∵ In the isosceles right angle △ABC, AB=AC, ∠ BAC = 90,

∴∠B=∠C=45。

Point o is the midpoint of BC,

∴OA=OC,∠EAO=∠C=45。

∫∠EOF = 90，

∴∠ AEO =∠b+∠ BOE, ∠ CFO =180-∠ C-(180-∠ BOE -90)= 45+∠.

∴∠AEO=CFO，

At △AEO and △CFO,

∠AEO =∠ CFO ∠EAO=∠COA=OC,

∴△AEO≌△CFO(AAS)，

∴ae=cf；

(2) The method of choosing Xiaoying.

Proof: As shown in Figure 2, connect EF.

According to the folding, ∠BAD=∠FAD, AB=AF, BD=DF,

∵∠BAD=∠FAD，

∴, ∠ CAE = ∠ FAE from (1).

At △AEF and △AEC,

AF = AC∠FAE =∠caae = AE，

∴△AEF≌△AEC(SAS)，

∴CE=FE,∠AFE=∠C=45。

∴∠DFE=∠AFD+∠AFE=90。

In Rt△DFE, DF2+FE2=DE2,

∴BD2+CE2=DE2.？

(3) Solution: When 135 < α < 180, the equivalence relation BD2+CE2=DE2 still holds. The evidence is as follows:

As shown in Figure 4, according to Xiaoying's method, let AB and EF intersect at G point.

∫ Fold △ABD in half along the straight line where AD is located to get △ADF,

∴AF=AB,∠AFD=∠ABD= 135, ∠ Bad = ∠ Fashion.

ac = ab，∴ AF = AC。

∠∠CAE = 90-∠BAE = 90-(45-∠bad)= 45+∠bad = 45+∠FAE。

∴∠CAE=∠FAE.

At △AEF and △AEC,

AF = AC∠FAE =∠caae = AE，

∴△AEF≌△AEC(SAS)，

∴CE=FE,∠AFE=∠C=45。

∴∠dfe=∠afd-∠afe=∠ 135-∠c = 135-45 = 90。

∴∠DFE=90。

In Rt△DFE, DF2+FE2=DE2, ∴ BD2+Ce2 = DE2.