∴ 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.