(2) Select one to search:

Proof: extend CB to k, make BK=DE, even AK, then △ AKB △ AED.

∠∠BAF+∠DAE = 45，

∴∠KAF=45，

∴∠KAF=∠FAE.

AK = AE，AF=AF，

∴△AKF≌△AEF.

∴KF=EF.

And ∵BK=DE,

∴EF=BF+DE

Select b to find:

Proof: extend CB to K, make BK=DE, even AK, then △ AKB △ AED.

∠∠BAF+∠DAE = 45，

∴∠KAF=45，

∴∠KAF=∠FAE.

AK = AE，AF=AF，

∴△AKF≌△AEF.

∴KF=EF.

And ∵BK=DE,

∴EF=BF+DE

△CEF perimeter =CF+CE+EF

=CF+CE+(BF+DE)

=(CF+BF)+(CE+DE)

=BC+DC=2a (fixed value)

Select c to find:

Proof: As shown in the figure, intercept AG=AM on AK and connect BG, gn.

AG = AM，AB=AD，∠KAB=∠EAD，

∴△ABG≌△ADM，

∴BG=DM,∠ABG=∠ADB=45。

∫∠Abd = 45，

∴∠GBD=90。

∠∠BAF+∠DAE = 45，

∴∠KAF=45，

∴∠KAF=∠FAE.

And ∵AG=AM, AN=AN,

∴△GAN≌△NAM.

∴NG=MN，

∫∠GBD = 90 degrees,

∴BG2+BN2=NG2，

∴BN2+DM2=MN2.

To sum up, the findings of students A, B and C are all correct.