证明：过点A作AG⊥AB,取AG＝BF,连接CG、EG （注：G与C在AB的同一侧）∵∠ACB＝90,AC＝BC∴∠CAB＝∠B＝45∵AG⊥AB∴∠GAB＝90∴∠CAG＝∠GAB-∠CAB＝45,EG²＝AE²+AG²∴∠CAG＝∠B∵AG＝BF∴△CBF≌△...