---
book: 10
number: 99
id: "X.99"
kind: "theorem"
uses: ["[[book-10/proposition-75]]", "[[book-10/proposition-15]]", "[[book-10/proposition-23]]", "[[book-10/proposition-22]]", "[[book-2/proposition-7]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-13]]", "[[book-10/proposition-73]]", "[[book-5/proposition-11]]", "[[book-10/proposition-17]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.99"
license: "CC-BY-SA-4.0"
---

# X.99

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

## Proof

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

For let *BG* be the annex to *AB*; therefore *AG*, *GB* are medial straight lines commensurable in square only which contain a medial rectangle. [[book-10/proposition-75|X. 75]]

Let *CH* equal to the square on *AG* be applied to *CD*, producing *CK* as breadth, and let *KL* equal to the square on *BG* be applied to *KH*, 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 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 *LF* is equal to twice the rectangle *AG*, *GB*. [[book-2/proposition-7|II. 7]]

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

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

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

And, since *AG*, *GB* are commensurable in square only, therefore *AG* is incommensurable in length with *GB*; therefore the square on *AG* is also incommensurable with the rectangle *AG*, *GB*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

But the squares on *AG*, *GB* are commensurable with the square on *AG*, and twice the rectangle *AG*, *GB* with the rectangle *AG*, *GB*; therefore the squares on *AG*, *GB* are incommensurable with twice the rectangle *AG*, *GB*. [[book-10/proposition-13|X. 13]]

But *CL* is equal to the squares on *AG*, *GB*, and *FL* is equal to twice the rectangle *AG*, *GB*; therefore *CL* is also 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 third apotome.

For, since the square on *AG* is commensurable with the square on *GB*, therefore *CH* is also commensurable with *KL*, so that *CK* is also commensurable with *KM*. [[book-6/proposition-1|VI. 1]], [[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*.

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, as *CK* is to *MN*, so is *MN* to *KM*; [[book-5/proposition-11|V. 11]] therefore the rectangle *CK*, *KM* is equal to [the square on *MN*, that is, to] the fourth part of the square on *FM*.

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 commensurable parts, therefore the square on *CM* is greater than the square on *MF* by the square on a straight line commensurable with *CM*. [[book-10/proposition-17|X. 17]]

And neither of the straight lines *CM*, *MF* is commensurable in length with the rational straight line *CD* set out; therefore *CF* is a third apotome. [[book-10/definitions#Definition 3 (part 3)|X. Deff. III. 3]]

Therefore etc. Q. E. D.