---
book: 10
number: 54
id: "X.54"
kind: "theorem"
uses: ["[[book-10/proposition-17]]", "[[book-2/proposition-14]]", "[[book-6/proposition-17]]", "[[book-6/proposition-1]]", "[[book-10/proposition-15]]", "[[book-10/proposition-12]]", "[[book-10/proposition-19]]", "[[book-10/proposition-13]]", "[[book-10/proposition-11]]", "[[book-10/proposition-36]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.54"
license: "CC-BY-SA-4.0"
---

# X.54

If an area be contained by a rational straight line and the first binomial, the side of the area is the irrational straight line which is called binomial.

## Proof

For let the area *AC* be contained by the rational straight line *AB* and the first binomial *AD*; I say that the side of the area *AC* is the irrational straight line which is called binomial.

For, since *AD* is a first binomial straight line, let it be divided into its terms at *E*, and let *AE* be the greater term.

It is then manifest that *AE*, *ED* are rational straight lines commensurable in square only, the square on *AE* is greater than the square on *ED* by the square on a straight line commensurable with *AE*, and *AE* is commensurable in length with the rational straight line *AB* set out. [[book-10/definitions#Definition 1 (part 2)|X. Deff. II. 1]]

Let *ED* be bisected at the point *F*.

Then, since the square on *AE* is greater than the square on *ED* by the square on a straight line commensurable with *AE*, therefore, if there be applied to the greater *AE* a parallelogram equal to the fourth part of the square on the less, that is, to the square on *EF*, and deficient by a square figure, it divides it into commensurable parts. [[book-10/proposition-17|X. 17]]

Let then the rectangle *AG*, *GE* equal to the square on *EF* be applied to *AE*; therefore *AG* is commensurable in length with *EG*.

Let *GH*, *EK*, *FL* be drawn from *G*, *E*, *F* parallel to either of the straight lines *AB*, *CD*; let the square *SN* be constructed equal to the parallelogram *AH*, and the square *NQ* equal to *GK*, [[book-2/proposition-14|II. 14]] and let them be placed so that *MN* is in a straight line with *NO*; therefore *RN* is also in a straight line with *NP*.

And let the parallelogram *SQ* be completed; therefore *SQ* is a square. [Lemma]

Now, since the rectangle *AG*, *GE* is equal to the square on *EF*, therefore, as *AG* is to *EF*, so is *FE* to *EG*; [[book-6/proposition-17|VI. 17]] therefore also, as *AH* is to *EL*, so is *EL* to *KG*; [[book-6/proposition-1|VI. 1]] therefore *EL* is a mean proportional between *AH*, *GK*.

But *AH* is equal to *SN*, and *GK* to *NQ*; therefore *EL* is a mean proportional between *SN*, *NQ*.

But *MR* is also a mean proportional between the same *SN*, *NQ*; [Lemma] therefore *EL* is equal to *MR*, so that it is also equal to *PO*.

But *AH*, *GK* are also equal to *SN*, *NQ*; therefore the whole *AC* is equal to the whole *SQ*, that is, to the square on *MO*; therefore *MO* is the side of *AC*.

I say next that *MO* is binomial.

For, since *AG* is commensurable with *GE*, therefore *AE* is also commensurable with each of the straight lines *AG*, *GE*. [[book-10/proposition-15|X. 15]]

But *AE* is also, by hypothesis, commensurable with *AB*; therefore *AG*, *GE* are also commensurable with *AB*. [[book-10/proposition-12|X. 12]]

And *AB* is rational; therefore each of the straight lines *AG*, *GE* is also rational; therefore each of the rectangles *AH*, *GK* is rational, [[book-10/proposition-19|X. 19]] and *AH* is commensurable with *GK*.

But *AH* is equal to *SN*, and *GK* to *NQ*; therefore *SN*, *NQ*, that is, the squares on *MN*, *NO*, are rational and commensurable.

And, since *AE* is incommensurable in length with *ED*, while *AE* is commensurable with *AG*, and *DE* is commensurable with *EF*, therefore *AG* is also incommensurable with *EF*, [[book-10/proposition-13|X. 13]] so that *AH* is also incommensurable with *EL*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

But *AH* is equal to *SN*, and *EL* to *MR*; therefore *SN* is also incommensurable with *MR*.

But, as *SN* is to *MR*, so is *PN* to *NR*; [[book-6/proposition-1|VI. 1]] therefore *PN* is incommensurable with *NR*. [[book-10/proposition-11|X. 11]]

But *PN* is equal to *MN*, and *NR* to *NO*; therefore *MN* is incommensurable with *NO*.

And the square on *MN* is commensurable with the square on *NO*, and each is rational; therefore *MN*, *NO* are rational straight lines commensurable in square only.

Therefore *MO* is binomial [[book-10/proposition-36|X. 36]] and the side of *AC*. Q. E. D.