Demonstration of the theorem



Source of this page

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.
I hope this translation is good, but if there are any errors, you can write me.
If this translation is successful, perhaps I will try to translate another document of Alphaquark.



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 90o in the following way :

For each rectangle, let us divide into two in the following way :




Let us make swivel of 90o the right-angled triangles in the following way yellow and purple :



We thus find ourselves with four right-angled triangles.




We note that one finds oneself with a square inside another.






Let us take again our last diagram to indicate each of with dimensions by the following letters:

One has four right-angled 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 right-angled triangle represent each one of with dimensions great square.


Surface squares

Small square : n2.

Great square : (a + b)2.


Surface right-angled triangles

If one associates two right-angled triangles, one finds oneself with a rectangle of surface : a * b.

The total surface of the four right-angled 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 right-angled triangles :

n2 = (a + b)2 - 2ab

from where

n2 = a2 + 2ab + b2 - 2ab

what gives

n2 = a2 + b2

Conclusion, the square on the hypotenuse is equal to the sum of the squares of with dimensions of the right angle.