∴∠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?