---
book: 6
number: 7
id: "VI.7"
kind: "theorem"
uses: ["[[book-1/proposition-23]]", "[[book-1/proposition-32]]", "[[book-6/proposition-4]]", "[[book-5/proposition-11]]", "[[book-5/proposition-9]]", "[[book-1/proposition-5]]", "[[book-1/proposition-13]]", "[[book-1/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.7"
license: "CC-BY-SA-4.0"
---

# VI.7

*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*, *the triangles will be equiangular and will have those angles equal*, *the sides about which are proportional*.

## Proof

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

For, if the angle *ABC* is unequal to the angle *DEF*, one of them is greater.

Let the angle *ABC* be greater; and on the straight line *AB*, and at the point *B* on it, let the angle *ABG* be constructed equal to the angle *DEF*. [[book-1/proposition-23|I. 23]]

Then, since the angle *A* is equal to *D*, and the angle *ABG* to the angle *DEF*, therefore the remaining angle *AGB* is equal to the remaining angle *DFE*. [[book-1/proposition-32|I. 32]]

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

Therefore, as *AB* is to *BG*, so is *DE* to *EF* [[book-6/proposition-4|VI. 4]]

But, as *DE* is to *EF*, so by hypothesis is *AB* to *BC*; therefore *AB* has the same ratio to each of the straight lines *BC*, *BG*; [[book-5/proposition-11|V. 11]] therefore *BC* is equal to *BG*, [[book-5/proposition-9|V. 9]] so that the angle at *C* is also equal to the angle *BGC*. [[book-1/proposition-5|I. 5]]

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. [[book-1/proposition-13|I. 13]]

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 is equal to it.

But the angle at *A* is also equal to the angle at *D*; therefore the remaining angle at *C* is equal to the remaining angle at *F*. [[book-1/proposition-32|I. 32]]

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

But, again, let each of the angles at *C*, *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*.

For, with the same construction, we can prove similarly that *BC* is equal to *BG*; so that the angle at *C* is also equal to the angle *BGC*. [[book-1/proposition-5|I. 5]]

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

Thus in the triangle *BGC* the two angles are not less than two right angles: which is impossible. [[book-1/proposition-17|I. 17]]

Therefore, once more, the angle *ABC* is not unequal to the angle *DEF*; therefore it is equal to it.

But the angle at *A* is also equal to the angle at *D*; therefore the remaining angle at *C* is equal to the remaining angle at *F*. [[book-1/proposition-32|I. 32]]

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

Therefore etc. Q. E. D.