As shown in the figure, at point M (2, 2), the right-angled vertex of a 90-degree square is placed at point M, and the two sides of the square intersect with the positive and semi-axis of X axis and Y axis at points A, B and AP respectively. Prove: (1) Let ME⊥x axis be in E, MF⊥y axis be in F,

∫M(2，2)，∠FOE=∠MEO=∠MFO=90，

∴OEMF is a square, OE=2, OF=2,

∴MF=ME，

∵ME⊥x axis is on the E axis, MF⊥y axis is on the F axis,

∴OM bisects ∠EOF, that is, OM bisects ∠ AOB;

(2)≈AMF+∠AME =∠AME+∠BME = 90，

∴∠AMF=∠BME，

At △AME and △BMF,

∠MEA=∠MFBME=MF∠EMA=∠BMF，

∴△AME≌△BMF(ASA)，

∴AE=BF，

∴oa+ob=oa+of+bf=oa+of+ae=oe+of=4；

(3) Solution: The value of ON+ 12AB remains unchanged.

The reason is:

Pass p into q as PQ⊥ME, extend PQ to r, make QR=PQ, connect MR,

∫△AEM?△BFM，

∴MB=MA，

∫∠AMB = 90 degrees,

∴∠MBA=∠MAB=45，

∫OM shares ∠AOB, AP shares ∠BAO, ∠ boa = 90,

∴∠∠MOA=45，∠BAP=∠PAO，

∴∠∠MOA+∠PAO=∠MAB+∠BAP，

That is ∠MAP=∠MPA,

∴MP=MA，

∫∠MOE = 45，ME=OE=2，

∴∠OME=45，

∵PR⊥ME,PQ=QR，

∴MP=MR，

∴MB=MP=MA=MR，

∴∠RMQ=∠PMQ=45，

∴∠PMR=90 =∠BMA，

At △BMA and △PMR,

MB=MP∠BMA=∠PMRMA=MR，

∴△BMA≌△PMR(SAS)，

∴AB=PR，

∴on+ 12ab=on+ 12pr=on+pq=oe=2，

That is, the value of ON+ 12AB will not change.