(related to Proposition: Prop. 10.061: Square on First Bimedial Straight Line applied to Rational Straight Line)

- Let $AB$ be a first bimedial (straight line) having been divided into its (component) medial (straight lines) at $C$, of which $AC$ (is) the greater.
- And let the rational (straight line) $DE$ be laid down.
- And let the parallelogram $DF$, equal to the (square) on $AB$, have been applied to $DE$, producing $DG$ as breadth.
- I say that $DG$ is a second binomial (straight line).

- For let the same construction have been made as in the (proposition) before this.
- And since $AB$ is a first bimedial (straight line), having been divided at $C$, $AC$ and $CB$ are thus medial (straight lines) commensurable in square only, and containing a rational (area) [Prop. 10.37].
- Hence, the (squares) on $AC$ and $CB$ are also medial [Prop. 10.21].
- Thus, $DL$ is medial [Prop. 10.15], [Prop. 10.23 corr.] .
- And it has been applied to the rational (straight line) $DE$.
- $MD$ is thus rational, and incommensurable in length with $DE$ [Prop. 10.22].
- Again, since twice the (rectangle contained) by $AC$ and $CB$ is rational, $MF$ is also rational.
- And it is applied to the rational (straight line) $ML$.
- Thus, $MG$ [is] also rational, and commensurable in length with $ML$ - that is to say, with $DE$ [Prop. 10.20].
- $DM$ is thus incommensurable in length with $MG$ [Prop. 10.13].
- And they are rational.
- $DM$ and $MG$ are thus rational, and commensurable in square only.
- $DG$ is thus a binomial (straight line) [Prop. 10.36].
- So, we must show that (it is) also a second (binomial straight line).
- For since (the sum of) the squares on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$ [Prop. 10.59], $DL$ (is) thus also greater than $MF$.
- Hence, $DM$ (is) also (greater) than $MG$ [Prop. 6.1].
- And since the (square) on $AC$ is commensurable with the (square) on on $CB$, $DH$ is also commensurable with $KL$.
- Hence, $DK$ is also commensurable (in length) with $KM$ [Prop. 6.1], [Prop. 10.11].
- And the (rectangle contained) by $DKM$ is equal to the (square) on $MN$.
- Thus, the square on $DM$ is greater than (the square on) $MG$ by the (square) on (some straight line) commensurable (in length) with ($DM$) [Prop. 10.17].
- And $MG$ is commensurable in length with $DE$.
- Thus, $DG$ is a second binomial (straight line) [Def. 10.6] .∎

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