∴∠BAD+∠DAM+∠MAE+∠EAC=90。

∫∠DAE = 45，

∴∠BAD+∠EAC=45。

∵∠BAD=∠DAM，

∴∠BAD+∠EAC=∠DAM+∠EAC=45，

∴∠BAD+∠MAE=∠DAM+∠EAC，

∴∠MAE=∠EAC, that is, AE shares ∠ MAC;

(2) The method of choosing Xiaoying.

Proof: As shown in Figure 2, connect EF.

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

ABDECF (Figure 2)

∵∠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.

(Prove the corresponding score by rotating.

(3) 135 when?