Proof: connection points c, f;

Because △ABC is an isosceles right triangle and f is the midpoint of AB;

So CF=FB, and ∠DCF=∠EBF=45 degrees.

And because AD=BC. You can get DC = EB；;

So △DCF and △EBF are the same;

So FD=FE, and ∠ DFC = ∠ EBF;

And ∠ EBF+∠ EFC = 90, you can get ∠ EFC+∠ DFC = ∠ DFE = 90;

So: △DFE is an isosceles right triangle

2 is wrong.

As a counterexample, when d and e move to the midpoint of their respective sides,

Quadrilateral CDFE is a square.

(3) is wrong.

Prove:

Because: △DFE is an isosceles right triangle;

So de = √ 2df;

And DF is the distance from the fixed point f to the moving point d, where d is the midpoint of AC, that is, DF is perpendicular to AC and DF is at least 4.

So the minimum value of DE is 4√2, not 4;

4 that's right.

Prove:

△DCF and△△ EBF are congruent (previously proved)

So the quadrilateral CDFE = s △ DCF+s △ CEF = s △ EBF+s △ CEF = s △ CFB =16 is a constant value.

⑤ It is right.

Prove:

S △ CDE = (DC * CE)/2 = CE * (AC-AD)/2 = CE * (8-CE)/2 = [-(CE-4) squared+16]/2; (0 & lt= CE & lt=8)

When CE=4, S△CDE is the largest, and the maximum value is 8.

To sum up, the correct conclusion is ① ④ ⑤, so choose B.