The side of the sum of two medial areas is divided at one point only.

Let *AB* be divided at *C*, so that *AC* and *CB* are incommensurable in square and let the sum of the squares on *AC* and *CB* be medial, and the rectangle *AC* by *CB* medial and also incommensurable with the sum of the squares on them.

I say that *AB* is not divided at another point so as to fulfill the given conditions.

If possible, let it be divided at *D*, so that again *AC* is of course not the same as *BD*, but *AC* is supposed greater.

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

Again, to *EP* apply *EL*, equal to the sum of the squares on *AD* and *DB*. Then the remainder, twice the rectangle *AD* by *DB*, equals the remainder *MK*.

And since, by hypothesis, the sum of the squares on *AC* and *CB* is medial, therefore *EG* is also medial.

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

For the same reason *HN* is also rational and incommensurable in length with *EF*. And, since the sum of the squares on *AC* and *CB* is incommensurable with twice the rectangle *AC* by *CB*, therefore *EG* is also incommensurable with *GN*, so that *EH* is also incommensurable with *HN*.

And they are rational, therefore *EH* and *HN* are rational straight lines commensurable in square only. Therefore *EN* is a binomial straight line divided at *H*.

Similarly we can prove that it is also divided at *M*. And *EH* is not the same with *MN*, therefore a binomial has been divided at different points, which is absurd.

Therefore a side of the sum of two medial areas is not divided at different points. Therefore it is divided at one point only.

Therefore, *the side of the sum of two medial areas is divided at one point only.*

Q.E.D.