如图,在正方形ABCD中,E为AD上一点,BF平分角CBE交CD于F.试说明BF=FC+AE

如图,在正方形ABCD中,E为AD上一点,BF平分角CBE交CD于F.试说明BF=FC+AE
如图,在正方形ABCD中,E为AD上一点,BF平分角CBE交CD于F.试说明BF=FC+AE

如图,在正方形ABCD中,E为AD上一点,BF平分角CBE交CD于F.试说明BF=FC+AE
证明：延长EA至H,使AH==CF,
∵AB=BC,∠HAB=∠FCB=90°
∴△HAB≌△FCB
∴∠AHB=∠CFB
∠ABH=∠FBC
∵∠CFB+∠FBC=90°
∠ABF+∠FBC=90°
∴∠CFB=∠ABF
∵BF是∠CBE的平分线
∴∠EBF=∠CBF
∴∠EBH=∠HBA+∠ABE
=∠CBF+∠ABE
=∠EBF+∠ABE
=∠ABF
=∠CFB
=∠AHB
=∠EHB
∴在△EHB中,∠EBH=∠EHB
∴BE=EH=AH+AE=CF+AE
∴BE=CF+AE