If two triangles have their sides proportional, then the triangles are equiangular with the equal angles opposite the corresponding sides.

Let *ABC* and *DEF* be two triangles having their sides proportional, so that *AB* is to *BC* as *DE* is to *EF, BC* is to *CA* as *EF* is to *FD,* and further *BA* is to *AC* as *ED* is to *DF.*

I say that the triangle *ABC* is equiangular with the triangle *DEF* where the equal angles are opposite the corresponding sides, namely the angle *ABC* equals the angle *DEF,* the angle *BCA* equals the angle *EFD,* and the angle *BAC* equals the angle *EDF.*

Construct the angle *FEG* equal to the angle *CBA* and the angle *EFG* equal to the angle *BCA* on the straight line *EF* and at the points *E* and *F* on it. Therefore the remaining angle at *A* equals the remaining angle at *G.*

Therefore the triangle *ABC* is equiangular with the triangle *GEF.*

Therefore in the triangles *ABC* and *GEF* the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles, therefore *AB* is to *BC* as *GE* is to *EF.*

But, by hypothesis, *AB* is to *BC* as *DE* to *EF,* therefore *DE* is to *EF* as *GE* is to *EF.*

Therefore each of the straight lines *DE* and *GE* has the same ratio to *EF,* therefore *DE* equals *GE.*

For the same reason *DF* also equals *GF.*

Then since *DE* equals *GE,* and *EF* is common, the two sides *DE* and *EF* equal the two sides *GE* and *EF,* and the base *DF* equals the base *GF,* therefore the angle *DEF* equals the angle *GEF,* and the triangle *DEF* equals the triangle *GEF,* and the remaining angles equal the remaining angles, namely those opposite the equal sides.

Therefore the angle *DFE* also equals the angle *GFE,* and the angle *EDF* equals the angle *EGF.*

And, since the angle *DEF* equals the angle *GEF,* and the angle *GEF* equals the angle *ABC,* therefore the angle *ABC* also equals the angle *DEF.*

For the same reason the angle *ACB* also equals the angle *DFE,* and further, the angle at *A* equals the angle at *D,* therefore the triangle *ABC* is equiangular with the triangle *DEF.*

Therefore, *if two triangles have their sides proportional, then the triangles are equiangular with the equal angles opposite the corresponding sides.*

Q.E.D.

As in VI.2, a certain order is assumed for the proportionality. It is not intended, for instance, that *AB* : *BC* = *DE* : *EF* while *BC* : *CA* = *FD* : *EF.* See the remark about VI.Def.1.

This proposition is used in the proof of proposition XII.12.