---
book: 10
number: 102
id: "X.102"
kind: "theorem"
uses: ["[[book-10/proposition-78]]", "[[book-10/proposition-22]]", "[[book-2/proposition-7]]", "[[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_3.102"
license: "CC-BY-SA-4.0"
---

# X.102

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

## Proof

Let *AB* be the straight line which produces with a medial 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 sixth 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, twice the rectangle *AG*, *GB* medial, and the squares on *AG*, *GB* incommensurable with twice the rectangle *AG*, *GB*. [[book-10/proposition-78|X. 78]]

Now to *CD* let there be applied *CH* equal to the square on *AG* and producing *CK* as breadth, and *KL* equal to the square on *BG*; therefore the whole *CL* is equal to the squares on *AG*, *GB*; therefore *CL* is also medial.

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

Since now *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]]

And twice the rectangle *AG*, *GB* is medial; therefore *FL* is also medial.

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

And, since the squares on *AG*, *GB* are incommensurable with twice the rectangle *AG*, *GB*, and *CL* is equal to the squares on *AG*, *GB*, and *FL* equal to twice the rectangle *AG*, *GB*, 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 sixth apotome.

For, since *FL* is equal to twice the rectangle *AG*, *GB*, let *FM* be bisected at *N*, and let *NO* be drawn through *N* parallel to *CD*; therefore each of the rectangles *FO*, *NL* is equal to the rectangle *AG*, *GB*.

And, since *AG*, *GB* are incommensurable in square, therefore the square on *AG* is incommensurable with the square on *GB*.

But *CH* is equal to the square on *AG*, and *KL* is equal to the square on *GB*; 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 with *KM*. [[book-10/proposition-11|X. 11]]

And, since the rectangle *AG*, *GB* is a mean proportional between the squares on *AG*, *GB*, and *CH* is equal to the square on *AG*, *KL* equal to the square on *GB*, and *NL* equal to the rectangle *AG*, *GB*, therefore *NL* is also a mean proportional between *CH*, *KL*; therefore, as *CH* is to *NL*, so is *NL* to *KL*.

And for the same reason as before 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 neither of them is commensurable with the rational straight line *CD* set out; therefore *CF* is a sixth apotome. [[book-10/definitions#Definition 6 (part 3)|X. Deff. III. 6]] Q. E. D.