---
book: 10
number: 21
id: "X.21"
kind: "theorem"
uses: ["[[book-6/proposition-1]]", "[[book-10/proposition-11]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.21"
license: "CC-BY-SA-4.0"
---

# X.21

*The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called* *medial.*

## Proof

For let the rectangle *AC* be contained by the rational straight lines *AB*, *BC* commensurable in square only; I say that *AC* is irrational, and the side of the square equal to it is irrational; and let the latter be called *medial.*

For on *AB* let the square *AD* be described; therefore *AD* is rational. [[book-10/definitions#Definition 4|X. Def. 4]]

And, since *AB* is incommensurable in length with *BC*, for by hypothesis they are commensurable in square only, while *AB* is equal to *BD*, therefore *DB* is also incommensurable in length with *BC*.

And, as *DB* is to *BC*, so is *AD* to *AC*; [[book-6/proposition-1|VI. 1]] therefore *DA* is incommensurable with *AC*. [[book-10/proposition-11|X. 11]]

But *DA* is rational; therefore *AC* is irrational, so that the side of the square equal to *AC* is also irrational. [[book-10/definitions#Definition 4|X. Def. 4]]

And let the latter be called *medial.* Q. E. D.

Lemma. If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

Let *FE*, *EG* be two straight lines.

I say that, as *FE* is to *EG*, so is the square on *FE* to the rectangle *FE*, *EG*.

For on *FE* let the square *DF* be described, and let *GD* be completed.

Since then, as *FE* is to *EG*, so is *FD* to *DG*, [[book-6/proposition-1|VI. 1]] and *FD* is the square on *FE*, and *DG* the rectangle *DE*, *EG*, that is, the rectangle *FE*, *EG*, therefore, as *FE* is to *EG*, so is the square on *FE* to the rectangle *FE*, *EG*.

Similarly also, as the rectangle *GE*, *EF* is to the square on *EF*, that is, as *GD* is to *FD*, so is *GE* to *EF*. Q. E. D.