The square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.

Let *AB* be the side of a rational plus a medial area, divided into its straight lines at *C*, so that *AC* is the greater, let a rational straight line *DE* be set out, and let there be applied to *DE* the parallelogram *DF* equal to the square on *AB*, producing *DG* as its breadth.

I say that *DG* is a fifth binomial straight line.

Make the same construction as before.

Since *AB* is the side of a rational plus a medial area, divided at *C*, therefore *AC* and *CB* are straight lines incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational.

Since, then, the sum of the squares on *AC* and *CB* is medial, therefore *DL* is medial, so that *DM* is rational and incommensurable in length with *DE*.

Again, since twice the rectangle *AC* by *CB*, that is *MF*, is rational, therefore *MG* is rational and commensurable with *DE*.

Therefore *DM* is incommensurable with *MG*. Therefore *DM* and *MG* are rational straight lines commensurable in square only. Therefore *DG* is binomial.

I say next that it is also a fifth binomial straight line.

For it can be proved similarly that the rectangle *DK* by *KM* equals the square on *MN*, and that *DK* is incommensurable in length with *KM*. Therefore the square on *DM* is greater than the square on *MG* by the square on a straight line incommensurable with *DM*.

And *DM* and *MG* are commensurable in square only, and the less, *MG*, is commensurable in length with *DE*.

Therefore *DG* is a fifth binomial.

Therefore, *the square on the side of a rational plus a medial area applied to a rational straight line produces as breadth the fifth binomial.*

Q.E.D.