A major (straight line) can only be divided (into its component terms) at the same point. * Let $AB$ be a major (straight line) which has been divided at $C$, so that $AC$ and $CB$ are incommensurable in square, making the sum of the squares on $AC$ and $CB$ rational, and the (rectangle contained) by $AC$ and $CD$ medial [Prop. 10.39]. * I say that $AB$ cannot be (so) divided at another point.
In other words, \[\sqrt{\frac{1+\alpha}{2\sqrt{1+\alpha^2}}} + \sqrt{\frac{1-\alpha}{2\sqrt{1+\alpha^2}}} =\sqrt{\frac{1+\beta}{2\sqrt{1+\beta^2}}} + \sqrt{\frac{1-\beta}{2\sqrt{1+\beta^2}}}\] has only one solution: i.e., \[\beta=\alpha,\] where \(\alpha,\beta\) denote positive rational numbers.
This proposition corresponds to [Prop. 10.82], with plus signs instead of minus signs.
Proofs: 1
Propositions: 1