If an area is contained by a rational straight line and the sixth binomial, then the side of the area is the irrational straight line called the side of the sum of two medial areas.

Let the area *ABCD* be contained by the rational straight line *AB* and the sixth binomial *AD*, divided into its terms at *E*, so that *AE* is the greater term.

I say that the side of *AC* is the side of the sum of two medial areas.

Make the same construction as shown before.

It is then clear that *MO* is the side of *AC*, and that *MN* is incommensurable in square with *NO*.

Now, since *EA* is incommensurable in length with *AB*, therefore *EA* and *AB* are rational straight lines commensurable in square only, therefore *AK*, that is, the sum of the squares on *MN* and *NO*, is medial.

Again, since *ED* is incommensurable in length with *AB*, therefore *FE* is also incommensurable with *EK*. Therefore *FE* and *EK* are rational straight lines commensurable in square only. Therefore *EL*, that is, *MR*, that is, the rectangle *MN* by *NO*, is medial.

Since *AE* is incommensurable with *EF*, therefore *AK* is also incommensurable with *EL*.

But *AK* is the sum of the squares on *MN* and *NO*, and *EL* is the rectangle *MN* by *NO*, therefore the sum of the squares on *MN* and *NO* is incommensurable with the rectangle *MN* by *NO*. And each of them is medial, and *MN* and *NO* are incommensurable in square.

Therefore *MO* is the side of the sum of two medial areas, and is the side of *AC*.

Therefore, *if an area is contained by a rational straight line and the sixth binomial, then the side of the area is the irrational straight line called the side of the sum of two medial areas.*

Q.E.D.