---
book: 10
number: 95
id: "X.95"
kind: "theorem"
uses: ["[[book-10/proposition-18]]", "[[book-10/proposition-21]]", "[[book-10/proposition-19]]", "[[book-6/proposition-26]]", "[[book-10/proposition-77]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_3.95"
license: "CC-BY-SA-4.0"
---

# X.95

If an area be contained by a rational straight line and a fifth apotome, the side of the area is a straight line which produces with a rational area a medial whole.

## Proof

For let the area *AB* be contained by the rational straight line *AC* and the fifth apotome *AD*; I say that the side of the area *AB* is a straight line which produces with a rational area a medial whole.

For let *DG* be the annex to *AD*; therefore *AG*, *GD* are rational straight lines commensurable in square only, the annex *GD* 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 with *AG*. [[book-10/definitions#Definition 5 (part 3)|X. Deff. III. 5]]

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 the point *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*.

Now, since *AG* is incommensurable in length with *CA*, and both are rational, therefore *AK* is medial. [[book-10/proposition-21|X. 21]]

Again, since *DG* is rational and commensurable in length with *AC*, *DK* is rational. [[book-10/proposition-19|X. 19]]

Now let the square *LM* be constructed equal to *AI*, and let the square *NO* equal to *FK* and about the same angle, the angle *LPM*, be subtracted; 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.

Similarly then we can prove that *LN* is the side of the area *AB*.

I say that *LN* is the straight line which produces with a rational area a medial whole.

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

Again, since *DK* is rational and is equal to twice the rectangle *LP*, *PN*, the latter is itself also rational.

And, since *AI* is 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 medial but twice the rectangle contained by them rational.

Therefore the remainder *LN* is the irrational straight line called that which produces with a rational area a medial whole; [[book-10/proposition-77|X. 77]] and it is the side of the area *AB*.

Therefore the side of the area *AB* is a straight line which produces with a rational area a medial whole. Q. E. D.