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.