---
book: 13
number: 1
id: "XIII.1"
kind: "theorem"
uses: ["[[book-6/proposition-17]]", "[[book-6/proposition-1]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.1"
license: "CC-BY-SA-4.0"
---

# XIII.1

*If a straight line be cut in extreme and mean ratio*, *the square on the greater segment added to the half of the whole is five times the square on the half.*

## Proof

For let the straight line *AB* be cut in extreme and mean ratio at the point *C*, and let *AC* be the greater segment; let the straight line *AD* be produced in a straight line with *CA*, and let *AD* be made half of *AB*; I say that the square on *CD* is five times the square on *AD*.

For let the squares *AE*, *DF* be described on *AB*, *DC*, and let the figure in *DF* be drawn; let *FC* be carried through to *G*.

Now, since *AB* has been cut in extreme and mean ratio at *C*, 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 *CE* is the rectangle *AB*, *BC*, and *FH* the square on *AC*; therefore *CE* is equal to *FH*.

And, since *BA* is double of *AD*, while *BA* is equal to *KA*, and *AD* to *AH*, therefore *KA* is also double of *AH*.

But, as *KA* is to *AH*, so is *CK* to *CH*; [[book-6/proposition-1|VI. 1]] therefore *CK* is double of *CH*.

But *LH*, *HC* are also double of *CH*.

Therefore *KC* is equal to *LH*, *HC*.

But *CE* was also proved equal to *HF*; therefore the whole square *AE* is equal to the gnomon *MNO*.

And, since *BA* is double of *AD*, the square on *BA* is quadruple of the square on *AD*, that is, *AE* is quadruple of *DH*.

But *AE* is equal to the gnomon *MNO*; therefore the gnomon *MNO* is also quadruple of *AP*; therefore the whole *DF* is five times *AP*.

And *DF* is the square on *DC*, and *AP* the square on *DA*; therefore the square on *CD* is five times the square on *DA*.

Therefore etc. Q. E. D.