Question

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

Answer 1

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