To find the second 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. Set out a rational straight line *D*, and let *EF* be commensurable in length with *D*, therefore *EF* is rational.

Let it be contrived then that as the number *CA* is to *AB*, so is the square on *EF*to the square on *FG*, therefore the square on *EF* is commensurable with the square on *FG*. Therefore *FG* is also rational.

Now, since the number *CA* does not have to *AB* the ratio which a square number has to a square number, 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* is incommensurable in length with *FG*. Therefore *EF* and *FG* are rational straight lines commensurable in square only. Therefore *EG* is binomial.

It is next to be proved that it is also a second binomial straight line.

Since, inversely, the number *BA* is to *AC* as the square on *GF* is to the square on *FE*, while *BA* is greater than *AC*, therefore the square on *GF* is greater than the square on *FE*.

Let the sum of the squares on *EF* and *H* equal the square on *GF*. Then, in conversion, *AB* is to *BC* as the square on *FG* 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 *FG* also has to the square on *H* the ratio which a square number has to a square number.

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

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

Therefore *EG* is a second binomial straight line.

Q.E.D.