---
book: 13
number: 4
id: "XIII.4"
kind: "theorem"
uses: ["[[book-6/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.4"
license: "CC-BY-SA-4.0"
---

# XIII.4

*If a straight line be cut in extreme and mean ratio*, *the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.*

## Proof

Let *AB* be a straight line, let it be cut in extreme and mean ratio at *C*, and let *AC* be the greater segment; I say that the squares on *AB*, *BC* are triple of the square on *CA*.

For let the square *ADEB* be described on *AB*, and let the figure be drawn.

Since then *AB* has been cut in extreme and mean ratio at *C*, and *AC* is the greater segment, therefore the rectangle *AB*, *BC* is equal to the square on *AC*. [[book-6/definitions#Definition 3|VI. Def. 3]], [[book-6/proposition-17|VI. 17]]

And *AK* is the rectangle *AB*, *BC*, and *HG* the square on *AC*; therefore *AK* is equal to *HG*.

And, since *AF* is equal to *FE*, let *CK* be added to each; therefore the whole *AK* is equal to the whole *CE*; therefore *AK*, *CE* are double of *AK*.

But *AK*, *CE* are the gnomon *LMN* and the square *CK*; therefore the gnomon *LMN* and the square *CK* are double of *AK*.

But, further, *AK* was also proved equal to *HG*; therefore the gnomon *LMN* and the squares *CK*, *HG* are triple of the square *HG*.

And the gnomon *LMN* and the squares *CK*, *HG* are the whole square *AE* and *CK*, which are the squares on *AB*, *BC*, while *HG* is the square on *AC*.

Therefore the squares on *AB*, *BC* are triple of the square on *AC*. Q. E. D.