Euclid, Elements I 41©
translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 41: If a parallelogram has a base that’s the same for a triangle and is in the same parallels, the parallelogram is double the triangle.

(general diagram)

Diagrams for book 1 proposition 41, following the display, construction, and demonstration(diagram 1) For let a parallelogram, ABGD, have a base, BG, that’s the same for triangle EBG and let it be positioned in the same parallels, BG, AE. I say that parallelogram ABGD is double triangle BEG. (diagram 2) For let AG be joined. (diagram 3) In fact, triangle ABG is equal to triangle EBG. For it is on the same base as it, BG, and in the same parallels, BG, AE. (diagram 4) But parallelogram ABGD is double triangle ABG. For diameter AG bisects it. (diagram 5) Thus parallelogram is also double triangle EBG. Therefore, if a parallelogram has a base that’s the same for a triangle and is in the same parallels, the parallelogram is double the triangle, just what it was required to show.


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