To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of squares on them medial but twice the rectangle contained by them rational.

Let *AB* be the straight line which produces with a rational area a medial whole, and let *BC* be an annex to *AB*. Then *AC* and *CB* are straight lines incommensurable in square which fulfill the given conditions.

I say that no other straight line can be annexed to *AB* which fulfills the same conditions.

If possible, let *BD* be so annexed. Then *AD* and *DB* are both straight lines incommensurable in square which fulfill the given conditions.

As in the preceding cases, the excess of the sum of the squares on *AD* and *DB* over the sum of the squares on *AC* and *CB* is also the excess of twice the rectangle *AD* by *DB* over twice the rectangle *AC* by *CB*, while twice the rectangle *AD* by *DB* exceeds twice the rectangle *AC* by *CB* by a rational area, for both are rational, therefore the sum of the squares on *AD* and *DB* also exceeds the sum of the squares on *AC* and *CB* by a rational area, which is impossible, for both are medial.

Therefore no other straight line can be annexed to *AB* which is incommensurable in square with the whole and which with the whole fulfills the aforesaid conditions, therefore only one straight line can be so annexed.

Q.E.D.