∴∠B=∠D,AB=AD=BC=CD，

∵ points e, f, g and h are on AB, BC, CD and DA respectively, AE=AH=BF=DG=x,

∴BE=DH，

At △BEF and △DHG,

BE=DH∠B=∠HBF=DG，

∴△bef≌△dhg(sas)；

(2) Solution: It is a constant,

Even HF, ∫AH∨. BF，

∴ Quadrilateral ABFH is a parallelogram,

Similarly, it can be concluded that the quadrilateral HFCD is also a parallelogram,

∴ s =12 (sabfh+shfcd) =12s diamond =12x3 = 3;

(3) Solution: Let HM⊥FG pass through point H at point M,

According to the symmetry of the graph and eh < fg, it can be seen that the quadrilateral EFGH is an isosceles trapezoid.

EH = EF = 5，FG= 1 1，

∴MG=3，

Using Pythagorean theorem, we can get the height HM=4 of trapezoid.

Ah. BF，

∴ Quadrilateral ABFH is a parallelogram,

Using Pythagorean Theorem, AB=FH=64+ 16=45,

AE = AH，AB∨FH，

∴∠ 1=∠2=∠3,

EH = EF，EH∨FG，

∴∠3=∠4=∠5,

∴∠ 1=∠2=∠4=∠5,

∴∠B=∠ 1+∠2=∠4+∠5=∠EFG

∴tanB=tan∠EFG=43.