---
book: 10
number: 98
id: "X.98"
kind: "theorem"
uses: ["[[book-10/proposition-74]]", "[[book-10/proposition-15]]", "[[book-10/proposition-23]]", "[[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-5/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_3.98"
license: "CC-BY-SA-4.0"
---

# X.98

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

## Proof

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

For let *BG* be the annex to *AB*;. therefore *AG*, *GB* are medial straight lines commensurable in square only which contain a rational rectangle. [[book-10/proposition-74|X. 74]]

To *CD* let there be applied *CH* equal to the square on *AG*, producing *CK* as breadth, and *KL* equal to the square on *GB*, producing *KM* as breadth; therefore the whole *CL* is equal to the squares on *AG*, *GB*; therefore *CL* is also medial. [[book-10/proposition-15|X. 15 and 23, Por.]]

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]]

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

But twice the rectangle *AG*, *GB* is rational; therefore *FL* is rational.

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

Now, since the sum of the squares on *AG*, *GB*, that is, *CL*, is medial, while twice the rectangle *AG*, *GB*, that is, *FL*, is rational, therefore *CL* is incommensurable with *FL*.

But, as *CL* 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 second apotome.

For 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*.

Now, since the rectangle *AG*, *GB* is a mean proportional between the squares on *AG*, *GB*, and the square on *AG* is equal to *CH*, the rectangle *AG*, *GB* to *NL*, and the square on *BG* to *KL*, 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 *MK*; [[book-6/proposition-1|VI. 1]] therefore, as *CK* is to *NM*, so is *NM*, so is *KM*; [[book-5/proposition-11|V. 11]] 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*.

Since the *CM*, *MF* are two unequal straight lines, and the rectangle *CK*, *KM* equal to the fourth part of the square on *MF* and deficient by a square figure has been applied to the greater, *CM*, and divides it into commensurable parts, 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 the annex *FM* is commensurable in length with the rational straight line *CD* set out; therefore *CF* is a second apotome. [[book-10/definitions#Definition 2 (part 3)|X. Deff. III. 2]]

Therefore etc. Q. E. D.