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=∠...