Solution: See the figure below: Make a right triangle ABC, ∠A=90D, and take A, B and C as the opposite sides of ∠A, ∠B and ∠C respectively.

Draw the following picture: A 2-B 2 = (A+B) (A-B) = C 2 (Pythagorean Theorem)

1. Make a circle with C as the center and B as the radius, and pass through BC in H and BC extension line in D respectively;

2. Make a circle with B as the center and BF as the radius, intersecting BC in F and the reverse extension line in E respectively.

3. Select the midpoint O on the line segment ED (the perpendicular line of ED intersects with O);

4. Make a circle with ED/2 as the radius, and intersect with ED at E and D respectively;

5. BG⊥ED, o intersecting in G;

6. Make a circle with B as the center and AB=c as the radius. O in g; Then BG = AB = C. This is the end of the graph.

Because BG is the middle value of EB and BD, there are:

c^2=bg^2=eb*bd=bf*bd=(a-b)(bf+fc+cd)=(a-b)(a+b)=a^2-b^2。 Complete the certificate.

Take the auxiliary line in the diagram with the side length of triangle ABC as the reference for proof.

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