The area of such a rectangle is a times b: ab. Therefore the four triangles together are equal to two such rectangles. Their area is 2ab.

As for the square whose side is c, its area is simply c. Therefore, the area of the entire square is

c + 2ab . . . . . .(1)

At the same time, an equal square with side a + b (Fig. 2) is made up of a square whose side is a, a square whose side is b, and two rectangles whose sides are a, b. Therefore the area of that square is

a + b + 2ab

But this is equal to the square formed by the triangles, line(1):

a + b + 2ab = c + 2ab.

Therefore, on subtracting the two rectangles -- 2ab -- from each square, we are left with

a + b = c.

This is the Pythagorean Theorem

Proof using similar triangles

The Pythagorean theorem, is based on the proportionality of the sides of two similar triangles.

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As

so

These can be written as

Summing these two equalities, we obtain

In other words, the Pythagorean theorem:

The Arabian mathematician Thabit ibn Kurrah

A clever proof by dissection which reassembles two small squares into one larger one was given by the Arabian mathematician Thabit ibn Kurrah (Ogilvy 1994, Frederickson 1997).

Proof by Perigal

Another proof by dissection is due to Perigal (left...

Therefore the four triangles together are equal to two such rectangles. Their area is 2ab.

At the same time, an equal square with side a + b (Fig. 2) is made up of a square whose side is a, a square whose side is b, and two rectangles whose sides are a, b. Therefore the area of that square is

Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As

Another proof by dissection is due to Perigal (left figure; Pergial 1873; Dudeney 1970; Madachy 1979; Steinhaus 1999, pp. 4-5; Ball and Coxeter 1987). A related proof is accomplished using the above figure at right, in which the area of the large square is four times the area of one of the triangles plus the area of the interior square. From the figure d=b-a, so

Perhaps the most famous proof of all times is Euclid's geometric proof , although it is neither the simplest nor the most obvious.
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