Let BC be joined.
Then, since AB is parallel to CD, and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29]
And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB;
and the angle ABC is equal to the angle BCD;
therefore the base AC is equal to the base BD
and the triangle ABC is equal to the triangle DCB,
and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]
therefore the angle ACB is equal to the angle CBD.
And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another, AC is parallel to BD. [I. 27]
And it was also proved equal to it.
Ducatur enim recta A D.
Quoniam igitur A D, incidit in parallelas A B, C D, erunt anguli alterni B A D, C D A, æquales.
Quare cum duo latera B A, A D, trianguli B A D, æqualia sint duobus lateribus C D, D A, trianguli C D A, utrumque utrique,
& anguli quoque dictis lateribus inclusi æquales;
erunt bases B D, A C, æquales,
& angulus A D B, angulo D A C, æqualis.
Cum igitur hi anguli sint alterni inter rectas A C, B D, erunt A C, B D, parallelæ:
Probatum autem iam fuit, eadem esse æquales.
Complete text
Book I