---
book: 10
number: 59
id: "X.59"
kind: "theorem"
uses: ["[[book-10/proposition-21]]", "[[book-10/proposition-13]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-41]]", "[[book-2/proposition-5]]", "[[book-2/proposition-9]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.59"
license: "CC-BY-SA-4.0"
---

# X.59

If an area be contained by a rational straight line and the sixth binomial, the side of the area is the irrational straight line called the side of the sum of two medial areas.

## Proof

For let the area *ABCD* be contained by the rational straight line *AB* and the sixth binomial *AD*, divided into its terms at *E*, so that *AE* is the greater term; I say that the side of *AC* is the side of the sum of two medial areas.

Let the same construction be made as before shown.

It is then manifest that *MO* is the side of *AC*, and that *MN* is incommensurable in square with *NO.*

Now, since *EA* is incommensurable in length with *AB*, therefore *EA*, *AB* are rational straight lines commensurable in square only; therefore *AK*, that is, the sum of the squares on *MN*, *NO*, is medial. [[book-10/proposition-21|X. 21]]

Again, since *ED* is incommensurable in length with *AB*, therefore *FE* is also incommensurable with *EK*; [[book-10/proposition-13|X. 13]] therefore *FE*, *EK* are rational straight lines commensurable in square only; therefore *EL*, that is, *MR*, that is, the rectangle *MN*, *NO*, is medial. [[book-10/proposition-21|X. 21]]

And, since *AE* is incommensurable with *EF*, *AK* is also incommensurable with *EL.* [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

But *AK* is the sum of the squares on *MN*, *NO*, and *EL* is the rectangle *MN*, *NO*; therefore the sum of the squares on *MN*, *NO* is incommensurable with the rectangle *MN*, *NO.*

And each of them is medial, and *MN*, *NO* are incommensurable in square.

Therefore *MO* is the side of the sum of two medial areas [[book-10/proposition-41|X. 41]], and is the side of *AC*. Q. E. D.

[LEMMA. If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

Let *AB* be a straight line, and let it be cut into unequal parts at *C*, and let *AC* be the greater; I say that the squares on *AC*, *CB* are greater than twice the rectangle *AC*, *CB*.

For let *AB* be bisected at *D*.

Since then a straight line has been cut into equal parts at *D*, and into unequal parts at *C*, therefore the rectangle *AC*, *CB* together with the square on *CD* is equal to the square on *AD*, [[book-2/proposition-5|II. 5]] so that the rectangle *AC*, *CB* is less than double of the square on *AD*.

But the squares on *AC*, *CB* are double of the squares on *AD*, *DC*; [[book-2/proposition-9|II. 9]] therefore the squares on *AC*, *CB* are greater than twice the rectangle *AC*, *CB*. Q. E. D.]