If two triangles have two angles equal to two angles, respectively,  and one side equal to one side---in fact, either that by the equal angles, or that subtending one of the equal angles---then (the triangles) will also have the remaining sides equal to the [corresponding] remaining sides, and the remaining angle (equal) to the remaining angle.Let $ABC$ and $DEF$ be two triangles having the two angles $ABC$ and $BCA$ equal to the two (angles) $DEF$ and $EFD$, respectively. (That is) $ABC$ (equal) to $DEF$, and $BCA$ to $EFD$. And let them also have one side equal to one side. First of all, the (side) by the equal angles. (That is) $BC$ (equal) to $EF$. I say that they will have the  remaining sides equal to the corresponding remaining sides. (That is) $AB$ (equal) to $DE$, and $AC$ to $DF$. And (they will have) the remaining angle (equal) to the remaining angle. (That is) $BAC$ (equal) to $EDF$. <prf> Thus, the base $AC$ is equal to the base $DF$, and triangle $ABC$ (is) equal to triangle $DEF$, and the remaining angle $BAC$ (is) equal to the remaining angle $EDF$ [Prop.~1.4].Thus, if two triangles have two angles equal to two angles, respectively, and one side equal to one side---in fact, either that by the equal angles, or that subtending one of the equal angles---then (the triangles) will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle (equal) to the remaining angle.