Euclid, Elements I 42©
translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 42: To construct a parallelogram equal to the given triangle in the given rectilinear angle.

(general diagram)

Diagrams for book 1 proposition 42, following the display, construction, and demonstration(diagram 1) Let the given triangle be ABG, the given rectilinear angle D. It is, in fact, required to construct a parallelogram equal to triangle ABG in rectilinear angle D.
(diagram 2) Let BG be bisected at E, (diagram 3) and let AE be joined, (diagram 4) and let an equal to angle D, that by GEZ, be constructed on straight-line EG and at the point on it, E. (diagram 5) Also let a parallel, AH, to EG be drawn through A, (diagram 6) and let a parallel, GH, to EZ be drawn through G. (diagram 7) Therefore, ZEGH is a parallelogram. (diagram 8) And since, BE is equal to EG, triangle ABE is also equal to triangle AEG. For they are on equal bases, BE, EG, and are in the same parallels, BG, AH. (diagram 9) Therefore, triangle ABG is double triangle AEG. (diagram 10) But parallelogram ZEGH is also double triangle AEG. For it has the same base as it and is in the same parallels as it. (diagram 11) Therefore, parallelogram GEZ is equal to triangle ABG. It also has the angle by GEZ equal to the given angle D. Therefore, a parallelogram, ZEGH, has been constructed equal to the given triangle, ABG, in an angle, that by GEZ, which is equal to D, just what it was required to make.