 (diagram 1) Let there be two triangles, ABG, DEZ, having two angles, those by ABG, BGA respectively equal to those by DEZ, EZD, the angle by ABG to that by DEZ and the angle by BGA to that by EZD. But let it have one side equal to one side, first that at the equal angles, BG, to EZ. (diagram 2) I say that it will have the remaining sides respectively equal to the remaining sides, AB to DE and AG to DZ, and the remaining angle to the remaining angle, that by BAG to that by EDZ. (diagram 3) For if AB is unequal to DE, one of them is larger. Let AB be larger, and let an equal, BH, to DE be positioned, (diagram 4) and let HG be joined. (diagram 5) And so, since BH is equal to DE and BG to EZ, in fact, two BH, BG, are respectively equal to two, DE, EZ. And an angle, that by HBG, is equal to an angle, that by DEZ. Therefore, a base, HG, is equal to a base, DZ, and triangle HBG is equal to triangle DEZ, (diagram 6) and the remaining angles will be equal to the remaining angles which the equal sides subtend. Therefore, the angle by HGB is equal to that by DZE. But the angle by DZE is supposed equal to that by BGA. Therefore, the angle by BGH is equal to that by BGA, the smaller to the larger, which is impossible. Therefore, AB is not unequal to DE. (diagram 7) Therefore, it is equal. But BG is also equal to EZ. In fact, two, AB, BG, are respectively equal to two, DE, EZ. And an angle, that by ABG is equal to an angle, that by DEZ. (diagram 2) Therefore, a base, AG, is equal to a base, DZ, and a remaining angle, that by BAG, is equal to the remaining angle, that by EDZ. (diagram 8) But, in fact, let the sides subtending the equal angles again be equal, as AB to DE. (diagram 9) I say again that the remaining sides will also be equal to the remaining sides, AG to DZ and BG to EZ, and furthermore the remaining angle, that by BAG, is equal to the remaining angle, that by EDZ. (diagram 10) For if BG is unequal to EZ, one of them is larger. Let BG, if possible, be larger, and let an equal, BQ, to EZ be positioned, (diagram 11) and let AQ be joined. (diagram 12) Since BQ is equal to EZ and AB to DE, in fact, two, AB, BQ, are respectively equal to two, DE, EZ. And they enclose equal angles. (diagram 13) Therefore, a base, AQ, is equal to a base, DZ, and triangle ABQ is equal to triangle DEZ, and the remaining angles will be equal to the remaining angles which the equal sides subtend. (diagram 14) Therefore, the angle by BQA is equal to that by EZD. But the angle by EZD is equal to that by BGA. (diagram 15) In fact, the outside angle, that by BQA, is equal to the interior and opposite, that by BGA, which is impossible. Therefore, BG is not unequal to EZ. (diagram 16) Therefore, it is equal. But AB is also equal to DE. In fact, two, AB, BG, are respectively equal to two, DE, EZ. And they contain equal angles. (diagram 9) Therefore, a base, AG, is equal to a base, DZ, and triangle ABG is equal to triangle DEZ, and a remaining angle, that by BAG, is equal to the remaining angle, that by EDZ. Therefore, if two triangles have two angles respectively equal to two angles and one side equal to one side, wither that at the equal angles or that subtending one of the equal angles, it will also have the remaining sides respectively equal to the remaining sides and the remaining angle to the remaining angle, just what it was required to show.