∴ eh = ad+DH-AE = a+b-2b = a-B. (2 points)

(2)∵AG=AB-BG=a-b，EH=a-b，

∴ Company = Uh. (3 points)

∵∠FAG=45 +90 = 135，

∠FEH= 180 -45 = 135，

∴∠ fag = ∠ feh。 (4 points)

∵△AFE is an isosceles right triangle,

∴ AF = Fe。 (5 points)

In △AGF and △EHF, af = ef ∠ fag = ∠ fehag = eh,

∴△AGF≌△EHF, that is, it can rotate△ AGF around F to the position of△ EHF. (7 points)

(3) for FI⊥AD, the vertical foot is me;

∫△AFE is an isosceles right triangle

Is the midline on the hypotenuse of ∴FI = ie =12ae =12? 2b=b

∴IH=IE+EH=b+a-b=a

∴FI=DH=b, IH=DC=a, and fih = hdc = 90.

∴△FIH≌△HDC(SAS)

∴FH=HC①( 10 integral)

∵△AGF?△EHF, △BCG rotates 90 clockwise around point C to △CDH.

∴FG=FH②,GC=HC③

FH=HC=CG=FG from ① ② ③.

∴ Quadrilateral FHCG is a diamond (12 points).

from△AGF?△EHF:∠ 1 =∠2。

∠ 1+∠GFE=∠2+∠GFE=90

∴ Quadrilateral FHCG is a square. (13)

In Rt△BCG, according to Pythagorean theorem: GC2=BC2+BG2=a2+b2.

The ∴ area of a square GCHF =GC2=a2+b2

Xiaoming's exploration can be successful. (14)

Note: Answer by other methods, as long as it is correct, score according to the standard.