---
book: 1
number: 41
id: "I.41"
kind: "theorem"
uses: ["[[book-1/proposition-37]]", "[[book-1/proposition-34]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:1.prop.41"
license: "CC-BY-SA-4.0"
---

# I.41

If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.

![I.41](figures/I-41.svg)

## Proof

For let the parallelogram *ABCD* have the same base *BC* with the triangle *EBC*, and let it be in the same parallels *BC*, *AE*;

I say that the parallelogram *ABCD* is double of the triangle *BEC*.

For let *AC* be joined.

Then the triangle *ABC* is equal to the triangle *EBC*; for it is on the same base *BC* with it and in the same parallels *BC*, *AE*. [[book-1/proposition-37|I. 37]]

But the parallelogram *ABCD* is double of the triangle *ABC*; for the diameter *AC* bisects it; [[book-1/proposition-34|I. 34]] so that the parallelogram *ABCD* is also double of the triangle *EBC*.

Therefore etc.

Q. E. D.