已知正方形ABCD中,E为AD上一点,BF平分∠EBC交CD于F,求证:BE=AE+CF
问题描述:
已知正方形ABCD中,E为AD上一点,BF平分∠EBC交CD于F,求证:BE=AE+CF
答
将△ABE绕B点旋转,使AB和BC重合,
设△BCG是旋转后的△ABE
∴△ABE≌△CBG
∴AE = CG,BE = BG,∠ABE = ∠CBG
∵BF是∠EBC的角平分线
∴∠EBF = ∠FBC
∴∠ABE+∠EBF =∠GBC + ∠FBC
∴∠ABF = ∠FBG
∵正方形ABCD
∴AB//CD
∴∠ABF = ∠BFG
∴∠GBF = ∠BFG
∴BG = GF
∵GF = CG+CF = AE+ CF,BG = BE
∴BE = AE+CF