Rotate △ADE clockwise by 90 degrees around point A to get △ABG. At this time, AB and AD overlap, which can be obtained by rotation:

AB=AD，BG=DE，∠ 1=∠2，∠ABG=∠D=90，

∴∠ABG+∠ABF=90 +90 = 180，

Therefore, point G, point B and point F are on the same straight line.

≈EAF = 45 ∴∠2+∠3=∠bad-∠eaf=90-45 = 45。

∵∠ 1=∠2,∴∠ 1+∠3=45 .

That is ∠ GAF = ∠ FAE.

And AG=AE, AF=AF.

∴△GAF≌△EAF.

∴GF=EF, so de+BF = ef.

2

Proof: Generalize CF, as ∠4=∠ 1,

∵ Rotate Rt△ABC along the hypotenuse to get △ADC, where point E and point F are points on the sides of DC and BC ∠EAF= 1/2∠DAB respectively.

∴∠ 1+∠2=∠3+∠5,

∠2+∠3=∠ 1+∠5,

∵∠4=∠ 1,

∴∠2+∠3=∠4+∠5,

∴∠GAF=∠FAE，

At △AGB and △AED,

∠4=∠ 1

AB=AD

ABG =∠ < ade

∴△AGB≌△AED(ASA)，

∴AG=AE,BG=DE，

∫ In △AGF and △AEF,

AG=AE

∠GAF=∠EAF

AF=AF

∴△AGF≌△AEF(SAS)，

∴GF=EF，

∴de+bf=ef；

three

When ∠B and ∠D satisfy ∠ b+∠ d = 180, DE+BF = ef.