---
book: 11
number: 39
id: "XI.39"
kind: "theorem"
uses: ["[[book-1/proposition-34]]", "[[book-11/proposition-31]]", "[[book-11/proposition-28]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.39"
license: "CC-BY-SA-4.0"
---

# XI.39

*If there be two prisms of equal height*, *and one have a parallelogram as base and the other a triangle*, *and if the parallelogram be double of the triangle*, *the prisms will be equal.*

## Proof

Let *ABCDEF*, *GHKLMN* be two prisms of equal height, let one have the parallelogram *AF* as base, and the other the triangle *GHK*, and let the parallelogram *AF* be double of the triangle *GHK*; I say that the prism *ABCDEF* is equal to the prism *GHKLMN*.

For let the solids *AO*, *GP* be completed.

Since the parallelogram *AF* is double of the triangle *GHK*, while the parallelogram *HK* is also double of the triangle *GHK*, [[book-1/proposition-34|I. 34]] therefore the parallelogram *AF* is equal to the parallelogram *HK*.

But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [[book-11/proposition-31|XI. 31]] therefore the solid *AO* is equal to the solid *GP*.

And the prism *ABCDEF* is half of the solid *AO*, and the prism *GHKLMN* is half of the solid *GP*; [[book-11/proposition-28|XI. 28]] therefore the prism *ABCDEF* is equal to the prism *GHKLMN*.

Therefore etc. Q. E. D.