---
book: 11
number: 32
id: "XI.32"
kind: "theorem"
uses: ["[[book-1/proposition-45]]", "[[book-11/proposition-31]]", "[[book-11/proposition-25]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:11.prop.32"
license: "CC-BY-SA-4.0"
---

# XI.32

*Parallelepipedal solids which are of the same height are to one another as their bases.*

## Proof

Let *AB*, *CD* be parallelepipedal solids of the same height; I say that the parallelepipedal solids *AB*, *CD* are to one another as their bases, that is, that, as the base *AE* is to the base *CF*, so is the solid *AB* to the solid *CD*.

For let *FH* equal to *AE* be applied to *FG*, [[book-1/proposition-45|I. 45]] and on *FH* as base, and with the same height as that of *CD*, let the parallelepipedal solid *GK* be completed.

Then the solid *AB* is equal to the solid *GK*; for they are on equal bases *AE*, *FH* and of the same height. [[book-11/proposition-31|XI. 31]]

And, since the parallelepipedal solid *CK* is cut by the plane *DG* which is parallel to opposite planes, therefore, as the base *CF* is to the base *FH*, so is the solid *CD* to the solid *DH*. [[book-11/proposition-25|XI. 25]]

But the base *FH* is equal to the base *AE*, and the solid *GK* to the solid *AB*; therefore also, as the base *AE* is to the base *CF*, so is the solid *AB* to the solid *CD*.

Therefore etc. Q. E. D.