---
book: 10
number: 61
id: "X.61"
kind: "theorem"
uses: ["[[book-10/proposition-37]]", "[[book-10/proposition-21]]", "[[book-10/proposition-15]]", "[[book-10/proposition-23]]", "[[book-10/proposition-22]]", "[[book-10/proposition-20]]", "[[book-10/proposition-13]]", "[[book-10/proposition-36]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.61"
license: "CC-BY-SA-4.0"
---

# X.61

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

## Proof

Let *AB* be a first bimedial straight line divided into its medials at *C*, of which medials *AC* is the greater; let a rational straight line *DE* be set out, and let there be applied to *DE* the parallelogram *DF* equal to the square on *AB*, producing *DG* as its breadth; I say that *DG* is a second binominal straight line.

For let the same construction as before be made.

Then, since *AB* is a first bimedial divided at *C*, therefore *AC*, *CB* are medial straight lines commensurable in square only, and containing a rational rectangle, [[book-10/proposition-37|X. 37]] so that the squares on *AC*, *CB* are also medial. [[book-10/proposition-21|X. 21]]

Therefore *DL* is medial. [[book-10/proposition-15|X. 15 and 23, Por.]]

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

Again, since twice the rectangle *AC*, *CB* is rational, *MF* is also rational.

And it is applied to the rational straight line *ML*; therefore *MG* is also rational and commensurable in length with *ML*, that is, *DE*; [[book-10/proposition-20|X. 20]] 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 second binomial straight line.

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

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 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*. [[book-10/proposition-17|X. 17]]

And *MG* is commensurable is length with *DE*.

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