 (diagram 1) For let a straight line, EZ, falling into two straight-lines, AB, GD, make the external angle, that by EΗΒ, equal the angle that’s internal and opposite and on the same sides, that by HQD, (diagram 2) or the angles that are interior and on the same sides, those by BHQ, HQD, equal to two right-angles. I say that AB is parallel to GD. (diagram 3) For since the angle by EHB is equal to that by HQD, but the angle by EHB is equal to that by AHQ, therefore, the angle by AHQ is also equal to that by HQD. And they are alternating. Therefore, AB is parallel to GD. (diagram 4) Again, since the angles by BHQ, HQD are equal to two right-angles, but those by AHQ, BHQ are also equal to two right-angles, therefore, the angles by AHQ, BHQ are equal to those by BHQ, HQD. (diagram 5) Let a common, that by BHQ, be taken away. Therefore, a remainder, the angle by AHQ, is equal to a remainder, that by HQD. And they are alternating. Therefore, AB is parallel to GD. Therefore, if a straight-line falling into two straight-lines makes the external angle equal to the angle that’s internal and opposite and on the same sides or the angles that are interior and on the same sides equal to two right-angles, the straight-lines will be parallel to one another, just what it was required to show.