To find the fourth binomial straight line.

Set out two numbers *AC* and *CB* such that *AB* has neither to *BC* nor 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*. Then *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*. Then the square on *EF* is commensurable with the square on *FG*. Therefore *FG* is also rational.

Now, since *BA* does not have to *AC* 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, so that *EG* is binomial.

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

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

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

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

And *EF* and *FG* are rational straight lines commensurable in square only, and *EF* is commensurable in length with *D*. Therefore *EG* is a fourth binomial straight line.

Q.E.D.