To find the fifth binomial line.

Set out two numbers *AC* and *CB* such that *AB* does not have to either of them the ratio which a square number has to a square number. Set out any rational straight line *D*, and let *EF* be commensurable with *D*. Then *EF* is rational.

Let it be contrived that *CA* is to *AB* as the square on *EF* is to the square on *FG*.

But *CA* does not have to *AB* the ratio which a square number has to a square number, therefore neither does the square on *EF* have to the square on *FG* the ratio which a square number has to a square number.

Therefore *EF* and *FG* are rational straight lines commensurable in square only, therefore *EG* is binomial.

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

Since *CA* is to *AB* as the square on *EF* is to the square on *FG*, therefore, inversely, *BA* is to *AC* as the square on *FG* is to the square on *FE*. Therefore the square on *GF* is greater than the square on *FE*.

Let then the sum of the squares on *EF* and *H* equal the square on *GF*. Then, in conversion, the number *AB* is to *BC* as the square on *GF* is to the square on *H*.

But *AB* does not have to *BC* the ratio which a square number has to a square number, therefore neither does the square on *FG* have to the square on *H* the ratio which a square number has to a square number.

Therefore *FG* is incommensurable in length with *H*, so that the square on *FG* is greater than the square on *FE* by the square on a straight line incommensurable with *FG*.

And *GF* and *FE* are rational straight lines commensurable in square only, and the lesser term *EF* is commensurable in length with the rational straight line *D* set out.

Therefore *EG* is a fifth binomial straight line.

Q.E.D.