Proposition: Prop. 10.044: Second Bimedial Straight Line is Divisible Uniquely

(Proposition 44 from Book 10 of Euclid's “Elements”)

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.

fig044e

Modern Formulation

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.