---
book: 10
number: 94
id: "X.94"
kind: "theorem"
uses: ["[[book-10/proposition-18]]", "[[book-10/proposition-19]]", "[[book-10/proposition-21]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-6/proposition-26]]", "[[book-6/proposition-17]]", "[[book-5/proposition-11]]", "[[book-10/proposition-76]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.94"
license: "CC-BY-SA-4.0"
---

# X.94

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

## Proof

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

For let *DG* be the annex to *AD*; therefore *AG*, *GD* are rational straight lines commensurable in square only, *AG* is commensurable in length with the rational straight line *AC* set out, and the square on the whole *AG* is greater than the square on the annex *DG* by the square on a straight line incommensurable in length with *AG*, [[book-10/definitions#Definition 4 (part 3)|X. Deff. III. 4]]

Since then the square on *AG* is greater than the square on *GD* by the square on a straight line incommensurable in length with *AG*, therefore, if there be applied to *AG* a parallelogram equal to the fourth part of the square on *DG* and deficient by a square figure, it will divide it into incommensurable parts. [[book-10/proposition-18|X. 18]]

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 incommensurable in length with *FG*.

Let *EH*, *FI*, *GK* be drawn through *E*, *F*, *G* parallel to *AC*, *BD*.

Since then *AG* is rational and commensurable in length with *AC*, therefore the whole *AK* is rational. [[book-10/proposition-19|X. 19]]

Again, since *DG* is incommensurable in length with *AC*, and both are rational, therefore *DK* is medial. [[book-10/proposition-21|X. 21]]

Again, since *AF* is incommensurable in length with *FG*, therefore *AI* is also incommensurable with *FK*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

Now let the square *LM* be constructed equal to *AI*, and let there be subtracted *NO* equal to *FK* and about the same angle, 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 the rectangle *AF*, *FG* is equal to the square on *EG*, therefore, proportionally, 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 *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 *EK* is also equal to *MN*.

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

Since, then, the whole *AK* is equal to the squares *LM*, *NO*, and, in these, *DK* is equal to the gnomon *UVW* and the square *NO*, therefore the remainder *AB* is equal to *ST*, that is, to the square on *LN*; therefore *LN* is the side of the area *AB*.

I say that *LN* is the irrational straight line called minor.

For, since *AK* is rational and is equal to the squares on *LP*, *PN*, therefore the sum of the squares on *LP*, *PN* is rational.

Again, since *DK* is medial, and *DK* is equal to twice the rectangle *LP*, *PN*, therefore twice the rectangle *LP*, *PN* is medial.

And, since *AI* was proved incommensurable with *FK*, therefore the square on *LP* is also incommensurable with the square on *PN*.

Therefore *LP*, *PN* are straight lines incommensurable in square which make the sum of the squares on them rational, but twice the rectangle contained by them medial.

Therefore *LN* is the irrational straight line called minor; [[book-10/proposition-76|X. 76]] and it is the side of the area *AB*.

Therefore the side of the area *AB* is minor. Q. E. D.