A binomial straight line is divided into its terms at one point only.

Let *AB* be a binomial straight line divided into its terms at *C*. Then *AC* and *CB* are rational straight lines commensurable in square only.

I say that *AB* is not divided at another point into two rational straight lines commensurable in square only.

For, if possible, let it be divided at *D* also, so that *AD* and *DB* are also rational straight lines commensurable in square only.

It is then manifest that *AC* is not the same as *DB*.

If possible, let it be so. Then *AB* is also the same as *CB*, and *AC* is to *CB* as *BD* is to *DA*. Thus *AB* is divided at *D* also in the same way as by the division at *C*, which is contrary to the hypothesis.

Therefore *AC* is not the same with *DB*.

For this reason also the points *C* and *D* are not equidistant from the point of bisection.

Therefore that by which the sum of the squares on *AC* and *CB* differs from the sum of the squares on *AD* and *DB* is also that by which twice the rectangle *AD* by *DB* differs from twice the rectangle *AC* by *CB*, because both the squares on *AC* and *CB* together with twice the rectangle *AC* by *CB*, and the squares on *AD* and *DB* together with twice the rectangle *AD* by *DB*, equal the square on *AB*.

But the sum of the squares on *AC* and *CB* differs from the sum of the squares on *AD* and *DB* by a rational area, for both are rational, therefore twice the rectangle *AD* by *DB* also differs from twice the rectangle *AC* by *CB* by a rational area, though they are medial, which is absurd, for a medial area does not exceed a medial by a rational area.

Therefore a binomial straight line is not divided at different points. Therefore it is divided at one point only.

Therefore, *a binomial straight line is divided into its terms at one point only.*

Q.E.D.