---
book: 13
number: 2
id: "XIII.2"
kind: "theorem"
uses: ["[[book-6/proposition-1]]", "[[book-2/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:13.prop.2"
license: "CC-BY-SA-4.0"
---

# XIII.2

*If the square on a straight line be five times the square on a segment of it*, *then*, *when the double of the said segment is cut in extreme and mean ratio*, *the greater segment is the remaining part of the original straight line.*

## Proof

For let the square on the straight line *AB* be five times the square on the segment *AC* of it, and let *CD* be double of *AC*; I say that, when *CD* is cut in extreme and mean ratio, the greater segment is *CB*.

Let the squares *AF*, *CG* be described on *AB*, *CD* respectively, let the figure in *AF* be drawn, and let *BE* be drawn through.

Now, since the square on *BA* is five times the square on *AC*, *AF* is five times *AH*.

Therefore the gnomon *MNO* is quadruple of *AH*.

And, since *DC* is double of *CA*, therefore the square on *DC* is quadruple of the square on *CA*, that is, *CG* is quadruple of *AH*.

But the gnomon *MNO* was also proved quadruple of *AH*; therefore the gnomon *MNO* is equal to *CG*.

And, since *DC* is double of *CA*, while *DC* is equal to *CK*, and *AC* to *CH*, therefore *KB* is also double of *BH*. [[book-6/proposition-1|VI. 1]]

But *LH*, *HB* are also double of *HB*; therefore *KB* is equal to *LH*, *HB*.

But the whole gnomon *MNO* was also proved equal to the whole *CG*; therefore the remainder *HF* is equal to *BG*.

And *BG* is the rectangle *CD*, *DB*, for *CD* is equal to *DG*; and *HF* is the square on *CB*; therefore the rectangle *CD*, *DB* is equal to the square on *CB*.

Therefore, as *DC* is to *CB*, so is *CB* to *BD*.

But *DC* is greater than *CB*; therefore *CB* is also greater than *BD*.

Therefore, when the straight line *CD* is cut in extreme and mean ratio, *CB* is the greater segment.

Therefore etc. Q. E. D.

Lemma. That the double of *AC* is greater than *BC* is to be proved thus.

If not, let *BC* be, if possible, double of *CA*.

Therefore the square on *BC* is quadruple of the square on *CA*; therefore the squares on *BC*, *CA* are five times the square on *CA*.

But, by hypothesis, the square on *BA* is also five times the square on *CA*; therefore the square on *BA* is equal to the squares on *BC*, *CA*: which is impossible. [[book-2/proposition-4|II. 4]]

Therefore *CB* is not double of *AC*.

Similarly we can prove that neither is a straight line less than *CB* double of *CA*; for the absurdity is much greater.

Therefore the double of *AC* is greater than *CB*. Q. E. D.