∵DC=AC, EC=BC, and there is a diagonal ∠ACB=∠DCE.

∴△ACB≌△DCE(SAS)

∴AB=DE

∴ The distance to measure DE is the length of AB.

Therefore, scheme (1) is feasible.

(2) Scheme (Ⅱ) is feasible;

∵AB⊥BC,DE⊥CD

∴∠ABC=∠EDC=90

BC = CD，∠ACB=∠ECD。

∴△ABC≌△EDC

∴AB=ED

The measuring length DE is the distance of AB.

Therefore, scheme (Ⅱ) is feasible.

(3) The purpose of ed⊥bf bf⊥ab in scheme (II) is ∠ Abd = ∠ BDE.

If only ∠ Abd = ∠ BDE ≠ 90 is satisfied, scheme (2) is not established;

Reason: If ∠ Abd = ∠ BDE ≠ 90, ∠ACB=∠ECD.

∴△ABC ∽ △EDC

∴ AB ED = BC CD,

As long as the lengths of ED, BC and CD are measured, the length of AB can be obtained.

However, there are no other conditions for this problem, and the lengths of other line segments may not be measured.

∴ Scheme (Ⅱ) is not established.