---
book: 6
number: 5
id: "VI.5"
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-8]]", "[[book-1/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.5"
license: "CC-BY-SA-4.0"
---

# VI.5

*If two triangles have their sides proportional*, *the triangles will be equiangular and will have those angles equal which the corresponding sides subtend*.

## Proof

Let *ABC*, *DEF* be two triangles having their sides proportional, so that, as *AB* is to *BC*, so is *DE* to *EF*, as *BC* is to *CA*, so is *EF* to *FD*, and further, as *BA* is to *AC*, so is *ED* to *DF*; I say that the triangle *ABC* is equiangular with the triangle *DEF*, and they will have those angles equal which the corresponding sides subtend, namely the angle *ABC* to the angle *DEF*, the angle *BCA* to the angle *EFD*, and further the angle *BAC* to the angle *EDF*.

For on the straight line *EF*, and at the points *E*, *F* on it, let there be constructed the angle *FEG* equal to the angle *ABC*, and the angle *EFG* equal to the angle *ACB*; [[book-1/proposition-23|I. 23]] therefore the remaining angle at *A* is equal to the remaining angle at *G*. [[book-1/proposition-32|I. 32]]

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

Therefore in the triangles *ABC*, *GEF* the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [[book-6/proposition-4|VI. 4]] therefore, as *AB* is to *BC*, so is *GE* to *EF*.

But, as *AB* is to *BC*, so by hypothesis is *DE* to *EF*; therefore, as *DE* is to *EF*, so is *GE* to *EF*. [[book-5/proposition-11|V. 11]]

Therefore each of the straight lines *DE*, *GE* has the same ratio to *EF*; therefore *DE* is equal to *GE*. [[book-5/proposition-9|V. 9]]

For the same reason *DF* is also equal to *GF*.

Since then *DE* is equal to *EG*, and *EF* is common, the two sides *DE*, *EF* are equal to the two sides *GE*, *EF*; and the base *DF* is equal to the base *FG*; therefore the angle *DEF* is equal to the angle *GEF*, [[book-1/proposition-8|I. 8]] and the triangle *DEF* is equal to the triangle *GEF*, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [[book-1/proposition-4|I. 4]]

Therefore the angle *DFE* is also equal to the angle *GFE*, and the angle *EDF* to the angle *EGF*.

And, since the angle *FED* is equal to the angle *GEF*, while the angle *GEF* is equal to the angle *ABC*, therefore the angle *ABC* is also equal to the angle *DEF*.

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

Therefore etc. Q. E. D.