A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.

Let *AB* be binomial, and let *CD* be commensurable in length with *AB*.

I say that *CD* is binomial and the same in order with *AB*.

Since *AB* is binomial, divide it into its terms at *E*, and let *AE* be the greater term, therefore *AE* and *EB* are rational straight lines commensurable in square only.

Let it be contrived that *AB* is to *CD* as *AE* is to *CF*. Then the remainder *EB* is to the remainder *FD* as *AB* is to *CD*.

But *AB* is commensurable in length with *CD*, therefore *AE* is also commensurable with *CF*, and *EB* with *FD*.

And *AE* and *EB* are rational, therefore *CF* and *FD* are also rational.

And *AE* is to *CF* as *EB* is to *FD*. Therefore, alternately, *AE* is to *EB* as *CF* is to *FD*.

But *AE* and *EB* are commensurable in square only, therefore *CF* and *FD* are also commensurable in square only.

And they are rational, therefore *CD* is binomial.

I say next that it is the same in order with *AB*.

For the square on *AE* is greater than the square on *EB* either by the square on a straight line commensurable with *AE* or by the square on a straight line incommensurable with it.

If then the square on *AE* is greater than the square on *EB* by the square on a straight line commensurable with *AE*, then the square on *CF* is also greater than the square on *FD* by the square on a straight line commensurable with *CF*.

And, if *AE* is commensurable with the rational straight line set out, then *CF* is also commensurable with it, and for this reason each of the straight lines *AB* and *CD* is a first binomial, that is, the same in order.

But, if *EB* is commensurable with the rational straight line set out, then *FD* is also commensurable with it, and for this reason again *CD* is the same in order with *AB*, for each of them is a second binomial.

But, if neither of the straight lines *AE* nor *EB* is commensurable with the rational straight line set out, then neither of the straight lines *CF* nor *FD* is commensurable with it, and each of the straight lines *AB* and *CD* is a third binomial.

But, if the square on *AE* is greater than the square on *EB* by the square on a straight line incommensurable with *AE*, then the square on *CF* is also greater than the square on *FD* by the square on a straight line incommensurable with *CF*.

And, if *AE* is commensurable with the rational straight line set out, then *CF* is also commensurable with it, and each of the straight lines *AB* and *CD* is a fourth binomial.

But, if *EB* is so commensurable, then *FD* is also, and each of the straight lines *AB* and *CD* is a fifth binomial.

But, if neither of the straight lines *AE* or *EB* is so commensurable, then neither of the straight lines *CF* or *FD* is commensurable with the rational straight line set out, and each of the straight lines *AB* and *CD* is a sixth binomial.

Hence a straight line commensurable in length with a binomial straight line is binomial and the same in order.

Therefore, *a straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.*

Q.E.D.