Euclid's Elements by Fitzpatrick R. (ed.)
By Fitzpatrick R. (ed.)
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Extra info for Euclid's Elements
So the two (straight-lines) AB, BC are equal to the two (straightlines) DE, EF , respectively. And they encompass equal angles. Thus, the base AC is equal to the base DF , and triangle ABC (is) equal to triangle DEF , and the remaining angle BAC (is) equal to the remaining angle EDF [Prop. 4]. Thus, if two triangles have two angles equal to two angles, respectively, and one side equal to one side—in fact, either that by the equal angles, or that subtending one of the equal angles—then (the triangles) will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle (equal) to the remaining angle.
4]. Thus, angle ACB is equal to CBD. Also, since the straight-line BC, (in) falling across the two straight-lines AC and BD, has made the alternate angles (ACB and CBD) equal to one another, AC is thus parallel to BD [Prop. 27]. And (AC) was also shown (to be) equal to (BD). Thus, straight-lines joining equal and parallel (straight- 35 ΣΤΟΙΧΕΙΩΝ α΄. ELEMENTS BOOK 1 lines) on the same sides are themselves also equal and parallel. (Which is) the very thing it was required to show. † The Greek text has “BC, CD”, which is obviously a mistake.
Thus, (the sum of) AGH and BGH is greater π ΒΗΘ, ΗΘ∆ µείζονές ε σιν. λλ α π ΑΗΘ, than (the sum of) BGH and GHD. But, (the sum of) ΒΗΘ δυσ ν ρθα ς σαι ε σίν. 13]. ΗΘ∆ δύο ρθ ν λάσσονές ε σιν. α δ π λασσόνων Thus, (the sum of) BGH and GHD is [also] less than δύο ρθ ν κβαλλόµεναι ε ς πειρον συµπίπουσιν· α two right-angles. But (straight-lines) being produced to ρα ΑΒ, Γ∆ κβαλλόµεναι ε ς πειρον συµπεσο νται· ο infinity from (internal angles whose sum is) less than two συµπίπτουσι δ δι τ παραλλήλους α τ ς ποκε σθαι· right-angles meet together [Post.