(A right triangle is one with a right angle; the legs are the two sides that make up the right angle; the hypotenuse is the third side opposite the right angle).
Visually, the theorem can be illustrated as follows: Given a right triangle, with legs a and b and hypotenuse c, (Figure 1)
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c / |
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a
Figure 1
then it follows that a2+b2=c2.
Proof: Draw a right triangle with sides a, b, and c as above. Then take a copy of this triangle and place its a side in line with the b side of the first, so that their c sides form a right angle (this is possible because the angles in any triangle add up to two right angles -- think it through). Then place the a side of a third triangle in line with the b side of the second, again in such a manner that the c sides form a right angle. Finally, complete a square of side (a+b) by placing the a side of a fourth triangle in line with the b side of the third. On the one hand, the area of this square is (a+b)2 because (a+b) is the length of its sides. On the other hand, the square is made up of four equal triangles each having area ab/2 plus one square in the middle of side length c. So the total area of the square can also be written as 4 · ab/2 + c2. We may set those two expressions equal to each other and simplify:
Actually, the sheer volume of distinct known proofs of this theorem is staggering. See [Pythagorean Theorem Proofs] for just a sampling.
The converse of the Pythagorean Theorem is also true: if you have a triangle with sides a, b, and c and such that a2+b2=c2, then there must be a right angle between the sides a and b. This can be proven using the Law of Cosines which is a generalization of the Pythagorean theorem applying to all triangles, not just right-angled ones.
Another generalization of the Pythagorean Theorem was already given by Euclid in his Elements: if one erects similar figures (see Geometry) on the sides of a right triangle, then the sum of the areas of the two smaller ones equals the area of the larger one.
Yet another generalization of the Pythagorean Theorem is Parseval's identity in inner product spaces.
a b c
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