Question 1:AE= 3cm, EF = 5cm；; Let AE=x, then EF = 8-x, AE=4, ∠ A = 90, ∴ x=3, ∴AE="3" cm, EF="5" cm.

Problem 2: solution: as shown in figure 1, ∫∠mfe = 90, ∴∠ DFM+∠ AFE = 90.

∠∠a =∠d = 90，∠AFE=∠DMF，∴△AEF∽△DFM，∴ AE = 3，AF=DF=4，EF。

∴△ circumference of ∴△FMD = 4++= 16 ...

Question 3: The result of ① B will not change.

Reason: As shown in Figure 2, let AF=x, EF = 8-AE, ∴ AE = 4-,

△AEF∽△DFM, =x+8, FD = 8-x can be obtained in the same way as above.

Then, = 16.

②? The conclusion of classmate c is still valid.

Proof: As shown in Figure 2, ∵B and F are symmetric about GE, ∴BF⊥EG is in P, G is GK⊥AB is in K, ∴ FBE = ∠ KGE.

In the square ABCD, GK=BC=AB, ∠ A = ∠ EKG = 90, ∴△AFB≌△KEG, ∴FB=GK. As can be seen from the above, AE = 4-, △ AFB △ keg, \.

S =，(0﹤x﹤8)

When x=4, that is, f coincides with the midpoint of AD, =24.

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