A second bimedial (straight line) can be divided (into its component terms) at one point only.

- Let $AB$ be a second bimedial (straight line) which has been divided at $C$, so that $AC$ and $BC$ are medial (straight lines), commensurable in square only, (and) containing a medial (area) [Prop. 10.38].
- So, (it is) clear that $C$ is not (located) at the point of bisection, since ($AC$ and $BC$) are not commensurable in length.
- I say that $AB$ cannot be (so) divided at another point.

In other words,

\[\alpha^{1/4}+\frac{\beta^{1/2}}{\alpha^{1/4}} = \gamma^{1/4}+\frac{\delta^{1/2}}{\gamma^{1/4}}\] has only one solution: i.e., \[\gamma=\alpha\quad\text{ and }\quad \delta=\beta,\] where \(\alpha,\beta,\gamma,\delta\) denote positive rational numbers.

This proposition corresponds to [Prop. 10.81], with plus signs instead of minus signs.

Proofs: 1

Propositions: 1

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

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