Students in Class 8 (1) went to the field for a math activity class. In order to measure the distance between the two ends of the pond, the following scheme is designed:

(1) As shown in Figure 1, firstly, take a point C on the flat ground that can directly reach A and B, connect AC and BC, respectively extend AC to D and BC to E, so that DC=AC and EC=BC, and finally measure the distance of DE as the length of AB;

(2) As shown in Figure 2, first pass through point B as the perpendicular BF of AB, then take two points C and D on BF to make BC=CD, then pass through point D as the perpendicular DE of BD, and pass through the extension line of AC at point E, then the length of DE is the distance of AB.

Answer the following questions after reading:

(1) Is Scheme (Ⅰ) feasible? Please explain the reasons;

(2) Is Scheme (Ⅱ) feasible? Please explain the reasons;

(3) The purpose of making ed⊥bf bf⊥ab in scheme (Ⅱ) is to make a right triangle; If only ∠ Abd = ∠ BDE ≠ 90 is satisfied, is Scheme (Ⅱ) established? It's not true.

If so, the answer should be

Solution: (1) Scheme (Ⅰ) is feasible;

∵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 making ed⊥bf bf⊥ab in scheme (Ⅱ) is to make a right triangle;

If ∠ Abd = ∠ BDE ≠ 90, ∠ACB=∠ECD.

∴△ABC∽△EDC

∴ Abed = BCCD

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

The length of ed is not equal to that of AB.

∴ Scheme (Ⅱ) is not established.