If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, then the triangles are equiangular and have those angles equal the sides about which are proportional.

Let *ABC* and *DEF* be two triangles having one angle equal to one angle, the angle *BAC* equal to the angle *EDF,* the sides about other angles *ABC* and *DEF* proportional, so that *AB* is to *BC* as *DE* is to *EF.* And, first, each of the remaining angles at *C* and *F* less than a right angle.

I say that the triangle *ABC* is equiangular with the triangle *DEF,* the angle *ABC* equals the angle *DEF,* and the remaining angle, namely the angle at *C,* equals the remaining angle, the angle at *F.*

If the angle *ABC* does not equal the angle *DEF,* then one of them is greater.

Let the angle *ABC* be greater. Construct the angle *ABG* equal to the angle *DEF* on the straight line *AB* and at the point *B* on it.

Then, since the angle *A* equals *D,* and the angle *ABG* equals the angle *DEF,* therefore the remaining angle *AGB* equals the remaining angle *DFE.*

Therefore the triangle *ABG* is equiangular with the triangle *DEF.*

Therefore *AB* is to *BG* as *DE* is to *EF.*

But, by hypothesis, *DE* is to *EF* as *AB* is to *BC,* therefore *AB* has the same ratio to each of the straight lines *BC* and *BG.* Therefore *BC* equals *BG,* so that the angle at *C* also equals the angle *BGC.*

But, by hypothesis, the angle at *C* is less than a right angle, therefore the angle *BGC* is also less than a right angle, so that the angle *AGB* adjacent to it is greater than a right angle.

And it was proved equal to the angle at *F,* therefore the angle at *F* is also greater than a right angle. But it is by hypothesis less than a right angle, which is absurd.

Therefore the angle *ABC* is not unequal to the angle *DEF.* Therefore it equals it.

But the angle at *A* also equals the angle at *D,* therefore the remaining angle at *C* equals the remaining angle at *F.*

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

Next let each of the angles at *C* and *F* be supposed not less than a right angle.

I say again that, in this case too, the triangle *ABC* is equiangular with the triangle *DEF.*

With the same construction, we can prove similarly that *BC* equals *BG,* so that the angle at *C* also equals the angle *BGC.*

But the angle at *C* is not less than a right angle, therefore neither is the angle *BGC* less than a right angle.

Therefore, once more, the angle *ABC* is not unequal to the angle *DEF.* Therefore it equals it.

But the angle at *A* also equals the angle at *D,* therefore the remaining angle at *C* equals the remaining angle at *F.*

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

Therefore, *if two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, then the triangles are equiangular and have those angles equal the sides about which are proportional.*

Q.E.D.