The Garden of Archimedes  A Museum for Mathematics

Pythagoras and his theorem

A simple proof of the Pythagorean theorem

	The first precise statement, and the first unequivocal proof of the theorem are found in the first book of the Elements by Euclid (circa 300 B.C.). In right-angle triangles, the square of the side opposite the right angle is equal to the squares on the sides containing the right angle. Today the "side opposite the right angle" is generally called the hypotenuse, while the "sides containing the right angle " take the name of cathets. Furthermore, instead of "equal" it's preferable to say equivalent to, or with the same area. Therefore a modern formulation of the theorem could be: in right-angle triangles, the area of the square built on the hypotenuse is equivalent to the sum of the areas of the squares built on the cathets, or also, given that the area of the square is equivalent to the square of the side, in right-angle triangles, the square of the hypotenuse is equivalent to the sum of the squares of the cathets. If the latter are indicated by a and b, and the hypotenuse with c, the theorem takes the algebric shape: a2 + b2 = c2
	Among classical theorems, the one that bears the name of Pythagoras may be the one proved in the most different ways. Among all of these, the simplest is probably that depicted to the side. The right-angled triangle in question is one of those coloured in red. The large square, having the sum of the cathets as one side, is formed by four triangles and by the two squares built on the cathets. In the second, it is formed by the same four triangles placed differently, and by the square of the hypotenuse. Since the area of the large square and that of the four triangles is the same in both cases, so are the areas of the remaining figures (the squares built on the cathets in the former, and the square on the hypotenuse in the latter). As it can be seen, the proof is very easy, and above all it's clear - the result can be seen even before the theorem is reasoned out. On the other hand, it's always important to be cautious before accepting the validity of a visual proof. Sight can be deceptive. What seems to be a square can be a rectangle with almost equal sides. Two apparently identical figures can actually differ, if only by a little. In this way, many geometrical paradoxes are constructed.
	In our case, the proof is correct, but not yet complete. We still need to demonstrate that the white parts of the two figures are actually square, and more precisely the squares on the cathets in the former figure and that of the hypotenuse in the latter. In the first figure this is evident by construction. In the second one, all of the sides of the quadrilateral in question are equal to the hypotenuse, and therefore it only needs to be verified that the angles are right angles. Let's consider, for example, the angle with its vertex in point A. Together with the two red ones that share its vertex, it forms two right angles. But the sum of the angles of a triangle is also equal to two right angles, and therefore the white angle with its vertex in A is equivalent to the third angle of the triangle, that is a right angle. Similarly it can be proved that the other angles are right angles - (this isn't be necessary, because a rhombus with a right-angle is a square), and therefore the figure is a square, with the hypotenuse as the side.