If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend.

Let $ABC$, $DEF$ be two triangles having the two sides $AB$,$AC$ equal to the two sides $DE$,$DF$ respectively, namely $AB$ to $DE$ and $AC$ to $DF$, and the angle $BAC$ equal to the angle $EDF$.I say that the base $BC$ is also equal to the base $EF$, the triangle $ABC$ will be equal to thr triangle $DEF$, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle $ABC$ to the angle $DEF$, and the angle $ACB$ to the angle $DFE$.For, if the triangle $ABC$ be applied to the triangle $DEF$, and if the point $A$ be placed on the point $D$and the straight line $AB$ on $DE$,then the point $B$ will also coincide with $E$ because $AB$ is equal to $DE$.Again, $AB$ coinciding with $DE$,the straight line $AC$ will also coincide with $DF$, because the angle $BAC$ is equal to the angle $EDF$;hence the point $C$ will also coincide with the point $F$, because $AC$ is again equal to $DF$But $B$ also coincided with $E$hence the base $BC$ will coincide with the base $EF$.[For if, when $B$ coincides with $E$ and $C$ with $F$, the base $BC$ does not coincide with the base $EF$, two straight lines will enclose a space : which is impossible. Therefore the base $BC$ will coincide with $EF$ and will be equal to it] [C.N. 4]Thus the whole triangle $ABC$ will coincide with the whole triangle $DEF$,and will be equal to it.And the remaining angles will also coincide with the remaining angles and will be equal to them,the angle $ABC$ to the angle $DEF$,and the angle $ACB$ to the angle $DFE$.Therefore etc.(Being) what it was required to prove.