SAS△AEF； ∠B+∠D= 180

Solution: (1)∵AB=CD,

∴ Rotate △ABE counterclockwise around point A by 90 to △ADG, so that AB and AD coincide.

∴∠BAE=∠DAG，

∠∠BAD = 90，∠EAF=45，

∴∠BAE+∠DAF=45，

∴∠EAF=∠FAG，

∠∠ADC =∠B = 90，

∴∠ FDG = 180, f, d, G***,

At △AFG and △AEF,

AE=AG

∠EAF=∠FAG

AF=AF，

∴△AFG≌△AEF(SAS)，

∴EF=FG，

Namely: ef = be+df.

(2) when ∠ b+∠ d = 180, ef = be+df;

AB = AD，

∴ Rotate △ABE counterclockwise around point A by 90 to △ADG, so that AB and AD coincide.

∴∠BAE=∠DAG，

∠∠BAD = 90，∠EAF=45，

∴∠BAE+∠DAF=45，

∴∠EAF=∠FAG，

∫∠ADC+∠B = 180，

∴∠ FDG = 180, f, d, G***,

At △AFG and △AEF,

AE=AG

∠EAF=∠FAG

AF=AF

∴△AFG≌△AEF(SAS)，

∴EF=FG，

Namely: ef = be+df.

(3) conjecture: DE2=BD2+EC2,

It is proved that △ Abe ′ is obtained by rotating △AEC clockwise by 90 degrees around point A,

∴△aec≌△abe′，

∴be′=ec,ae′=ae，

∠C =∠ABE′，∠EAC =∠E′AB，

In Rt△ABC,

AB = AC，

∴∠ABC=∠ACB=45，

∴∠abc+∠abe′=90，

That is, ∠ e 'bd = 90,

∴e′b2+bd2=e′d2，

∫∠DAE = 45，

∴∠BAD+∠EAC=45，

∴∠e′ab+∠bad=45，

That is, ∠ e' ad = 45,

In AE' d and δ△AE' d,

AE′= AE

∠E′AD =∠DAE

AD=AD

∴△ae′d≌△aed(sas)，

∴de=de′，

∴DE2=BD2+EC2.

analyse

: (1) Rotate △ABE counterclockwise by 90 to △ADG to make AB and AD coincide, then prove △ AFG △ AEF, then get EF=FG, and then get EF = BE+DF;

(2) When ∠ B+∠ D = 180, EF=BE+DF, which is similar to the proof of (1);

(3) According to the fact that △AEC rotates 90 clockwise around point A, △Abe' is obtained. According to the nature of rotation, it is obtained that △ AEC △ Abe ′ is be ′ = EC, AE ′ = AE, ∠ C = ∠ Abe ′, ∠EAC =∞.

Comments: This question mainly examines geometric transformation, and the key is to correctly draw and prove △ AFG △ AEF. This is a comprehensive problem, which is very difficult. The ideas of the examples given in the topic have made good preparations for solving this problem.