To find a second binomial (straight line).

- Let the two numbers $AC$ and $CB$ be laid down such that their sum $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number, and does not have to $AC$ the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I] .
- And let the rational (straight line) $D$ be laid down.
- And let $EF$ be commensurable in length with $D$.
- $EF$ is thus a rational (straight line).
- So, let it also have been contrived that as the number $CA$ (is) to $AB$, so the (square) on $EF$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] .
- Thus, the (square) on $EF$ is commensurable with the (square) on on $FG$ [Prop. 10.6].
- Thus, $FG$ is also a rational (straight line).
- And since the number $CA$ does not have to $AB$ the ratio which (some) square number (has) to (some) square number, the (square) on $EF$ does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either.
- Thus, $EF$ is incommensurable in length with $FG$ [Prop. 10.9].
- $EF$ and $FG$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $EG$ is a binomial (straight line) [Prop. 10.36].
- So, we must show that (it is) also a second (binomial straight line).

If the rational straight line has unit length then the length of a second binomial straight line is \[\frac{\alpha}{\sqrt{1-\beta^{\,2}}}+\alpha,\]

where \(\alpha,\beta\) denote positive rational numbers.

This, and the second apotome, whose length according to [Prop. 10.86] is \[\frac{\alpha}{\sqrt{1-\beta^{\,2}}}-\alpha,\] are the roots of the quadratic function \[x^2- \frac{2\,\alpha}{\sqrt{1-\beta^{\,2}}}\,x+\frac{\alpha^2\,\beta^{\,2}}{1-\beta^{\,2}}=0, \]

where \(\alpha,\beta\) denote positive rational numbers.

Proofs: 1

Propositions: 1

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

**Prime.mover and others**: "Pr∞fWiki", https://proofwiki.org/wiki/Main_Page, 2016