---
book: 6
number: 1
id: "VI.1"
kind: "theorem"
uses: ["[[book-1/proposition-38]]", "[[book-1/proposition-41]]", "[[book-5/proposition-15]]", "[[book-5/proposition-11]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.1"
license: "CC-BY-SA-4.0"
---

# VI.1

*Triangles and parallelograms which are under the same height are to one another as their bases*.

## Proof

Let *ABC*, *ACD* be triangles and *EC*, *CF* parallelograms under the same height; I say that, as the base *BC* is to the base *CD*, so is the triangle *ABC* to the triangle *ACD*, and the parallelogram *EC* to the parallelogram *CF*.

For let *BD* be produced in both directions to the points *H*, *L* and let [any number of straight lines] *BG*, *GH* be made equal to the base *BC*, and any number of straight lines *DK*, *KL* equal to the base *CD*; let *AG*, *AH*, *AK*, *AL* be joined.

Then, since *CB*, *BG*, *GH* are equal to one another, the triangles *ABC*, *AGB*, *AHG* are also equal to one another. [[book-1/proposition-38|I. 38]]

Therefore, whatever multiple the base *HC* is of the base *BC*, that multiple also is the triangle *AHC* of the triangle *ABC*.

For the same reason, whatever multiple the base *LC* is of the base *CD*, that multiple also is the triangle *ALC* of the triangle *ACD*; and, if the base *HC* is equal to the base *CL*, the triangle *AHC* is also equal to the triangle *ACL*, [[book-1/proposition-38|I. 38]] if the base *HC* is in excess of the base *CL*, the triangle *AHC* is also in excess of the triangle *ACL*, and, if less, less.

Thus, there being four magnitudes, two bases *BC*, *CD* and two triangles *ABC*, *ACD*, equimultiples have been taken of the base *BC* and the triangle *ABC*, namely the base *HC* and the triangle *AHC*, and of the base *CD* and the triangle *ADC* other, chance, equimultiples, namely the base *LC* and the triangle *ALC*; and it has been proved that, if the base *HC* is in excess of the base *CL*, the triangle *AHC* is also in excess of the triangle *ALC*; if equal, equal; and, if less, less.

Therefore, as the base *BC* is to the base *CD*, so is the triangle *ABC* to the triangle *ACD*. [[book-5/definitions#Definition 5|V. Def. 5]]

Next, since the parallelogram *EC* is double of the triangle *ABC*, [[book-1/proposition-41|I. 41]] and the parallelogram *FC* is double of the triangle *ACD*, while parts have the same ratio as the same multiples of them, [[book-5/proposition-15|V. 15]] therefore, as the triangle *ABC* is to the triangle *ACD*, so is the parallelogram *EC* to the parallelogram *FC*.

Since, then, it was proved that, as the base *BC* is to *CD*, so is the triangle *ABC* to the triangle *ACD*, and, as the triangle *ABC* is to the triangle *ACD*, so is the parallelogram *EC* to the parallelogram *CF*, therefore also, as the base *BC* is to the base *CD*, so is the parallelogram *EC* to the parallelogram *FC*. [[book-5/proposition-11|V. 11]]

Therefore etc. Q. E. D.