 (diagram 1) Let there be two triangles, ABG, DEZ, having the two sides, AB, AG, equal to the two sides, DE, DZ, respectively, AB to DE and AG to DZ, but let the angle at A be larger than the angle at D. (diagram 2) I say that a base, BG, is also larger than a base, EZ. (diagram 3) For since the angle by BAG is larger than the angle by EDZ, let an equal, the angle by EDH, to the angle by BAG be constructed at straight-line DE and the point on it, D. (diagram 4) And let an equal, DH, to either of AG, DZ be positioned, (diagram 5) and let EH, ZH be joined. (diagram 6) And so, since AB is equal to DE and AG to DH, in fact, two, BA, AG, are respectively equal to two, ED, DH. And an angle, that by BAG is equal to an angle, that by EDH. Therefore, a base, BG, is equal to a base, EH. (diagram 7) Again, since DZ is equal to DH, the angle by DHZ is also equal to that by DZH. (diagram 8) Therefore, the angle by DZH is larger than that by EHZ. (diagram 9) Therefore, the angle by EZH is much larger than that by EHZ. (diagram 10) And since a triangle, EZH, having the angle by EZH larger than that by EHZ, but the larger side subtends the larger angle, therefore, a side, EH, is also larger than EZ. (diagram 11) But EH is equal to BG. Therefore, BG is also larger than EZ. Therefore, if two triangles have the two sides respectively equal to the two sides but the angle enclosed by the equal straight-lines larger than the angle, then they will have the base larger than the base, just what it was required to show.