(related to Proposition: Prop. 10.079: Construction of Apotome is Unique)

- Let $AB$ be an apotome, with $BC$ (so) attached to it.
- $AC$ and $CB$ are thus rational (straight lines which are) commensurable in square only [Prop. 10.73].
- I say that another rational (straight line), which is commensurable in square only with the whole, cannot be attached to $AB$.

- For, if possible, let $BD$ be (so) attached (to $AB$).
- Thus, $AD$ and $DB$ are also rational (straight lines which are) commensurable in square only [Prop. 10.73].
- And since by whatever (area) the (sum of the squares) on $AD$ and $DB$ exceeds twice the (rectangle contained) by $AD$ and $DB$, the (sum of the squares) on $AC$ and $CB$ also exceeds twice the (rectangle contained) by $AC$ and $CB$ by this (same area).
- For both exceed by the same (area) - (namely), the (square) on $AB$ [Prop. 2.7].
- Thus, alternately, 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 the (sum of the squares) on $AD$ and $DB$ exceeds the (sum of the squares) on $AC$ and $CB$ by a rational (area).
- For both (are) rational (areas).
- Thus, twice the (rectangle contained) by $AD$ and $DB$ also exceeds twice the (rectangle contained) by $AC$ and $CB$ by a rational (area).
- The very thing is impossible.
- For both are medial (areas) [Prop. 10.21], and a medial (area) cannot exceed a(nother) medial (area) by a rational (area) [Prop. 10.26].
- Thus, another rational (straight line), which is commensurable in square only with the whole, cannot be attached to $AB$.
- Thus, only one rational (straight line), which is commensurable in square only with the whole, can be attached to an apotome.
- (Which is) the very thing it was required to show.∎

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