---
book: 10
number: 55
id: "X.55"
kind: "theorem"
uses: ["[[book-10/proposition-17]]", "[[book-10/proposition-13]]", "[[book-10/proposition-15]]", "[[book-10/proposition-21]]", "[[book-6/proposition-1]]", "[[book-10/proposition-11]]", "[[book-10/proposition-12]]", "[[book-10/proposition-19]]", "[[book-10/proposition-37]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_2.55"
license: "CC-BY-SA-4.0"
---

# X.55

If an area be contained by a rational straight line and the second binomial, the side of the area is the irrational straight line which is called a first bimedial.

## Proof

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

For, since *AD* is a second binomial straight line, let it be divided into its terms at *E*, so that *AE* is the greater term; therefore *AE*, *ED* are rational straight lines commensurable in square only, the square on *AE* is greater than the square on *ED* by the square on a straight line commensurable with *AE*, and the lesser term *ED* is commensurable in length with *AB*. [[book-10/definitions#Definition 2 (part 2)|X. Deff. II. 2]]

Let *ED* be bisected at *F*, and let there be applied to *AE* the rectangle *AG*, *GE* equal to the square on *EF* and deficient by a square figure; therefore *AG* is commensurable in length with *GE*. [[book-10/proposition-17|X. 17]]

Through *G*, *E*, *F* let *GH*, *EK*, *FL* be drawn parallel to *AB*, *CD*, let the square *SN* be constructed equal to the parallelogram *AH*, and the square *NQ* equal to *GK*, and let them be placed so that *MN* is in a straight line with *NO*; therefore *RN* is also in a straight line with *NP*.

Let the square *SQ* be completed.

It is then manifest from what was proved before that *MR* is a mean proportional between *SN*, *NQ* and is equal to *EL*, and that *MO* is the side of the area *AC*.

It is now to be proved that *MO* is a first bimedial straight line.

Since *AE* is incommensurable in length with *ED*, while *ED* is commensurable with *AB*, therefore *AE* is incommensurable with *AB*. [[book-10/proposition-13|X. 13]]

And, since *AG* is commensurable with *EG*, *AE* is also commensurable with each of the straight lines *AG*, *GE*. [[book-10/proposition-15|X. 15]]

But *AE* is incommensurable in length with *AB*; therefore *AG*, *GE* are also incommensurable with *AB*. [[book-10/proposition-13|X. 13]]

Therefore *BA*, *AG* and *BA*, *GE* are pairs of rational straight lines commensurable in square only; so that each of the rectangles *AH*, *GK* is medial. [[book-10/proposition-21|X. 21]]

Hence each of the squares *SN*, *NQ* is medial.

Therefore *MN*, *NO* are also medial.

And, since *AG* is commensurable in length with *GE*, *AH* is also commensurable with *GK*, [[book-6/proposition-1|VI. 1]]. [[book-10/proposition-11|X. 11]] that is, *SN* is commensurable with *NQ*, that is, the square on *MN* with the square on *NO*.

And, since *AE* is incommensurable in length with *ED*, while *AE* is commensurable with *AG*, and *ED* is commensurable with *EF*, therefore *AG* is incommensurable with *EF*; [[book-10/proposition-13|X. 13]] so that *AH* is also incommensurable with *EL*, that is, *SN* is incommensurable with *MR*, that is, *PN* with *NR*, [[book-6/proposition-1|VI. 1]], [[book-10/proposition-11|X. 11]] that is, *MN* is incommensurable in length with *NO*.

But *MN*, *NO* were proved to be both medial and commensurable in square; therefore *MN*, *NO* are medial straight lines commensurable in square only.

I say next that they also contain a rational rectangle.

For, since *DE* is, by hypothesis, commensurable with each of the straight lines *AB*, *EF*, therefore *EF* is also commensurable with *EK*. [[book-10/proposition-12|X. 12]]

And each of them is rational; therefore *EL*, that is, *MR* is rational, [[book-10/proposition-19|X. 19]] and *MR* is the rectangle *MN*, *NO*.

But, if two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational and is called a first bimedial straight line. [[book-10/proposition-37|X. 37]]

Therefore *MO* is a first bimedial straight line. Q. E. D.