1.) Start with a right triangle whose sides are a and b and with hypotenuse c ==>
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2.) Make a large square consisting of 4 of the right triangles with an inscribed square whose sides are c in length:
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3.) Now, the sides of the larger square are (a + b) in length. The area of this big square is:
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    Big square area = (a + b)2
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4.) The area of this large square is also the sum of the areas of the 4 right triangles (which is ab/2) plus the area of the inner square, which is c2:
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    Big square area = 4 (ab/2) + c2
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5.) Since these 2 areas are the same we can set them equal to each other:
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    4 (ab/2) + c2 = (a + b)2
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6.) By the distributive property:   (a + b)2 = (a + b) (a + b)                 = a (a + b) + b ( a + b)                 = a2 + ab + ab + b2                 = a2     +2ab +     b2
| Putting this back in gives us ===>
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    4 (ab/2) + c2 = a2 + 2 ab + b2
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7.) Simplifying the 4(ab/2) term, we have:
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    2ab + c2 = a2 + 2ab + b2
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8.) Finally, subtracting 2ab from both sides of the equation gives us the theorem of Pythagoras: |
      2ab + c2 = a2 + 2ab + b2
    -2ab                   -2ab
          c2 = a2   +     b2      
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