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

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.

Since *AD* is a second binomial straight line, divide it into its terms at *E*, so that *AE* is the greater term. Then *AE* and *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*.

Bisect *ED* at *F*, and apply to *AE* the rectangle *AG* by *GE* equal to the square on *EF* and deficient by a square figure. Then *AG* is commensurable in length with *GE*.

Draw *GH*, *EK*, and *FL* through *G*, *E*, and *F* parallel to *AB* and *CD*. Construct the square *SN* equal to the parallelogram *AH*, and the square *NQ* equal to *GK*, and place them so that *MN* is in a straight line with *NO*. Then *RN* is also in a straight line with *NP*. Complete the square *SQ*.

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

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

Since *AE* is commensurable in length with *ED*, while *ED* is commensurable with *AB*, therefore *AE* is incommensurable with *AB*.

Since *AG* is commensurable with *EG*, therefore *AE* is also commensurable with each of the straight lines *AG* and *GE*.

But *AE* is incommensurable in length with *AB*, therefore *AG* and *GE* are also incommensurable with *AB*.

Therefore *BA* and *AG*, and *BA* and *GE*, are pairs of rational straight lines commensurable in square only, so that each of the rectangles *AH* and *GK* is medial.

Hence, each of the squares *SN* and *NQ* is medial. Therefore *MN* and *NO* are also medial.

Since *AG* is commensurable in length with *GE*, therefore *AH* is also commensurable with *GK*, that is, *SN* is commensurable with *NQ*, that is, the square on *MN* with the square on *NO*.

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*, so that *AH* is also incommensurable with *EL*, that is, *SN* is incommensurable with *MR*, that is, *PN* with *NR*, that is, *MN* is incommensurable in length with *NO*.

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

I say next that they also contain a rational rectangle.

Since *DE* is, by hypothesis, commensurable with each of the straight lines *AB* and *EF*, therefore *EF* is also commensurable with *EK*.

And each of them is rational, therefore *EL*, that is, *MR* is rational, and *MR* is the rectangle *MN* by *NO*.

But, if two medial straight lines commensurable in square only and containing a rational rectangle are added together, then the whole is irrational and is called a first bimedial straight line. Therefore *MO* is a first bimedial straight line.

Therefore, *if an area is contained by a rational straight line and the second binomial, then the side of the area is the irrational straight line which is called a first bimedial.*

Q.E.D.