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

# VI.6

*If two triangles have one angle equal to one angle and the sides about the equal angles 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 one angle *BAC* equal to one angle *EDF* and the sides about the equal angles proportional, so that, as *BA* is to *AC*, so is *ED* to *DF*; I say that the triangle *ABC* is equiangular with the triangle *DEF*, and will have the angle *ABC* equal to the angle *DEF*, and the angle *ACB* to the angle *DFE*.

For on the straight line *DF*, and at the points *D*, *F* on it, let there be constructed the angle *FDG* equal to either of the angles *BAC*, *EDF*, and the angle *DFG* equal to the angle *ACB*; [[book-1/proposition-23|I. 23]] therefore the remaining angle at *B* is equal to the remaining angle at *G*. [[book-1/proposition-32|I. 32]]

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

Therefore, proportionally, as *BA* is to *AC*, so is *GD* to *DF*. [[book-6/proposition-4|VI. 4]]

But, by hypothesis, as *BA* is to *AC*, so also is *ED* to *DF*; therefore also, as *ED* is to *DF*, so is *GD* to *DF*. [[book-5/proposition-11|V. 11]]

Therefore *ED* is equal to *DG*; [[book-5/proposition-9|V. 9]] and *DF* is common; therefore the two sides *ED*, *DF* are equal to the two sides *GD*, *DF*; and the angle *EDF* is equal to the angle *GDF*; therefore the base *EF* is equal to the base *GF*, and the triangle *DEF* is equal to the triangle *DGF*, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [[book-1/proposition-4|I. 4]]

Therefore the angle *DFG* is equal to the angle *DFE*, and the angle *DGF* to the angle *DEF*.

But the angle *DFG* is equal to the angle *ACB*; therefore the angle *ACB* is also equal to the angle *DFE*.

And, by hypothesis, the angle *BAC* is also equal to the angle *EDF*; therefore the remaining angle at *B* is also equal to the remaining angle at *E*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABC* is equiangular with the triangle *DEF*.

Therefore etc. Q. E. D.