∫△ABC?△FDE，∴ ∠ FDE = ∠ B
∴∠FDE=∠DCB。 ∴DG∥BC。 ∴∠AGD=∠ACB=90 .∴DG⊥AC。
And ∵DC=DA, ∴G is the midpoint of AC.
∴CG= AC= ×8=4，DG= BC= ×6=3 .
∴S DCG = CG DG= ×4×3=6 .
(2)∵△ABC≌△FDE,∴∠B=∠ 1。
∠∠c = 90° ,ed⊥ab,∴∠a+∠b=90，∠a+∠2 = 90°。 ∴∠B=∠2。
∴∠ 1=∠2。 ∴GH=GD。
∠∠a+∠2 = 90，∠ 1+∠3=90 ,∴∠A=∠3。 ∴AG=GD。 ∴AG=GH。
Point g is the midpoint of AH.
In Rt△ABC, AB= 10,
∫d is the midpoint of AB, ∴AD= AB=5.
In △ADH and △ACB, ∫∠A =∠A, ∠ ADH = ∠ ACB = 90,
∴△ADH∽△ACB。 ∴ is the solution.
∴s△dgh = s△ADH =×DH ad =×5 = .
(3)① 。
(2) As shown in Figure 4, rotate △DEF around point D to make DE⊥BC intersect with point M and DF intersect with point A C, and find the area of the overlapping part (quadrangle DMCN). (The answer is not unique)
Test and analysis: (1) By proving that DG is the midline of Rt△ABC, the area of overlapping part (△DCG) can be obtained.
(2) By proving that S △DGH = S △ ADH, the area of overlapping part (△DGH) can be obtained.
(3)① As shown in the figure, rotate △DEF around point D to the △ de ′ f ′ position of (2), MP⊥DM crosses point M when it crosses point P, NQ⊥DM crosses point N when it crosses point Q, and then
∠∠NDQ =∠QDM (equal rotation angles), DM=MN, ∠ dnq =∠ dqm = 90,
∴△ABC≌△FDE(ASA)。
∴S △ DDMN =S △ DGH == .
2 Open (the answer is not unique).