---
book: 10
number: 101
id: "X.101"
kind: "theorem"
uses: ["[[book-10/proposition-77]]", "[[book-10/proposition-22]]", "[[book-2/proposition-7]]", "[[book-10/proposition-20]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-73]]", "[[book-10/proposition-18]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.101"
license: "CC-BY-SA-4.0"
---

# X.101

*The square on the straight line which produces with a rational area a medial whole, if applied to a rational straight line, produces as breadth a fifth apotome.*

## Proof

Let *AB* be the straight line which produces with a rational area a medial whole, and *CD* a rational straight line, and to *CD* let *CE* be applied equal to the square on *AB* and producing *CF* as breadth; I say that *CF* is a fifth apotome.

For let *BG* be the annex to *AB*; therefore *AG*, *GB* are straight lines incommensurable in square which make the sum of the squares on them medial but twice the rectangle contained by them rational. [[book-10/proposition-77|X. 77]]

To *CD* let there be applied *CH* equal to the square on *AG*, and *KL* equal to the square on *GB*; therefore the whole *CL* is equal to the squares on *AG*, *GB*.

But the sum of the squares on *AG*, *GB* together is medial; therefore *CL* is medial.

And it is applied to the rational straight line *CD*, producing *CM* as breadth; therefore *CM* is rational and incommensurable with *CD*. [[book-10/proposition-22|X. 22]]

And, since the whole *CL* is equal to the squares on *AG*, *GB*, and, in these, *CE* is equal to the square on *AB*, therefore the remainder *FL* is equal to twice the rectangle *AG*, *GB*. [[book-2/proposition-7|II. 7]]

Let then *FM* be bisected at *N*, and through *N* let *NO* be drawn parallel to either of the straight lines *CD*, *ML*; therefore each of the rectangles *FO*, *NL* is equal to the rectangle *AG*, *GB*:

And, since twice the rectangle *AG*, *GB* is rational and equal to *FL*, therefore *FL* is rational.

And it is applied to the rational straight line *EF*, producing *FM* as breadth; therefore *FM* is rational and commensurable in length with *CD*. [[book-10/proposition-20|X. 20]]

Now, since *CL* is medial, and *FL* rational, therefore *CL* is incommensurable with *FL*.

But, as *CL* is to *FL*, so is *CM* to *MF*; [[book-6/proposition-1|VI. 1]] therefore *CM* is incommensurable in length with *MF*. [[book-10/proposition-11|X. 11]]

And both are rational; therefore *CM*, *MF* are rational straight lines commensurable in square only; therefore *CF* is an apotome. [[book-10/proposition-73|X. 73]]

I say next that it is also a fifth apotome.

For we can prove similarly that the rectangle *CK*, *KM* is equal to the square on *NM*, that is, to the fourth part of the square on *FM*.

And, since the square on *AG* is incommensurable with the square on *GB*, while the square on *AG* is equal to *CH*, and the square on *GB* to *KL*, therefore *CH* is incommensurable with *KL*.

But, as *CH* is to *KL*, so is *CK* to *KM*; [[book-6/proposition-1|VI. 1]] therefore *CK* is incommensurable in length with *KM*. [[book-10/proposition-11|X. 11]]

Since then *CM*, *MF* are two unequal straight lines, and a parallelogram equal to the fourth part of the square on *FM* and deficient by a square figure has been applied to *CM*, and divides it into incommensurable parts, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line incommensurable with *CM*. [[book-10/proposition-18|X. 18]]

And the annex *FM* is commensurable with the rational straight line *CD* set out; therefore *CF* is a fifth apotome. [[book-10/definitions#Definition 5 (part 3)|X. Deff. III. 5]] Q. E. D.