---
book: 10
number: 97
id: "X.97"
kind: "theorem"
uses: ["[[book-10/proposition-73]]", "[[book-2/proposition-7]]", "[[book-10/proposition-20]]", "[[book-10/proposition-22]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-6/proposition-17]]", "[[book-10/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.97"
license: "CC-BY-SA-4.0"
---

# X.97

*The square on an apotome applied to a rational straight line produces as breadth a first apotome.*

## Proof

Let *AB* be an apotome, and *CD* rational, and to *CD* let there be applied *CE* equal to the square on *AB* and producing *CF* as breadth; I say that *CF* is a first apotome.

For let *BG* be the annex to *AB*; therefore *AG*, *GB* are rational straight lines commensurable in square only. [[book-10/proposition-73|X. 73]]

To *CD* let there be applied *CH* equal to the square on *AG*, and *KL* equal to the square on *BG*.

Therefore 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 *FM* be bisected at the point *N*, and let *NO* be drawn through *N* parallel to *CD*; therefore each of the rectangles *FO*, *LN* is equal to the rectangle *AG*, *GB*.

Now, since the squares on *AG*, *GB* are rational, and *DM* is equal to the squares on *AG*, *GB*,. therefore *DM* is rational.

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

Again, since twice the rectangle *AG*, *GB* is medial, and *FL* is equal to twice the rectangle *AG*, *GB*, therefore *FL* is medial.

And it is applied to the rational straight line *CD*, 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 rational, while twice the rectangle *AG*, *GB* is medial, therefore 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* to twice the rectangle *AG*, *GB*; therefore *DM* is incommensurable with *FL*.

But, as *DM* is to *FL*, so is *CM* to *FM*; [[book-6/proposition-1|VI. 1]] therefore *CM* is incommensurable in length with *FM*. [[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 first apotome.

For, 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 *BG*, 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*.

But, as *CH* is to *NL*, so is *CK* to *NM*, and, as *NL* is to *KL*, so is *NM* to *KM*; [[book-6/proposition-1|VI. 1]] therefore the rectangle *CK*, *KM* is equal to the square on *NM* [[book-6/proposition-17|VI. 17]], that is, to the fourth part of the square on *FM*.

And, since the square on *AG* is commensurable with the square on *GB*, *CH* is also commensurable with *KL*.

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

Since then *CM*, *MF* are two unequal straight lines, and to *CM* there has been applied the rectangle *CK*, *KM* equal to the fourth part of the square on *FM* and deficient by a square figure, while *CK* is commensurable with *KM*, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line commensurable in length with *CM*. [[book-10/proposition-17|X. 17]]

And *CM* is commensurable in length with the rational straight line *CD* set out; therefore *CF* is a first apotome. [[book-10/definitions#Definition 1 (part 3)|X. Deff. III. 1]]

Therefore etc. Q. E. D.