translated by Henry Mendell (Cal. State U., L.A.) (diagram 1= gen. diag.) For let two straight-lines, AB, GD, cut one another at point E. (diagram 2) I say that the angle by AEG is equal to that by DEB and that by GEB to that by AED. (diagram 3) For since a straight-line, AE, stands-on a straight-line, GD, making angles, those by GEA, AED, therefore, the angles by GEA, AED are equal to two right-angles. (diagram 4) Again, since a straight-line, DE, stands-on a straight-line, AB, making angles, those by AED, DEB, therefore, the angles by AED, DEB are equal to two right-angles. (diagram 5) But the angles by GEA, AED, were also proved equal to two right-angles. Therefore, the angles by GEA, AED are equal to those by AED, DEB. (diagram 6) Let a common be taken away, that by AED. Therefore, a remainder, that by GEA is equal to a remainder, that by BED. (diagram 7) (diagram 8) Similarly, in fact, it will be shown that the angles by GEB, DEA are equal. Therefore, if two straight-lines cut one another, they make the angles at the vertex equal to one another, just what it was required to show.