Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]
But, if not, let BE be drawn from the point B at right angles to CD; [I. 11]
therefore the angles CBE, EBD are two right angles.
Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each;
therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2]
Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each;
therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2]
But the angles CBE, EBD were also proved equal to the same three angles;
and things which are equal to the same thing are also equal to one another; [C. N. 1]
therefore the angles CBE, EBD are also equal to the angles DBA, ABC.
But the angles CBE, EBD are two right angles;
therefore the angles DBA, ABC are also equal to two right angles.
Si igitur A B, fuerit perpendicularis ad C D, erunt dicti 10. def anguli duo recti.
Si vero A B, non fuerit perpendicularis, faciet unum quidem angulum obtusum, alterum vero acutum. Dico igitur ipsos duobus esse rectis æquales. Educatur enim B E, ex B, perpendicularis ad C D,
ut sint duo anguli E B C, E B D, recti.
Quoniam vero angulus rectus E B D, æqualis est duobus angulis D B A, A B E; erunt, apposito communi angulo recto E B C,
duo recti E B D, E B C, tribus angulis D B A, A B E, E B C, æquales.
Rursus quia angulus A B C, duobus angulis A B E, E B C, æqualis est; apposito communi angulo A B D,
erunt duo anguli A B C, A B D, tribus angulis D B A, A B E, E B C, æquales.
Sed illis tribus ostensum fuit esse etiam æquales duos rectos E B D, E B C;
quæ autem eidem æqualia, inter se sunt æqualia.
Duo igitur anguli A B C, A B D, æquales sunt duobus rectis E B D, E B C.