---
book: 6
number: 2
id: "VI.2"
kind: "theorem"
uses: ["[[book-1/proposition-38]]", "[[book-5/proposition-7]]", "[[book-6/proposition-1]]", "[[book-5/proposition-11]]", "[[book-5/proposition-9]]", "[[book-1/proposition-39]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.2"
license: "CC-BY-SA-4.0"
---

# VI.2

*If a straight line be drawn parallel to one of the sides of a triangle*, *it will cut the sides of the triangle proportionally; and*, *if the sides of the triangle be cut proportionally*, *the line joining the points of section will be parallel to the remaining side of the triangle*.

## Proof

For let *DE* be drawn parallel to *BC*, one of the sides of the triangle *ABC*; I say that, as *BD* is to *DA*, so is *CE* to *EA*.

For let *BE*, *CD* be joined.

Therefore the triangle *BDE* is equal to the triangle *CDE*; for they are on the same base *DE* and in the same parallels *DE*, *BC*. [[book-1/proposition-38|I. 38]]

And the triangle *ADE* is another area.

But equals have the same ratio to the same; [[book-5/proposition-7|V. 7]] therefore, as the triangle *BDE* is to the triangle *ADE*, so is the triangle *CDE* to the triangle *ADE*.

But, as the triangle *BDE* is to *ADE*, so is *BD* to *DA*; for, being under the same height, the perpendicular drawn from *E* to *AB*, they are to one another as their bases. [[book-6/proposition-1|VI. 1]]

For the same reason also, as the triangle *CDE* is to *ADE*, so is *CE* to *EA*.

Therefore also, as *BD* is to *DA*, so is *CE* to *EA*. [[book-5/proposition-11|V. 11]]

Again, let the sides *AB*, *AC* of the triangle *ABC* be cut proportionally, so that, as *BD* is to *DA*, so is *CE* to *EA*; and let *DE* be joined.

I say that *DE* is parallel to *BC*.

For, with the same construction, since, as *BD* is to *DA*, so is *CE* to *EA*, but, as *BD* is to *DA*, so is the triangle *BDE* to the triangle *ADE*, and, as *CE* is to *EA*, so is the triangle *CDE* to the triangle *ADE*, [[book-6/proposition-1|VI. 1]] therefore also, as the triangle *BDE* is to the triangle *ADE*, so is the triangle *CDE* to the triangle *ADE*. [[book-5/proposition-11|V. 11]]

Therefore each of the triangles *BDE*, *CDE* has the same ratio to *ADE*.

Therefore the triangle *BDE* is equal to the triangle *CDE*; [[book-5/proposition-9|V. 9]] and they are on the same base *DE*.

But equal triangles which are on the same base are also in the same parallels. [[book-1/proposition-39|I. 39]]

Therefore *DE* is parallel to *BC*.

Therefore etc. Q. E. D.