If two triangles have one angle equal to one angle and the sides about the equal angles proportional, then the triangles are equiangular and have those angles equal opposite the corresponding sides.

Let *ABC* and *DEF* be two triangles having one angle *BAC* equal to one angle *EDF* and the sides about the equal angles proportional, so that *BA* is to *AC* as *ED* is to *DF.*

I say that the triangle *ABC* is equiangular with the triangle *DEF,* and has the angle *ABC* equal to the angle *DEF,* and the angle *ACB* equal to the angle *DFE.*

On the straight line *DF* and at the points *D* and *F* on it, construct the angle *FDG* equal to either of the angles *BAC* or *EDF,* and the angle *DFG* equal to the angle *ACB.*

Therefore the remaining angle at *B* equals the remaining angle at *G.* Therefore the triangle *ABC* is equiangular with the triangle DGF.

Therefore, proportionally *BA* is to *AC* as *GD* is to *DF.*

But, by hypothesis, *BA* is to *AC* also as *ED* is to *DF,* therefore also *ED* is to *DF* as *GD* is to *DF.*

Therefore *ED* equals *GD.* And *DF* is common, therefore the two sides *ED* and *DF* equal the two sides *GD* and *DF,* and the angle *EDF* equals the angle *GDF,* therefore the base *EF* equals the base *GF,* the triangle *DEF* equals the triangle *DGF,* and the remaining angles equal the remaining angles, namely those opposite the equal sides.

Therefore the angle *DFG* equals the angle *DFE,* and the angle *DGF* equals the angle *DEF.*

But the angle *DFG* equals the angle *ACB,* therefore the angle *ACB* also equals the angle *DFE.*

And, by hypothesis, the angle *BAC* also equals the angle *EDF,* therefore the remaining angle at *B* also equals the remaining angle at *E.* Therefore the triangle *ABC* is equiangular with the triangle *DEF.*

Therefore, *if two triangles have one angle equal to one angle and the sides about the equal angles proportional, then the triangles are equiangular and have those angles equal opposite the corresponding sides.*

Q.E.D.

Here’s a summary of the proof. Construct a triangle *DGF* equiangular with triangle *ABC.* Then triangle *DGF* is similar to triangle *ABC* (
VI.4), and that gives us the proportion

But we have assumed the proportion

and these two proportions together give us

(V.11), from which it follows that *GD* = *ED* (V.9). Therefore triangles *DEF* and *DGF* are congruent, and the rest follows easily.