If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals 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, 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* equal to *EF.*

I say that the remaining sides equal the remaining sides respectively, namely *AB* equals *DE* and *AC* equals *DF,* and the remaining angle equals the remaining angle, namely the angle *BAC* equals the angle *EDF.*

If *AB* does not equal *DE,* then one of them is greater.

Let *AB* be greater. Make *BG* equal to *DE,* and join *GC.*

Since *BG* equals *DE,* and *BC* equals *EF,* the two sides *GB* and *BC* equal the two sides *DE* and *EF* respectively, and the angle *GBC* equals the angle *DEF,* therefore the base *GC* equals the base *DF,* the triangle *GBC* equals the triangle *DEF,* and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle *GCB* equals the angle *DFE.* But the angle *DFE* equals the angle *ACB* by hypothesis. Therefore the angle *BCG* equals the angle *BCA,* the less equals the greater, which is impossible.

Therefore *AB* is not unequal to *DE,* and therefore equals it.

But *BC* also equals *EF.* Therefore the two sides *AB* and *BC* equal the two sides *DE* and *EF* respectively, and the angle *ABC* equals the angle *DEF.* Therefore the base *AC* equals the base *DF,* and the remaining angle *BAC* equals the remaining angle *EDF.*

Next, let sides opposite equal angles be equal, as *AB* equals *DE.*

I say again that the remaining sides equal the remaining sides, namely *AC* equals *DF* and *BC* equals *EF,* and further the remaining angle *BAC* equals the remaining angle *EDF.*

If *BC* is unequal to *EF,* then one of them is greater.

Let *BC* be greater, if possible. Make *BH* equal to *EF,* and join *AH.*

Since *BH* equals *EF,* and *AB* equals *DE,* the two sides *AB* and *BH* equal the two sides *DE* and *EF* respectively, and they contain equal angles, therefore the base *AH* equals the base *DF,* the triangle *ABH* equals the triangle *DEF,* and the remaining angles equal the remaining angles, namely those opposite the equal sides. Therefore the angle *BHA* equals the angle *EFD.*

But the angle *EFD* equals the angle *BCA,* therefore, in the triangle *AHC,* the exterior angle *BHA* equals the interior and opposite angle *BCA,* which is impossible.

Therefore *BC* is not unequal to *EF,* and therefore equals it.

But *AB* also equals *DE.* Therefore the two sides *AB* and *BC* equal the two sides *DE* and *EF* respectively, and they contain equal angles. Therefore the base *AC* equals the base *DF,* the triangle *ABC* equals the triangle *DEF,* and the remaining angle *BAC* equals the remaining angle *EDF.*

Therefore *if two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle.*

Q.E.D.

The remaining congruence theorem, side-side-angle, includes some ambiguous cases. Suppose triangles *ABC* and *DEF* are such that sides *AB* and *BC* are equal to sides *DE* and *EF* respectively, and angle *A* equals angle *D.* If it is also known that *AB* is less than or equal to *BC,* then it follows that the two triangles are congruent. If, however, *AB* is greater than *BC,* then the two triangles need not be congruent. Euclid does not include any form of a side-side-angle congruence theorem, but he does prove one special case, side-side-right angle, in the course of the proof of proposition III.14.

Although Euclid does not include a side-side-angle congruence theorem, he does have a side-side-angle similarity theorem, namely proposition VI.7. The analogous congruence theorem could be stated as follows: If two triangles have one angle equal to one angle, two sides adjoining the equal angles equal, namely, one side adjoining the equal angles, and one opposite the equal angles, and the remaining angles either both less or both not less than a right angle, then the remaining side equals the remaining side and the remaining angles equal the remaining angles.

As in propositions I.4 and I.8, it appears that the triangles are in the same plane, but, again, that is not necessary. Indeed, this proposition is invoked in proposition XI.35 when two triangles do not lie in the same plane.