---
book: 10
number: 91
id: "X.91"
kind: "theorem"
uses: ["[[book-10/proposition-73]]", "[[book-10/proposition-17]]", "[[book-10/proposition-15]]", "[[book-10/proposition-12]]", "[[book-10/proposition-19]]", "[[book-10/proposition-13]]", "[[book-10/proposition-21]]", "[[book-6/proposition-26]]", "[[book-6/proposition-17]]", "[[book-6/proposition-1]]", "[[book-5/proposition-11]]", "[[book-10/proposition-53]]", "[[book-10/proposition-11]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.91"
license: "CC-BY-SA-4.0"
---

# X.91

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

## Proof

For let the area *AB* be contained by the rational straight line *AC* and the first apotome *AD*;

I say that the side of the area *AB* is an apotome.

For, since *AD* is a first apotome, let *DG* be its annex; therefore *AG*, *GD* are rational straight lines commensurable in square only. [[book-10/proposition-73|X. 73]]

And the whole *AG* is commensurable with the rational straight line *AC* set out, and the square on *AG* is greater than the square on *GD* by the square on a straight line commensurable in length with *AG*; [[book-10/definitions#Definition 1 (part 3)|X. Deff. III. 1]] if therefore there be applied to *AG* a parallelogram equal to the fourth part of the square on *DG* and deficient by a square figure, it divides it into commensurable parts. [[book-10/proposition-17|X. 17]]

Let *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 with *FG*.

And through the points *E*, *F*, *G* let *EH*, *FI*, *GK* be drawn parallel to *AC*.

Now, since *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 commensurable with *AC*; therefore each of the straight lines *AF*, *FG* is commensurable in length with *AC*. [[book-10/proposition-12|X. 12]]

And *AC* is rational; therefore each of the straight lines *AF*, *FG* is also rational, so that each of the rectangles *AI*, *FK* is also rational. [[book-10/proposition-19|X. 19]]

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

But *DG* is rational and incommensurable in length with *AC*; therefore each of the straight lines *DE*, *EG* is also rational and incommensurable in length with *AC*; [[book-10/proposition-13|X. 13]] therefore each of the rectangles *DH*, *EK* is medial. [[book-10/proposition-21|X. 21]]

Now let the square *LM* be made equal to *AI*, and let there be subtracted the square *NO* having a common angle with it, the angle *LPM*, and equal to *FK*; 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 the rectangle contained by *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]]

But, as *AF* is to *EG*, so is *AI* to *EK*, and, as *EG* is to *FG*, so is *EK* to *KF*; [[book-6/proposition-1|VI. 1]] therefore *EK* is a mean proportional between *AI*, *KF*. [[book-5/proposition-11|V. 11]]

But *MN* is also a mean proportional between *LM*, *NO*, as was before proved, [[book-10/proposition-53|Lemma after X. 53]] and *AI* is equal to the square *LM*, and *KF* to *NO*; therefore *MN* is also equal to *EK*.

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

But *AK* is also equal to the squares *LM*, *NO*; therefore the remainder *AB* is equal to *ST*.

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

I say next that *LN* is an apotome.

For, since each of the rectangles *AI*, *FK* is rational, and they are equal to *LM*, *NO*, therefore each of the squares *LM*, *NO*, that is, the squares on *LP*, *PN* respectively, is also rational; therefore each of the straight lines *LP*, *PN* is also rational.

Again, since *DH* is medial and is equal to *LO*, therefore *LO* is also medial.

Since then *LO* is medial, while *NO* is rational, 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* is incommensurable in length with *PN*. [[book-10/proposition-11|X. 11]]

And both are rational; therefore *LP*, *PN* are rational straight lines commensurable in square only; therefore *LN* is an apotome. [[book-10/proposition-73|X. 73]]

And it is the side of the area *AB*; therefore the side of the area *AB* is an apotome.

Therefore etc.