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

Let the area *ABCD* be contained by the rational straight line *AB* and the third binomial *AD* divided into its terms at *E*, of which terms *AE* is the greater.

I say that the side of the area *AC* is the irrational straight line called a second bimedial.

Make the same construction as before.

Now, since *AD* is a third binomial straight line, therefore *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 neither of the terms *AE* and *ED* is commensurable in length with *AB*.

Then, in manner similar to the foregoing, we shall prove that *MO* is the side of the area *AC*, and *MN* and *NO* are medial straight lines commensurable in square only, so that *MO* is bimedial.

It is next to be proved that it is also a second bimedial straight line.

Since *DE* is incommensurable in length with *AB*, that is, with *EK*, and *DE* is commensurable with *EF*, therefore *EF* is incommensurable in length with *EK*.

And they are rational, therefore *FE* and *EK* are rational straight lines commensurable in square only. Therefore *EL*, that is, *MR*, is medial.

And it is contained by *MN* and *NO*, therefore the rectangle *MN* by *NO* is medial. Therefore *MO* is a second bimedial straight line.

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

Q.E.D.