translated by Henry Mendell (Cal. State U., L.A.) (diagram 1) For at some line AB and the point on it, B, let two straight-lines, BG, BD, not positioned on the same sides, make successive angles, those by ABG, ABD, equal to two right-angles. I say that BD is on a straight-line with GB. (diagram 2) For if BD is not on a straight-line with BG, let BE be on a straight-line with GB. (diagram 3) And so, since a straight-line, AB, stands-on a straight-line, GBE, therefore, the angles by ABG, ABE are equal to two right-angles. But the angles by ABG, ABD are also equal to two right-angles. Therefore, the angles by GBA, ABE are equal to those by GBA, ABD. (diagram 4) Let a common be taken away, that by GBA. Therefore, a remainder, the angle by ABE, is equal to a remainder, that by ABD, the smaller to the larger, which is impossible. Therefore, BE is not on a straight-line with GB. Similarly, in fact, we will show that not any other straight-line except BD is either. Therefore, GB is on a straight-line with BD. Therefore, if at some line and the point on it two straight-lines, not positioned on the same sides, make successive angles equal to two right-angles, the straight-lines will be on a straight-line with one another, just what it was required to show.