To find the first 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 *CA* the ratio which a square number has to a square number.

Set out any rational straight line *D*, and let *EF* be commensurable in length with *D*. Therefore *EF* is also rational.

Let it be contrived that the number *BA* is to *AC* as the square on *EF* is to the square on *FG*.

But *AB* has to *AC* the ratio which a number has to a number, therefore the square on *EF* also has to the square on *FG* the ratio which a number has to a number, so that the square on *EF* is commensurable with the square on *FG*.

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

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

I say that it is also a first binomial straight line.

Since the number *BA* is to *AC* as the square on *EF* is to the square on *FG*, while *BA* is greater than *AC*, therefore the square on *EF* is also greater than the square on *FG*.

Let then the sum of the squares on *FG* and *H* equal the square on *EF*.

Now since *BA* is to *AC* as the square on *EF* is to the square on *FG*, therefore, in conversion, *AB* is to *BC* as the square on *EF* is to the square on *H*.

But *AB* has to *BC* the ratio which a square number has to a square number, therefore the square on *EF* also has to the square on *H* the ratio which a square number has to a square number.

Therefore *EF* is commensurable in length with *H*. Therefore the square on *EF* is greater than the square on *FG* by the square on a straight line commensurable with *EF*.

And *EF* and *FG* are rational, and *EF* is commensurable in length with *D*.

Therefore *EF* is a first binomial straight line.

Q.E.D.