Euclid, Elements I 44©
translated by Henry Mendell (Cal. State U., L.A.)
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Prop. 44: To apply along the given straight-line a parallelogram equal to the given triangle in the given rectilinear angle.
(diagram 1) Let the given straight line be AB, the given triangle G, the given rectilinear angle D. It is, in fact, required to apply along the given straight-line, AB, a parallelogram equal to the given triangle, G, and in an angle equal to angle D.
(diagram 2) Let a parallelogram, BEZH, be constructed equal to triangle G in an angle, that by EBH, which is equal to D. And let it be positioned so that BE is in a straight-line with AB, (diagram 3) and let ZH be drawn to Q, (diagram 4) and let a parallel, AQ, to either of BH, EZ, be drawn through A. (diagram 5) And let QB be joined. (diagram 6) And since a straight-line falls into parallels, AQ, EZ, therefore, the angles by AQZ, QZE are equal to two right-angles. Therefore, the angles by BQH, HZE are smaller than two right-angles. But straight-lines from angles smaller than two right-angles in being extended to infinity fall-together. Therefore, QB, ZE, in being extended will fall-together. (diagram 7) Let them be extended and meet at K, (diagram 8) and through point K, let a parallel, KL, to either of EA, ZQ be drawn through K, (diagram 9) and let QA, HB be extended to points L, M. (diagram 10) Therefore, QLKZ is a parallelogram, (diagram 11) and QK its diameter, and parallelograms, AH, ME, about QK, and the so-called complements, LB, BZ. Therefore, LB is equal to BZ. But BZ is equal to triangle G. Therefore, LB is also equal to G. (diagram 12) And since the angle by HBE is equal to that by ABM, but that by HBE is equal to D, therefore, the angle by ABM is also equal to angle D. Therefore, a parallelogram equal to the given triangle, G, in an angle, that by ABM, which is equal to D, has been applied along the given straight-line, AB, just what it was required to make.