∴∠ 1+∠ 3 = 90, that is ∠ 3 = 90-∠ 1

∴∠ 2+∠ 4 = 90, that is ∠∠ 4 = 90-∠ 2.

∵∠ 1=∠2,

∴∠3=∠4,

∴AE=EF，

∫ AD ∨ BC,

∴∠2=∠5,

∵∠ 1=∠2,

∴∠ 1=∠5,

∴AE=AD，

∴ef=ad(2 points)

∫AD∨EF，

∴ Quadrilateral AEFD is a parallelogram, (1 point)

AE = AD，

∴ The quadrilateral AEFD is a diamond, (1 point)

(method 2) ∵AD∨BC,

∴∠2=∠5,

∵∠ 1=∠2,

∴∠ 1=∠5,

∵AF⊥DE，

∴∠AOE=∠AOD=90，

In △AEO and △ ADO, ∠ 1 = ∠ 5 ∠ AOE = ∠ Aodao = AO.

∴△AEO≌△ADO，

∴EO=OD

In △AEO and △FEO, ∠ 1 = ∠ 2eo = EO ∠ AOE = ∠ foe.

∴△AEO≌△FEO，

∴ao=fo(2 points)

∴AF and Ed are equally divided, (1 minute)

∴ Quadrilateral AEFD is a parallelogram,

And ∵AF⊥DE,

∴ The quadrilateral AEFD is a diamond; ( 1)

(2)(5 points) ∵ diamond AEFD,

∴AD=EF，

BE = EF，

∴AD=BE，

And ∵ AD ∨ BC,

∴ Quadrilateral ABED is a parallelogram, (1 min)

∴AB∥DE，

∴∠BAF=∠EOF，

Similarly, it can be seen that the quadrilateral AFCD is a parallelogram,

∴AF∥DC，

∴∠EDC=∠EOF，

∵AF⊥ED again,

∴∠EOF=∠AOD=90，

∴∠ BAF =∠ EDC =∠ EOF = 90 degrees, (2 points)

∴∠ 5+∠ 6 = 90, ( 1)

∴∠bad+∠adc=∠baf+∠6+∠5+∠edc=270； ( 1)

(3)(3 points) According to (2), the parallelogram AFCD with BAF = 90,

∴AF=CD=n，

AB = m，s △ abf = 12ab？ Af = 12mn，( 1)

As can be seen from (2),

∴DE=AB=m，

According to (1), OD = 12DE = MS quadrilateral AFCD = AF? OD = 12mn，( 1)

S quadrilateral ABCD=S△ABF+S quadrilateral AFCD = Mn. (1 min)