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A New Proof of the Pythagorean Theorem Using a Compass and Unmarked Straight Edge

A New Proof of the Pythagorean Theorem Using a Compass and Unmarked Straight Edge. Rahim, M. H., 2003, Int. J. Math. Educ. Sci. Technol., 34 , 144-150.

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A New Proof of the Pythagorean Theorem Using a Compass and Unmarked Straight Edge

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  1. A New Proof of the Pythagorean Theorem Using a Compass and Unmarked Straight Edge Rahim, M. H., 2003, Int. J. Math. Educ. Sci. Technol.,34, 144-150

  2. Mathematics possesses not only the truth, but supreme beauty-a beauty cold and austere, like that of sculpture, without any appeal to our weaker nature…sublimely pure and capable of stern perfection such as only the greatest art can show, Bertrand Russell (1872-1969) A Babylonian clay cuneiform tablet dating from about 1500 B.C. is one of the oldest known documents dealing with number theory. The tablet lists several sets of Pythagorean triples, or distinct positive whole numbers x, y, and z for which x2+y2=z2 Rahim, M. H., 2003, Int. J. Math. Educ. Sci. Technol.,34, 144-150

  3. Thus, BKJI is a rectangle (opposite sides are parallel with 90o angle at B). The area of the rectangle BKJI is equal to the area of the square ABCD: triangles OJP and ODP are congruent so as triangles BKP and BCP and BIO and BAO implying that the rectangle BKJI and the square ABCD are of equal area. Rahim, M. H., 1986, Int. J. Math. Educ. Sci. Technol.,17, 425-447

  4. Creating an area equivalent rectangle to the union of the two squares:

  5. Creating an area equivalent square to the rectangle BSQI Area of BSQI = BI*BS = n*(w+n) =n*w+n2 =m2+n2

  6. Given a right angle triangle and the altitude to the hypotenuse, then the altitude is the geometric mean of the segments into which it separates the hypotenuse a = x x b TB/UB = BS/TB (TB)2 = BS*UB x2 = n*(w+n) x2 = n*w+n2 x2 = m2+n2

  7. The square of side TB is area equivalent to the union of the tow squares ABCD and EFGH

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