(related to Proposition: Prop. 10.083: Construction of that which produces Medial Whole with Rational Area is Unique)

- Let $AB$ be a (straight line) which with a rational (area) makes a medial whole, and let $BC$ be (so) attached to $AB$.
- Thus, $AC$ and $CB$ are (straight lines which are) incommensurable in square, fulfilling the (other) proscribed (conditions) [Prop. 10.77].
- I say that another (straight line) fulfilling the same (conditions) cannot be attached to $AB$.

- For, if possible, let $BD$ be (so) attached (to $AB$).
- Thus, $AD$ and $DB$ are also straight lines (which are) incommensurable in square, fulfilling the (other) prescribed (conditions) [Prop. 10.77].
- Therefore, analogously to the (propositions) before this, since by whatever (area) the (sum of the squares) on $AD$ and $DB$ exceeds the (sum of the squares) on $AC$ and $CB$, twice the (rectangle contained) by $AD$ and $DB$ also exceeds twice the (rectangle contained) by $AC$ and $CB$ by this (same area).
- And twice the (rectangle contained) by $AD$ and $DB$ exceeds twice the (rectangle contained) by $AC$ and $CB$ by a rational (area).
- For they are (both) rational (areas).
- Thus, the (sum of the squares) on $AD$ and $DB$ also exceeds the (sum of the squares) on $AC$ and $CB$ by a rational (area).
- The very thing is impossible.
- For both are medial (areas) [Prop. 10.26].
- Thus, another straight line cannot be attached to $AB$, which is incommensurable in square with the whole, and fulfills the (other) aforementioned (conditions) with the whole.
- Thus, only one (such straight line) can be (so) attached.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"