Pythagore
Demonstration of the theorem
Author : Thibaut Bernard 
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Update: Sunday 30 May 2001.
Alphaquark author's Note :
This page is a translation of
Démonstration du théorème de Pythagore
with the help of
Altavista translation.
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Construction of the geometrical figure which will be used for the demonstration
Let us take a rectangle of width A and height B. This rectangle which we make swivel of 90^{o} in the following way :
For each rectangle, let us divide into two in the following way : 



Let us make swivel of 90^{o} the rightangled triangles in the following way yellow and purple : 





We thus find ourselves with four rightangled triangles. 



We note that one finds oneself with a square inside another. 
Demonstration
Notation
Let us take again our last diagram to indicate each of with dimensions by the following letters:
One has four rightangled triangles of which :
the with dimensions one opposed by a,
the base is indicated by b,
and the hypotenuse by n.
Each hypotenuse n thus represent dimensions small square.
With dimensions opposite more the base (a + b) of each rightangled triangle represent each one of with dimensions great square.
Surface squares
Small square : n^{2}.
Great square : (a + b)^{2}.
Surface rightangled triangles
If one associates two rightangled triangles, one finds oneself with a rectangle of surface : a * b.
The total surface of the four rightangled triangles is : 2ab.
Difference in surface
The surface of the small square is equal to the surface of the great square minus the surface of the four rightangled triangles :
n^{2} = (a + b)^{2}  2ab
from where
n^{2} = a^{2} + 2ab + b^{2}  2ab
what gives
n^{2} = a^{2} + b^{2}
Conclusion, the square on the hypotenuse is equal to the sum of the squares of with dimensions of the right angle.