---
book: 10
number: 92
id: "X.92"
kind: "theorem"
uses: ["[[book-10/proposition-73]]", "[[book-10/proposition-17]]", "[[book-10/proposition-15]]", "[[book-10/proposition-13]]", "[[book-10/proposition-21]]", "[[book-10/proposition-19]]", "[[book-6/proposition-26]]", "[[book-6/proposition-17]]", "[[book-6/proposition-1]]", "[[book-5/proposition-11]]", "[[book-10/proposition-11]]", "[[book-10/proposition-74]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.92"
license: "CC-BY-SA-4.0"
---

# X.92

If an area be contained by a rational straight line and a second apotome, the side of the area is a first apotome of a medial straight line.

## Proof

For let the area *AB* be contained by the rational straight line *AC* and the second apotome *AD*; I say that the side of the area *AB* is a first apotome of a medial straight line.

For let *DG* be the annex to *AD*; therefore *AG*, *GD* are rational straight lines commensurable in square only, [[book-10/proposition-73|X. 73]] and the annex *DG* is commensurable with the rational straight line *AC* set out, while the square on the whole *AG* is greater than the square on the annex *GD* by the square on a straight line commensurable in length with *AG*. [[book-10/definitions#Definition 2 (part 3)|X. Deff. III. 2]]

Since then the square on *AG* is greater than the square on *GD* by the square on a straight line commensurable with *AG*, therefore, if there be applied to *AG* a parallelogram equal to the fourth part of the square on *GD* and deficient by a square figure, it divides it into commensurable parts. [[book-10/proposition-17|X. 17]]

Let then *DG* be bisected at *E*, let there be applied to *AG* a parallelogram equal to the square on *EG* and deficient by a square figure, and let it be the rectangle *AF*, *FG*; therefore *AF* is commensurable in length with *FG*.

Therefore *AG* is also commensurable in length with each of the straight lines *AF*, *FG*. [[book-10/proposition-15|X. 15]]

But *AG* is rational and incommensurable in length with *AC*; therefore each of the straight lines *AF*, *FG* is also rational and incommensurable in length with *AC*; [[book-10/proposition-13|X. 13]] therefore each of the rectangles *AI*, *FK* is medial. [[book-10/proposition-21|X. 21]]

Again, since *DE* is commensurable with *EG*, therefore *DG* is also commensurable with each of the straight lines *DE*, *EG*. [[book-10/proposition-15|X. 15]]

But *DG* is commensurable in length with *AC*.

Therefore each of the rectangles *DH*, *EK* is rational. [[book-10/proposition-19|X. 19]]

Let then the square *LM* be constructed equal to *AI*, and let there be subtracted *NO* equal to *FK* and being about the same angle with *LM*, namely the angle *LPM*; therefore the squares *LM*, *NO* are about the same diameter. [[book-6/proposition-26|VI. 26]]

Let *PR* be their diameter, and let the figure be drawn.

Since then *AI*, *FK* are medial and are equal to the squares on *LP*, *PN*, the squares on *LP*, *PN* are also medial; therefore *LP*, *PN* are also medial straight lines commensurable in square only.

And, since the rectangle *AF*, *FG* is equal to the square on *EG*, therefore, as *AF* is to *EG*, so is *EG* to *FG*, [[book-6/proposition-17|VI. 17]] while, as *AF* is to *EG*, so is *AI* to *EK*, and, as *EG* is to *FG*, so is *EK* to *FK*; [[book-6/proposition-1|VI. 1]] therefore *EK* is a mean proportional between *AI*, *FK*. [[book-5/proposition-11|V. 11]]

But *MN* is also a mean proportional between the squares *LM*, *NO*, and *AI* is equal to *LM*, and *FK* to *NO*; therefore *MN* is also equal to *EK*.

But *DH* is equal to *EK*, and *LO* equal to *MN*; therefore the whole *DK* is equal to the gnomon *UVW* and *NO*.

Since then the whole *AK* is equal to *LM*, *NO*, and, in these, *DK* is equal to the gnomon *UVW* and *NO*, therefore the remainder *AB* is equal to *TS*.

But *TS* is the square on *LN*; therefore the square on *LN* is equal to the area *AB*; therefore *LN* is the side of the area *AB*.

I say that *LN* is a first apotome of a medial straight line.

For, since *EK* is rational and is equal to *LO*, therefore *LO*, that is, the rectangle *LP*, *PN*, is rational.

But *NO* was proved medial; therefore *LO* is incommensurable with *NO*.

But, as *LO* is to *NO*, so is *LP* to *PN*; [[book-6/proposition-1|VI. 1]] therefore *LP*, *PN* are incommensurable in length. [[book-10/proposition-11|X. 11]]

Therefore *LP*, *PN* are medial straight lines commensurable in square only which contain a rational rectangle; therefore *LN* is a first apotome of a medial straight line. [[book-10/proposition-74|X. 74]]

And it is the side of the area *AB*.

Therefore the side of the area *AB* is a first apotome of a medial straight line. Q. E. D.