Euclid,

ElementsI 40©

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

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Prop. 40: Equal triangles that are on equal bases and are on the same sides are also in the same parallels.

(diagram 1) Let here be equal triangles, ABG, GDE on equal bases, BG, GE and on the same sides. I say that they are also in the same parallels. (diagram 2) For let AD be joined AD. I say that AD is parallel to BE. (diagram 3) For if not, let a parallel, AZ, to BE, be drawn through A, and let ZE be joined. (diagram 4) Therefore, triangle ABG is equal to triangle ZGE. For they are on equal bases, BG, GE, and are in the same parallels, BE, AZ. (diagram45 = 3) But triangle ABG is equal to triangle DGE. (diagram 6) Therefore, triangle DGE is also equal to triangle ZGE, the larger to the smaller, which is impossible. Therefore, AZ is not parallel to BE. Similarly, in fact, we will show that not any other except AD is. (diagram 7) Therefore, AD is parallel to BE. Therefore, equal triangles that are on equal 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.