A second bimedial straight line is divided at one point only.

Let *AB* be a second bimedial straight line divided at *C*, so that *AC* and *CB* are medial straight lines commensurable in square only and containing a medial rectangle. It is then manifest that *C* is not at the point of bisection, because the segments are not commensurable in length.

I say that *AB* is not so divided at another point.

If possible, let it also be divided at *D*, so that *AC* is not the same with *DB*, but *AC* is supposed greater. It is then clear that the sum of the squares on *AD* and *DB* is also, as we proved above, less than the sum of the squares on *AC* and *CB*. Suppose that *AD* and *DB* are medial straight lines commensurable in square only and containing a medial rectangle.

Set out a rational straight line *EF*, apply to *EF* the rectangular parallelogram *EK* equal to the square on *AB*, and subtract *EG*, equal to the sum of the squares on *AC* and *CB*. Then the remainder *HK* equals twice the rectangle *AC* by *CB*.

Again, subtract *EL*, equal to the sum of the squares on *AD* and *DB*, which were proved less than the sum of the squares on *AC* and *CB*. Then the remainder *MK* also equals twice the rectangle *AD* by *DB*.

Now, since the squares on *AC* and *CB* are medial, therefore *EG* is medial.

And it is applied to the rational straight line *EF*, therefore *EH* is rational and incommensurable in length with *EF*.

For the same reason *HN* is also rational and incommensurable in length with *EF*.

And, since *AC* and *CB* are medial straight lines commensurable in square only, therefore *AC* is incommensurable in length with *CB*.

But *AC* is to *CB* as the square on *AC* is to the rectangle *AC* by *CB*, therefore the square on *AC* is incommensurable with the rectangle *AC* by *CB*.

But the sum of the squares on *AC* and *CB* is commensurable with the square on *AC*, for *AC* and *CB* are commensurable in square.

And twice the rectangle *AC* by *CB* is commensurable with the rectangle *AC* by *CB*.

Therefore the sum of the squares on *AC* and *CB* is also incommensurable with twice the rectangle *AC* by *CB*.

But *EG* equals the sum of the squares on *AC* and *CB*, and *HK* equals twice the rectangle *AC* by *CB*, therefore *EG* is incommensurable with *HK*, so that *EH* is also incommensurable in length with *HN*.

And they are rational, therefore *EH* and *HN* are rational straight lines commensurable in square only.

But, if two rational straight lines commensurable in square only are added together, then the whole is the irrational which is called binomial.

Therefore *EN* is a binomial straight line divided at *H*.

In the same way *EM* and *MN* is also proved to be rational straight lines commensurable in square only, and *EN* is a binomial straight line divided at different points, *H* and *M*.

And *EH* is not the same with *MN*, for the sum of the squares on *AC* and *CB* is greater than the sum of the squares on *AD* and *DB*.

But the sum of the squares on *AD* and *DB* is greater than twice the rectangle *AD* by *DB*, therefore the sum of the squares on *AC* and *CB*, that is, *EG*, is much greater than twice the rectangle *AD* by *DB*, that is, *MK*, so that *EH* is also greater than *MN*.

Therefore *EH* is not the same with *MN*.

Therefore, *a second bimedial straight line is divided at one point only.*

Q.E.D.