Question

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

Answer 1

∵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