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By Richard Bellman

ISBN-10: 080530360X

ISBN-13: 9780805303605

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29]. So ABC and BCD are two triangles having the two angles ABC and BCA equal to the two (angles) BCD and CBD, respectively, and one side equal to one side—the (one) by the equal angles and common to them, (namely) BC. Thus, they will also have the remaining sides equal to the corresponding remaining (sides), and the remaining angle (equal) to the remaining angle [Prop. 26]. Thus, side AB is equal to CD, and AC to BD. Furthermore, angle BAC is equal to CDB. And since angle ABC is equal to BCD, and CBD to ACB, the whole (angle) ABD is thus equal to the whole (angle) ACD.

The Greek text has “CD, BC”, which is obviously a mistake. ‡ The Greek text has “ABCD”, which is obviously a mistake. Ð . Proposition 35 Τὰ παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ Parallelograms which are on the same base and beἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. tween the same parallels are equal† to one another. Α Ε ∆ Ζ A D Η Β F G Γ B ῎Εστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΒΓΖ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΖ, ΒΓ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ τῷ ΕΒΓΖ παραλληλογράμμῳ.

To [triangle] DCE. Thus, [triangle] DCE is also equal to triangle F CE, the greater to the lesser. The very thing is impossible. Thus, AF is not parallel to BE. Similarly, we can show that neither (is) any other (straight-line) than AD. Thus, AD is parallel to BE. Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show. This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text.

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