---
book: 10
number: 93
id: "X.93"
kind: "theorem"
uses: ["[[book-10/proposition-17]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-15]]", "[[book-10/proposition-13]]", "[[book-10/proposition-21]]", "[[book-6/proposition-26]]", "[[book-6/proposition-17]]", "[[book-5/proposition-11]]", "[[book-10/proposition-75]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.93"
license: "CC-BY-SA-4.0"
---

# X.93

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

## Proof

For let the area *AB* be contained by the rational straight line *AC* and the third apotome *AD*; I say that the side of the area *AB* is a second 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, and neither of the straight lines *AG*, *GD* is commensurable in length with the rational straight line *AC* set out, while the square on the whole *AG* is greater than the square on the annex *DG* by the square on a straight line commensurable with *AG*. [[book-10/definitions#Definition 3 (part 3)|X. Deff. III. 3]]

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 *DG* and deficient by a square figure, it will divide 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*.

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

Therefore *AF*, *FG* are commensurable; therefore *AI* is also commensurable with *FK*. [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]]

And, since *AF*, *FG* are commensurable in length, 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*; so that *AF*, *FG* are so also. [[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 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 *GD* 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]]

And, since *AG*, *GD* are commensurable in square only, therefore *AG* is incommensurable in length with *GD*.

But *AG* is commensurable in length with *AF*, and *DG* with *EG*; therefore *AF* is incommensurable in length with *EG*. [[book-10/proposition-13|X. 13]]

But, as *AF* is to *EG*, so is *AI* to *EK*; [[book-6/proposition-1|VI. 1]] therefore *AI* is incommensurable with *EK*. [[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 being about the same angle with *LM*; therefore *LM*, *NO* are about the same diameter. [[book-6/proposition-26|VI. 26]]

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

Now, 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]]

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 also, as *AI* is to *EK*, so is *EK* to *FK*; [[book-5/proposition-11|V. 11]] therefore *EK* is a mean proportional between *AI*, *FK*.

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

But *AK* is also equal to *LM*, *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 a second apotome of a medial straight line.

For, since *AI*, *FK* were proved medial, and are equal to the squares on *LP*, *PN*, therefore each of the squares on *LP*, *PN* is also medial; therefore each of the straight lines *LP*, *PN* is medial.

And, since *AI* is commensurable with *FK*, [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]] therefore the square on *LP* is also commensurable with the square on *PN*.

Again, since *AI* was proved incommensurable with *EK*, therefore *LM* is also incommensurable with *MN*, that is, the square on *LP* with the rectangle *LP*, *PN*; so that *LP* is also incommensurable in length with *PN*; [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]] therefore *LP*, *PN* are medial straight lines commensurable in square only.

I say next that they also contain a medial rectangle.

For, since *EK* was proved medial, and is equal to the rectangle *LP*, *PN*, therefore the rectangle *LP*, *PN* is also medial, so that *LP*, *PN* are medial straight lines commensurable in square only which contain a medial rectangle.

Therefore *LN* is a second apotome of a medial straight line; [[book-10/proposition-75|X. 75]] and it is the side of the area *AB*.

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