If an area is contained by a rational straight line and the fifth binomial, then the side of the area is the irrational straight line called the side of a rational plus a medial area.

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

I say that the side of the area *AC* is the irrational straight line called the side of a rational plus a medial area.

Make the same construction shown before. It is then manifest that *MO* is the side of the area *AC*. It is then to be proved that *MO* is the side of a rational plus a medial area.

Since *AG* is incommensurable with *GE*, therefore *AH* is also commensurable with *HE*, that is, the square on *MN* with the square on *NO*. Therefore *MN* and *NO* are incommensurable in square.

Since *AD* is a fifth binomial straight line, and *ED* the lesser segment, therefore *ED* is commensurable in length with *AB*.

But *AE* is incommensurable with *ED*, therefore *AB* is also incommensurable in length with *AE*. Therefore *AK*, that is, the sum of the squares on *MN* and *NO*, is medial.

Since *DE* is commensurable in length with *AB*, that is, with *EK*, while *DE* is commensurable with *EF*, therefore *EF* is also commensurable with *EK*.

And *EK* is rational, therefore *EL*, that is, *MR*, that is, the rectangle *MN* by *NO*, is also rational.

Therefore *MN* and *NO* are straight lines incommensurable in square which make the sum of the squares on them medial, but the rectangle contained by them rational.

Therefore *MO* is the side of a rational plus a medial area and is the side of the area *AC*.

Therefore, *if an area is contained by a rational straight line and the fifth binomial, then the side of the area is the irrational straight line called the side of a rational plus a medial area.*

Q.E.D.