The square root of (the sum of) two medial (areas) can be divided (into its component terms) at one point only.

- Let $AB$ be [the square root of (the sum of) two medial (areas)] which has been divided at $C$, such that $AC$ and $CB$ are incommensurable in square, making the sum of the (squares) on $AC$ and $CB$ medial, and the (rectangle contained) by $AC$ and $CB$ medial, and, moreover, incommensurable with the sum of the (squares) on ($AC$ and $CB$) [Prop. 10.41].
- I say that $AB$ cannot be divided at another point fulfilling the prescribed (conditions).

In other words, \[\beta^{1/4}\sqrt{\frac{1+\alpha}{2\sqrt{1+\alpha^2}}}+\beta^{1/4}\sqrt{\frac{1-\alpha}{2\sqrt{1+\alpha^2}}}=\gamma^{1/4}\sqrt{\frac{1+\delta}{2\sqrt{1+\delta^2}}} + \gamma^{1/4}\sqrt{\frac{1-\delta}{2\sqrt{1+\delta^2}}}\] has only one solution: i.e., \[\delta=\alpha\quad\text{ and }\quad\gamma=\beta,\]

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

This proposition corresponds to [Prop. 10.84], 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