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Consider a square with a smaller square inside of it, where each of the corners of the smaller square touch the midpoints of the sides of the larger square. ( I would suggest drawing a picture of this. I tried to post one, but I can't for WikiAnswers) There are 4 congruent right triangles formed from this picture. There are two legs to these triangles, of length a and b. The hypotenuse of each of these triangles will be called c. Let us add up one side of the big square. This quantity is (a+b). Let us square this, so the area of the big square is (a+b)2. Thus the area of the big square is a2 + 2ab + b2. We can now subtract the area of the smaller circle which is c2. So we now have a2 + 2ab + b2-c2. We now need to subtract the 4 congruent triangles. The area of a triangle is one half the base times height. In this case, one triangle is .5ab. Multiply this by 4, and we have 2ab. Now we have the entire expression a2 + 2ab + b2-c2 - 2ab = 0, since we have taken the full area of the square, and subtracted out all of the individual parts. The 2ab and the -2ab add up to zero, so we now have a2 + b2-c2 = 0. We can add the -c2 to the other side, thus giving us a2 + b2 = c2 This is one of the many ways to prove the Pythagorean Theorem.
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