To find a fifth binomial straight line.

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

If the rational straight line has unit length then the length of a fifth binomial straight line is \[\alpha\,(\sqrt{1+\beta}+1),\]

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

This, and the fifth apotome, whose length according to [Prop. 10.89] is \[\alpha\,(\sqrt{1+\beta}-1),\] are the roots of the quadratic function \[x^2- 2\,\alpha\,\sqrt{1+\beta}\,x+\alpha^2\,\beta=0,\]

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