(related to Proposition: Prop. 10.042: Binomial Straight Line is Divisible into Terms Uniquely)

- Let $AB$ be a binomial (straight line) which has been divided into its (component) terms at $C$.
- $AC$ and $CB$ are thus rational (straight lines which are) commensurable in square only [Prop. 10.36].
- I say that $AB$ cannot be divided at another point into two rational (straight lines which are) commensurable in square only.

- For, if possible, let it also have been divided at $D$, such that $AD$ and $DB$ are also rational (straight lines which are) commensurable in square only.
- So, (it is) clear that $AC$ is not the same as $DB$.
- For, if possible, let it be (the same).
- So, $AD$ will also be the same as $CB$.
- And as $AC$ will be to $CB$, so $BD$ (will be) to $DA$.
- And $AB$ will (thus) also be divided at $D$ in the same (manner) as the division at $C$.
- The very opposite was assumed.
- Thus, $AC$ is not the same as $DB$.
- So, on account of this, points $C$ and $D$ are not equally far from the point of bisection.
- Thus, by whatever (amount the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$, twice the (rectangle contained) by $AD$ and $DB$ also differs from twice the (rectangle contained) by $AC$ and $CB$ by this (same amount) - on account of both (the sum of) the (squares) on $AC$ and $CB$, plus twice the (rectangle contained) by $AC$ and $CB$, and (the sum of) the (squares) on $AD$ and $DB$, plus twice the (rectangle contained) by $AD$ and $DB$, being equal to the (square) on $AB$ [Prop. 2.4].
- But, (the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$ by a rational (area).
- For (they are) both rational (areas).
- Thus, twice the (rectangle contained) by $AD$ and $DB$ also differs from twice the (rectangle contained) by $AC$ and $CB$ by a rational (area, despite both) being medial (areas) [Prop. 10.21].
- The very thing is absurd.
- For a medial (area) cannot exceed a medial (area) by a rational (area) [Prop. 10.26].
- Thus, a binomial (straight line) cannot be divided (into its component terms) at different points.
- Thus, (it can be so divided) at one point only.
- (Which is) the very thing it was required to show.∎

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