The thing that makes the proof work is that in both cases, the side of
the big square is the long and short sides of the triangle put together.
The four triangles are just moved from one place to another between the
two, so they cover the same area. That means the area remaining - blue in
one case, red in the other - is the same. The blue area is exactly the
square on the hypotenuse, and the red area is the sum of the squares on
the other two sides.
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timeasmymeasure
no subject
Date: 2009-02-21 11:05 am (UTC)The thing that makes the proof work is that in both cases, the side of the big square is the long and short sides of the triangle put together. The four triangles are just moved from one place to another between the two, so they cover the same area. That means the area remaining - blue in one case, red in the other - is the same. The blue area is exactly the square on the hypotenuse, and the red area is the sum of the squares on the other two sides.
Is that any clearer?