Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively,
namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD;
and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF;
I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE and AC to DF,
and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF.
SINT duo anguli B, & C, trianguli A B C, æquales duobus angulis E, & E F D, trianguli D E F, uterque utrique,
hoc est, B, ipsi E, & C, ipsi E F D;
Sitque primo latus B C, quod angulis B, & C, adiacet, lateri E F, quod angulis E, & E F D, adiacet, æquale.
Dico, reliqua quoque latera A B, A C, reliquis lateribus D E, D F, æqualia esse, utrumque utrique, hoc est, A B, ipsi D E, & A C, ipsi D F, ea nimirum, quæ æqualibus angulis subtenduntur;
reliquumque angulum A, reliquo angulo D.