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A right angle can be defined here as the angle formed when two straight lines cross each other in such a way that all 4 angles produced are equal. The theorem also works the other way around: if the lengths of the three sides (a,b,c) of a triangle satisfy the above relation, then the angle between sides a and b must be of 90 degrees. For instance, a triangle with sides a = 3, b = 4, c = 5 (inches, feet, meterswhatever) is rightangled, because a^{2} + b^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = c^{2} Ancient Egyptian builders may have known the (3,4,5) triangle and used it (with measured rods or strings) to construct right angles; even today builders may still nail together boards of those lengths to help align a corner. Many proofs exist and the easiest ones are probably the ones based on algebra, using the elementary identities discussed in the preceding section, namely (a + b)^{2} = a^{2} + 2ab + b^{2} (recall that 2ab means 2 times a times b). For example
= 10^{2} + (2)(10)(5) + 5^{2} = 100 + 100 + 25 = 225 and (a – b)^{ 2} = a^{2} – 2ab + b^{2} For example:
= 10^{2} – (2)(10)(5) + 5^{2} = 100 – 100 + 25 = 25 It is also necessary to know some simple areas: the area of a rectangle is (length) times (width), so the area of the one drawn above is ab. A diagonal cut divides it into two rightangled triangles with short sides a and b, and the area of such a triangle is therefore (1/2) ab. Now look at the square on the left constructed out of four (a,b,c) triangles. The length of each side is (a+b) and therefore the entire square has an area (a+b)^{2}. However, the square can also be divided into four (a,b,c) triangles plus a square of side c in the middle (strictly speaking, we also ought to prove it is a square, but we will skip that). The area of each triangle, as shown earlier, is (1/2)ab, and the area of the square is c^{2}. Since the big square is equal to the sum of all its parts (a + b)^{ 2} = (4)(1/2)(a)(b) + c^{2} Using the identity for (a + b)^{2} and multiplying (4)(1/2) = 2 a^{2} + 2ab + b^{2} = 2ab + c^{2} Subtract 2ab from both sides and you are left with a^{2} + b^{2} = c^{2} The same result can also be shown using a different square, of area c^{2}. As the drawing on the right shows, that area can be divided into 4 triangles like the ones before, plus a small square of side (a– b). We get c^{2} = (4)(1/2)(a)(b) + (a– b)^{ 2} = 2ab + (a^{2} – 2ab + b^{2}) = a^{2} + b^{2} Q.E.D. Q.E.D. stands for "quod erat demonstrandum," Latin for "which was to be demonstrated," and in traditional geometry books those letters mark the end of a proof. The importance of the work of Pythagoras and of later masters of Greek geometry (especially Euclid) was not only in what they proved, but of the method they developed: start from some basic statements which are assumed to be valid ("axioms") and deduce by logic their more complicated consequences ("theorems"). Mathematics still follows the same pattern. For a practical application of the Pythagoras theoremderiving the distance to the horizon (neglecting atmospheric effects)see here. 
Next Stop: #M7 Trigonometry: What is it good for?
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated 25 November 2001