Euclid, Elements I 39©
translated by Henry Mendell (Cal. State U., L.A.)
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Prop. 39: Equal triangles that are on the same bases and are on the same sides are also in the same parallels.
(diagram 1) Let there be equal triangles, ABG, DBG that are on the same base and on the same sides of BG I say that they are also in the same parallels. (diagram 2) For Let AD be joined. I say that AD is parallel to BG. (diagram 3) For it not, let a parallel, AE, to straight-line BG be drawn through A, (diagram 4) and let EG be joined. (diagram 5) Therefore, triangle ABG is equal to triangle EBG. For it is on the same base as it, BG, and in the same parallels. (diagram 6) But ABG is equal to DBG. (diagram 7) Therefore, DBG is also equal to EBG, the larger to the smaller, which is impossible. Therefore, AE is not parallel to BG. Similarly, in fact, we will show that not any other except AD is. (diagram 8) Therefore, AD is parallel to BG. Therefore, equal triangles that are on the same bases and are on the same sides are also in the same parallels, just what it was required to show.
Note: it is a requirement of the proof that there is at least one parallel to BG through A. This is guaranteed by the construction I 31.