
∴EF=EB-BF=CF-AE
禁止抄袭

∵CF⊥BE AE⊥BE
∴∠CFE=∠AEF=90
∵AB=BC
∴ ∠BAC= ∠BCA
∴∠BAC= ∠BCA=(180- ∠ABC)/2=90/2=45
∵ ∠BAE+ ∠ABE=90 ∠FBC=FCB=90 ∠ABE+∠FBC=90
∴ ∠BAE=∠FCB
在△ABE与△BFC中
∠CFE=∠AEF
∠BAE=∠FCB
AB=AC
△ABE≌△BFC
∴AE=BF BE=CF
∴EF=EB-BF=CF-AE

∵CF⊥BE AE⊥BE
∴∠CFE=∠AEF=90°
∵AB=BC
∴ ∠BAC= ∠BCA
∴∠BAC= ∠BCA=(180- ∠ABC)/2=90°/2=45°
∵ ∠BAE+ ∠ABE=90° ∠FBC=FCB=90° ∠ABE+∠FBC=90°
∴ ∠BAE=∠FCB
在△ABE与△BFC中
∠CFE=∠AEF
∠BAE=∠FCB
AB=AC
△ABE≌△BFC
∴AE=BF BE=CF
∴EF=EB-BF=CF-AE
这道题其实蛮简单，最好先自己想想，在来看答案！

如图,角ABC=90度,AB=BC,D为AC上一点,分别过AC作BD的垂线,垂足分别为E,F 求证EF=CF-AE ∵CF⊥BE AE⊥BE ∴∠CFE=∠AEF=90 ∵AB=BC ∴

∵CF⊥BE AE⊥BE
∴∠CFE=∠AEF=90
∵AB=BC
∴ ∠BAC= ∠BCA
∴∠BAC= ∠BCA=(180- ∠ABC)/2=90/2=45
∵ ∠BAE+ ∠ABE=90 ∠FBC=FCB=90 ∠ABE+∠FBC=90
∴ ∠BAE=∠FCB
在△ABE与△BFC中
∠CFE=∠AEF
∠BAE=∠FCB
AB=AC
△ABE≌△BFC
∴AE=BF BE=CF
∴EF=EB-BF=CF-AE