∵ 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.