题目内容：

E,F是正方形ABCD的边AB,BC上的点,AE+CF=EF.求证：∠EDF=45°

优质解答

延长BC到点G,使得 CG = AE ,则 FG = CF+CG = CF+AE = EF .在△ADE和△CDG中,AD = CD ,∠DAE = 90°= ∠DCG ,AE = CG ,所以,△ADE ≌ △CDG ,可得：∠ADE = ∠CDG ,DE = DG .所以,∠EDG = ∠CDE+∠CDG = ∠CDE+∠ADE ...