---
book: 6
number: 4
id: "VI.4"
kind: "theorem"
uses: ["[[book-1/proposition-17]]", "[[book-1/proposition-28]]", "[[book-1/proposition-34]]", "[[book-6/proposition-2]]", "[[book-5/proposition-16]]", "[[book-5/proposition-22]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.4"
license: "CC-BY-SA-4.0"
---

# VI.4

*In equiangular triangles the sides about the equal angles are proportional*, *and those are corresponding sides which subtend the equal angles*.

## Proof

Let *ABC*, *DCE* be equiangular triangles having the angle *ABC* equal to the angle *DCE*, the angle *BAC* to the angle *CDE*, and further the angle *ACB* to the angle *CED*; I say that in the triangles *ABC*, *DCE* the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.

For let *BC* be placed in a straight line with *CE*.

Then, since the angles *ABC*, *ACB* are less than two right angles, [[book-1/proposition-17|I. 17]] and the angle *ACB* is equal to the angle *DEC*, therefore the angles *ABC*, *DEC* are less than two right angles; therefore *BA*, *ED*, when produced, will meet. [[book-1/postulates#Postulate 5|I. Post. 5]]

Let them be produced and meet at *F*.

Now, since the angle *DCE* is equal to the angle *ABC*, *BF* is parallel to *CD*. [[book-1/proposition-28|I. 28]]

Again, since the angle *ACB* is equal to the angle *DEC*, *AC* is parallel to *FE*. [[book-1/proposition-28|I. 28]]

Therefore *FACD* is a parallelogram; therefore *FA* is equal to *DC*, and *AC* to *FD*. [[book-1/proposition-34|I. 34]]

And, since *AC* has been drawn parallel to *FE*, one side of the triangle *FBE*, therefore, as *BA* is to *AF*, so is *BC* to *CE*. [[book-6/proposition-2|VI. 2]]

But *AF* is equal to *CD*; therefore, as *BA* is to *CD*, so is *BC* to *CE*, and alternately, as *AB* is to *BC*, so is *DC* to *CE*. [[book-5/proposition-16|V. 16]]

Again, since *CD* is parallel to *BF*, therefore, as *BC* is to *CE*, so is *FD* to *DE*. [[book-6/proposition-2|VI. 2]]

But *FD* is equal to *AC*; therefore, as *BC* is to *CE*, so is *AC* to *DE*, and alternately, as *BC* is to *CA*, so is *CE* to *ED*. [[book-5/proposition-16|V. 16]]

Since then it was proved that, as *AB* is to *BC*, so is *DC* to *CE*, and, as *BC* is to *CA*, so is *CE* to *ED*; therefore, *ex aequali*, as *BA* is to *AC*, so is *CD* to *DE*. [[book-5/proposition-22|V. 22]]

Therefore etc. Q. E. D.