解:设A(x1,y1),B(x2,y2),代入中,得x1?y1=x2?y2=k, 联立,得x2-bx+k=0, 则x1?x2=k,又x1?y1=k, ∴x2=y1, 同理x2?y2=k, 可得x1=y2,
∴ON=OM,AM=BN, ∴①OA=OB,②△AOM≌△BON,正确; ③作OH⊥AB,垂足为H, ∵OA=OB,∠AOB=45°, ∵②△AOM≌△BON,正确; ∴∠MOA=∠BON=22.5°, ∠AOH=∠BOH=22.5°, ∴△OAM≌△OAH≌△OBH≌△OBN, ∴S△AOB=S△AOH+S△BOH=S△AOM+S△BON=k+k=k,正确; ④延长MA,NB交于G点,
∵NG=OM=ON=MG,BN=AM, ∴GB=GA, ∴△ABG为等腰直角三角形, 当AB=时,GA=GB=1, ∴ON-BN=GN-BN=GB=1,正确. 正确的结论有4个. 故选D. |