 <(diagram 1) Let there be a triangle, ABG, and let one side of it, BG, be extended to D. I say that the exterior angle, that by AGD, is larger than each of the interior and opposite angles, the angles by GBA, BAG. (diagram 2) Let AG be bisected at E, (diagram 3) and let BE, being joined, be extended on a straight-line to Z, and let an equal, EZ, to BE be positioned, (diagram 4) and let ZG be joined, and let AG be drawn through to H. (diagram 5) And so since AE is equal to EG and BE to EZ, two, in fact, AE, EB, are respectively equal to two, GE, EZ. And an angle, that by AEB, is equal to an angle, that by ZEG. For it is at a vertex. (diagram 6) Therefore, a base, AB, is equal to a base, ZG, and triangle ABE is equal to triangle ZEG, and the remaining angles are respectively equal to the remaining angles, which the equal sides subtend. (diagram 7) Therefore, the angle by BAE equal to that by EGZ. But the angle by EGD is larger than that by EGZ. Therefore, the angle by AGD is larger than that by BAE. (diagram 8) Similarly, in fact, upon BG having been bisected, the angle by BGH, that is the angle by AGD, will be shown also larger than that by ABG. Therefore, upon one of the sides of any triangle being extended, the external angle is larger than each of the interior and opposite angles, just what it was required to show.