 (diagram 1) Let there be two triangles, ABG, DEZ, having the two sides, AB, AG, respectively equal to the two sides, DE, DZ, AB to AG and AG to DZ, but let it also have a base, BG, equal to a base EZ.(diagram 2) I say that an angle, that by BAG, is also equal to an angle, that by EDZ. (diagram 3) For upon triangle ABG being fitted onto triangle DEZ and point B positioned onto point E and BG onto EZ, point G will also fit onto Z since BG is equal to EZ. In fact, upon BG fitting onto EZ, also BA, GA will fit onto ED, DZ. For if a base, BG, will fit onto a base, EZ, but sides BA, AG will not fit onto ED, DZ, but overlap as EH, HZ (παραλλάξουσιν ὡς αἱ ΕΗ, ΗΖ), there will be constructed on the same straight-line two different straight-lines respectively equal to the same straight-lines at one and another point on the same sides while having the same limits as the initial lines. But they are not constructed. (diagram 4) Therefore, it is not the case that upon base BG fitting onto base EZ, sides BA, AG will not also fit onto ED, DZ. Therefore, they will fit. Thus, an angle, that by BAG, will also fit onto an angle, that by EDZ, and will be equal to it. Therefore, if two triangles have the two sides equal to the two sides respectively, but also have the base equal to the base, then they will have the angle that’s enclosed by the equal straight lines equal to the angle, just what it was required to show.