The square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.

Let *AB* be a second bimedial straight line divided into its medials at *C*, so that *AC* is the greater segment, let *DE* be any rational straight line, and to *DE* let there be applied the parallelogram *DF* equal to the square on *AB* and producing *DG* as its breadth.

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

Make the same construction as shown before.

Then, since *AB* is a second bimedial divided at *C*, therefore *AC* and *CB* are medial straight lines commensurable in square only and containing a medial rectangle, so that the sum of the squares on *AC* and *CB* is also medial.

And it equals *DL*, therefore *DL* is also medial. And it is applied to the rational straight line *DE*, therefore *MD* is also rational and incommensurable in length with *DE*.

For the same reason, *MG* is also rational and incommensurable in length with *ML*, that is, with *DE*, therefore each of the straight lines *DM* and *MG* is rational and incommensurable in length with *DE*.

Since *AC* is incommensurable in length with *CB*, and *AC* is to *CB* as the square on *AC* is to the rectangle *AC* by *CB*, therefore the square on *AC* is also incommensurable with the rectangle *AC* by *CB*.

Hence the sum of the squares on *AC* and *CB* is incommensurable with twice the rectangle *AC* by *CB*, that is, *DL* is incommensurable with *MF*, so that *DM* is also incommensurable with *MG*.

And they are rational, therefore *DG* is binomial.

It is to be proved that it is a third binomial straight line.

In manner similar to the foregoing we may conclude that *DM* is greater than *MG*, and that *DK* is commensurable with *KM*.

And the rectangle *DK* by *KM* equals the square on *MN*, therefore the square on *DM* is greater than the square on *MG* by the square on a straight line commensurable with *DM*. And neither of the straight lines *DM* nor *MG* is commensurable in length with *DE*.

Therefore *DG* is a third binomial straight line.

Therefore, *the square on the second bimedial straight line applied to a rational straight line produces as breadth the third binomial.*

Q.E.D.