Euclid,

ElementsI 2 (problem)©

translated by Henry Mendell (Cal. State U., L.A.)

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Go to Prop 1. Go to Prop. 3

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Prop. 2: To position at the given point a straight-line equal to the given line.

(diagram 1). Let the given point be A, and the given straight-line BG. It is, in fact, required to place at point A a straight-line equal to the given straight-line, BG.

(diagram 2)
Let a straight-line, AB, be joined from A to point B, (diagram 3) and let an equilateral triangle, DAB, be constructed on it (prop. 1), (diagram 4) and let straight-lines, AE, BZ, be extended in a straight-line with DA, DB,(diagram 5) and with a center, B, and a distance, BG, let a circle be described, GHQ, (diagram 6) and again with a center, D, and a distance, DH, let a circle be described, HKL. And so, since point B is center of circle GHQ, BG is equal to BH. Again, since point D is center of circle KLH, DL is equal to DH, where DA is equal to DB. Therefore, a remainder, AL, is equal to a remainder, BH. But BG was also shown equal to BH. Therefore, each of AL, BG is equal to BH. But things equal to the same are also equal to one another. Therefore, AL is also equal to BG. Therefore, at the given point, A, a straight-line, AL, has been positioned equal to the given line, BG, just what it was required to make.

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