---
book: 10
number: 62
id: "X.62"
kind: "theorem"
uses: ["[[book-10/proposition-38]]", "[[book-10/proposition-15]]", "[[book-10/proposition-23]]", "[[book-10/proposition-22]]", "[[book-10/proposition-11]]", "[[book-10/proposition-12]]", "[[book-10/proposition-13]]", "[[book-6/proposition-1]]", "[[book-10/proposition-36]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.62"
license: "CC-BY-SA-4.0"
---

# X.62

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

## Proof

Let *AB* be a second bimedial straight line divided into its medials at *C*, so that *AC* is the greater segment; let *DE* be any rational straight line, and to *DE* let there be applied the parallelogram *DF* equal to the square on *AB* and producing *DG* as its breadth; I say that *DG* is a third binomial straight line.

Let the same construction be made as before shown.

Then, since *AB* is a second bimedial divided at *C*, therefore *AC*, *CB* are medial straight lines commensurable in square only and containing a medial rectangle, [[book-10/proposition-38|X. 38]] so that the sum of the squares on *AC*, *CB* is also medial. [[book-10/proposition-15|X. 15 and 23 Por.]]

And it is equal to *DL*; therefore *DL* is also medial.

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

For the same reason, *MG* is also rational and incommensurable in length with *ML*, that is, with *DE*; therefore each of the straight lines *DM*, *MG* is rational and incommensurable in length with *DE*.

And, since *AC* is incommensurable in length with *CB*, and, as *AC* is to *CB*, so is the square on *AC* to the rectangle *AC*, *CB*, therefore the square on *AC* is also incommensurable with the rectangle *AC*, *CB*. [[book-10/proposition-11|X. 11]]

Hence the sum of the squares on *AC*, *CB* is incommensurable with twice the rectangle *AC*, *CB*, [[book-10/proposition-12|X. 12, 13]] that is, *DL* is incommensurable with *MF*, so that *DM* is also incommensurable with *MG*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

And they are rational; therefore *DG* is binomial. [[book-10/proposition-36|X. 36]]

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

In manner similar to the foregoing we may conclude that *DM* is greater than *MG*, and that *DK* is commensurable with *KM*.

And the rectangle *DK*, *KM* is equal to the square on *MN*; therefore the square on *DM* is greater than the square on *MG* by the square on a straight line commensurable with *DM*.

And neither of the straight lines *DM*, *MG* is commensurable in length with *DE*.

Therefore *DG* is a third binomial straight line. [[book-10/definitions#Definition 3 (part 2)|X. Deff. II. 3]] Q. E. D.