---
book: 10
number: 60
id: "X.60"
kind: "theorem"
uses: ["[[book-10/proposition-36]]", "[[book-10/proposition-15]]", "[[book-10/proposition-20]]", "[[book-10/proposition-21]]", "[[book-10/proposition-22]]", "[[book-10/proposition-13]]", "[[book-10/proposition-53]]", "[[book-6/proposition-1]]", "[[book-6/proposition-17]]", "[[book-10/proposition-11]]", "[[book-10/proposition-59]]", "[[book-10/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.60"
license: "CC-BY-SA-4.0"
---

# X.60

*The square on the binomial straight line applied to a rational straight line produces as breadth the first binomial*.

## Proof

Let *AB* be a binomial straight line divided into its terms at *C*, so that *AC* is the greater term; let a rational straight line *DE* be set out, and let *DEFG* equal to the square on *AB* be applied to *DE* producing *DG* as its breadth; I say that *DG* is a first binomial straight line.

For let there be applied to *DE* the rectangle *DH* equal to the square on *AC*, and *KL* equal to the square on *BC*; therefore the remainder, twice the rectangle *AC*, *CB*, is equal to *MF*.

Let *MG* be bisected at *N*, and let *NO* be drawn parallel [to *ML* or *GF*].

Therefore each of the rectangles *MO*, *NF* is equal to once the rectangle *AC*, *CB*.

Now, since *AB* is a binomial divided into its terms at *C*, therefore *AC*, *CB* are rational straight lines commensurable in square only; [[book-10/proposition-36|X. 36]] therefore the squares on *AC*, *CB* are rational and commensurable with one another, so that the sum of the squares on *AC*, *CB* is also rational. [[book-10/proposition-15|X. 15]]

And it is equal to *DL*; therefore *DL* is rational.

And it is applied to the rational straight line *DE*; therefore *DM* is rational and commensurable in length with *DE*. [[book-10/proposition-20|X. 20]]

Again, since *AC*, *CB* are rational straight lines commensurable in square only, therefore twice the rectangle *AC*, *CB*, that is *MF*, is medial. [[book-10/proposition-21|X. 21]]

And it is applied to the rational straight line *ML*; therefore *MG* is also rational and incommensurable in length with *ML*, that is, *DE*. [[book-10/proposition-22|X. 22]]

But *MD* is also rational and is commensurable in length with *DE*; therefore *DM* is incommensurable in length with *MG*. [[book-10/proposition-13|X. 13]]

And they are rational; therefore *DM*, *MG* are rational straight lines commensurable in square only; therefore *DG* is binomial. [[book-10/proposition-36|X. 36]]

It is next to be proved that it is also a first binomial straight line.

Since the rectangle *AC*, *CB* is a mean proportional between the squares on *AC*, *CB*, cf. [[book-10/proposition-53|Lemma after X. 53]] therefore *MO* is also a mean proportional between *DH*, *KL*.

Therefore, as *DH* is to *MO*, so is *MO* to *KL*, that is, as *DK* is to *MN*, so is *MN* to *MK*; [[book-6/proposition-1|VI. 1]] therefore the rectangle *DK*, *KM* is equal to the square on *MN*. [[book-6/proposition-17|VI. 17]]

And, since the square on *AC* is commensurable with the square on *CB*, *DH* is also commensurable with *KL*, so that *DK* is also commensurable with *KM*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

And, since the squares on *AC*, *CB* are greater than twice the rectangle *AC*, *CB*, [[book-10/proposition-59|Lemma]] therefore *DL* is also greater than *MF*, so that *DM* is also greater than *MG*. [[book-6/proposition-1|VI. 1]]

And the rectangle *DK*, *KM* is equal to the square on *MN*, that is, to the fourth part of the square on *MG*, and *DK* is commensurable with *KM*.

But, if there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into commensurable parts, the square on the greater is greater than the square on the less by the square on a straight line commensurable with the greater; [[book-10/proposition-17|X. 17]] therefore the square on *DM* is greater than the square on *MG* by the square on a straight line commensurable with *DM*.

And *DM*, *MG* are rational, and *DM*, which is the greater term, is commensurable in length with the rational straight line *DE* set out.

Therefore *DG* is a first binomial straight line. [[book-10/definitions#Definition 1 (part 2)|X. Deff. II. 1]] Q. E. D.