To find the third binomial line.

Set out two numbers *AC* and *CB* such that the sum of them *AB* has to *BC* the ratio which a square number has to a square number, but does not have to *AC* the ratio which a square number has to a square number.

Also set out any other number *D*, not square, and let it not have to either of the numbers *BA* and *AC* the ratio which a square number has to a square number.

Set out any rational straight line *E*, and let it be contrived that *D* is to *AB* as the square on *E* is to the square on *FG*. Then the square on *E* is commensurable with the square on *FG*.

And *E* is rational, therefore *FG* is also rational. And, since *D* does not have to *AB* the ratio which a square number has to a square number, neither does the square on *E* have to the square on *FG* the ratio which a square number has to a square number, therefore *E* is incommensurable in length with *FG*.

Next let it be contrived that the number *BA* is to *AC* as the square on *FG* is to the square on *GH*. Then the square on *FG* is commensurable with the square on *GH*.

But *FG* is rational, therefore *GH* is also rational. And, since *BA* does not have to *AC* the ratio which a square number has to a square number, neither does the square on *FG* have to the square on *HG* the ratio which a square number has to a square number, therefore *FG* is incommensurable in length with *GH*.

Therefore *FG* and *GH* are rational straight lines commensurable in square only. Therefore *FH* is binomial.

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

Since *D* is to *AB* as the square on *E* is to the square on *FG*, and *BA* is to *AC* as the square on *FG* is to the square on *GH*, therefore, *ex aequali*, *D* is to *AC* as the square on *E* is to the square on *GH*.

But *D* does not have to *AC* the ratio which a square number has to a square number, therefore neither does the square on *E* have to the square on *GH* the ratio which a square number has to a square number. Therefore *E* is incommensurable in length with *GH*.

Since *BA* is to *AC* as the square on *FG* is to the square on *GH*, therefore the square on *FG* is greater than the square on *GH*.

Let then the sum of the squares on *GH* and *K* equal the square on *FG*. Then, in conversion, *AB* is to *BC* as the square on *FG* is to the square on *K*.

But *AB* has to *BC* the ratio which a square number has to a square number, therefore the square on *FG* also has to the square on *K* the ratio which a square number has to a square number. Therefore *FG* is commensurable in length with *K*.

Therefore the square on *FG* is greater than the square on *GH* by the square on a straight line commensurable with *FG*.

And *FG* and *GH* are rational straight lines commensurable in square only, and neither of them is commensurable in length with *E*.

Therefore *FH* is a third binomial straight line.

Q.E.D.