Figure 3 Conclusion: AF-BF = 2oe.

Prove to Figure 2 that point B is the extension line of BG⊥OE and OE at point G,

The quadrilateral BGEF is a rectangle,

∴EF=BG,BF=GE，

In the square ABCD, OA=OB, ∠ AOB = 90,

∵BG⊥OE，

∴∠OBG+∠BOE=90，

∠∠AOE+∠BOE = 90，

∴∠AOE=∠OBG，

At △AOE and △OBG,

∴△AOE≌△OBG(AAS)，

∴OG=AE,OE=BG，

∵af﹣ef=ae,ef=bg=oe,ae=og=oe+ge=oe+bf，

∴AF﹣OE=OE+BF，

∴af﹣bf=2oe；

Fig. 3 proves to be consistent with fig. 2.