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C² = (a - b)² + 2ab   = a² - 2ab + b² + 2ab   = a² + b² Proof #4 The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side ( a + b) and a hole with the side c. We can compute the area of the big square in two ways. Thus ( a + b)² = 4 ab/2 + c² simplifying which we get the needed identity. A proof which combines this with is credited to the 12th century Hindu mathematician Bhaskara (Bhaskara II): (p.

4) gives Bhaskara credit also for. Here we add the two identities c² = ( a - b)² + 4 ab/2 and c² = ( a + b)² - 4 ab/2 which gives 2 c² = 2 a² + 2 b². The latter needs only be divided by 2.

This is the algebraic proof # 36 in. Its variant, specifically applied to the, has featured in the Chinese classic Chou Pei Suan Ching dated somewhere between 300 BC and 200 AD and which Loomis refers to as proof 253. Proof #5 This proof, discovered by President J. Garfield in 1876 [], is a variation on the previous one. But this time we draw no squares at all.

The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - ( a + b)/2( a + b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c c/2. As before, simplifications yield a² + b² = c². (There is more to that.) Two copies of the same trapezoid can be combined in two ways by attaching them along the slanted side of the trapezoid. One leads to the, the other to.: it also invokes an image of trapezoid but under in a different light.

Proof #6 We start with the original right triangle, now denoted ABC, and need only one additional construct - the AD. The triangles ABC, DBA, and DAC are similar which leads to two ratios: AB/BC = BD/AB and AC/BC = DC/AC. Written another way these become ABAB = BDBC and ACAC = DCBC Summing up we get.

ABAB + ACAC = BDBC + DCBC = (BD+DC)BC = BCBC. In a little different form, this proof appeared in the Mathematics Magazine, 33 (March, 1950), p. 210, in the Mathematical Quickies section, see, by C. Taking AB = a, AC = b, BC = c and denoting BD = x, we obtain as above a² = cx and b² = c(c - x), which perhaps more transparently leads to the same identity. In a private correspondence, Dr.

France Dacar, Ljubljana, Slovenia, has suggested that the diagram on the right may serve two purposes. First, it gives an additional graphical representation to the present proof #6.

In addition, it highlights the relation of the latter to. Mentock has observed that a little trick makes the proof more succinct. In the common notations, c = b cos A + a cos B. But, from the original triangle, it's easy to see that cos A = b/c and cos B = a/c so c = b (b/c) + a (a/c). This variant immediately brings up a question: are we getting in this manner a trigonometric proof? I do not think so, although a trigonometric function (cosine) makes here a prominent appearance.

The ratio of two lengths in a figure is a meaning that it remains fixed in passing between similar figures, i.e., figures of the same shape. That a particular ratio used in the proof happened to play a sufficiently important role in trigonometry and, more generally, in mathematics, so as to deserve a special notation of its own, does not cause the proof to depend on that notation.

(However, check where trigonometric identities are used in a significant way.) Michael Brozinsky came up with a that I believe could be properly referred to as lipogrammatic. Finally, it must be mentioned that the configuration exploited in this proof is just a specific case of the one from the next proof - Euclid's second and less known proof of the Pythagorean proposition. A separate page is devoted to a proof by the.

Proof #7 The next proof is taken verbatim from Euclid VI.31 in translation by Sir Thomas L. Polya analyzes it in his Induction and Analogy in Mathematics (II.5) which is a recommended reading to students and teachers of Mathematics. In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle. Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.