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
