 The right triangle theorem claims a relationship between these sides and hypotenuse of a right triangle. It's usually called the Pythagorean theorem, but we should stop doing that because the actual Pythagorean theorem has nothing to do with triangles. The right triangle theorem first appeared in Mesopotamia, around 1800 BC, and it was explicitly stated by ancient Indian geometers by about 800 BC. There's a claim that ancient Egyptian rope stretchers form 3, 4, 5 right triangles. The evidence is based on the existence of ropes of length 3, plus 4, plus 5, 12. Since the only conceivable use of a marked rope is to form right triangles, and surveyors have no other method of forming right angles, and the ancient Egyptians actually cared about right angles, this must be true. Or it could be wild speculation based on equivocal evidence and dubious claims. There are over 100 different proofs of the right triangle theorem, the exact number depending on how you define different. The earliest, a proof of a special case, appeared in Plato's Meno. The general case and converse would be proven by Euclid in the last two propositions of Book I of the Elements. The proof in the Elements is believed to be original with Euclid, and Book I seems to be organized with the proof as a goal. The Euclidean proof is based on constructing squares, that's Book I, Proposition 46, what we now called side-angle-side congruence, Book I, Proposition 4, and the equality of parallelogramic regions, Book I, Proposition 41. The construction procedure as follows. Let A, B, C be a right triangle with our right angle at A, construct the squares on the sides A, B, A, C, B, C, join B, K, A, E, and then drop A, L perpendicular to B, C. Now, because a diagram is evocative of a windmill, Euclid's proof is sometimes known as the windmill proof, although Euclid himself wouldn't have used that term since windmills first appeared in the Islamic world a thousand years after Euclid wrote the Elements. So, let's consider the triangles B, C, K, and E, C, A. B, C is C, E since they're sides of the same square, A, C is C, K since they're sides of the same square, and angle E, C, A is a right angle plus angle A, C, B. B, C, K is a right angle plus angle A, C, B. So, angle E, C, A is equal to angle B, C, K, and so now we have two sides in the included angle of one triangle equal to the two sides and included angle of another triangle, so triangles B, C, K, and E, C, A are equal. Now, notice that triangle B, C, K is on base C, K with vertex on A, H parallel to C, K. So, it's going to be half of the square A, C, K, H. Likewise, triangle E, C, A is on base C, E with vertex on A, L parallel to C, E. So, it's half the rectangle C, E, L, M. So, the square A, C, K, H is equal to the rectangle C, E, L, M. By essentially the same argument, the square A, B, F, G is equal to the rectangle B, M, L, D. So, the squares on the sides A, B, A, C are together equal to the square of a hypotenuse, B, C. And that completes the proof. It's worth noting that in this form, we're talking about the literal squares on the sides of the hypotenuse of our triangle.