Only one straight line, which is incommensurable in square with the whole, and (together) with the whole makes the sum of the squares on them medial, and twice the (rectangle contained) by them medial, and, moreover, incommensurable with the sum of the (squares) on them, can be attached to that (straight line) which with a medial (area) makes a medial whole. * Let $AB$ be a (straight line) which with a medial (area) makes a medial whole, $BC$ being (so) attached to it. * Thus, $AC$ and $CB$ are incommensurable in square, fulfilling the (other) aforementioned (conditions) [Prop. 10.78]. * I say that a(nother) (straight line) fulfilling the aforementioned (conditions) cannot be attached to $AB$.
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. 10.47], with minus signs instead of plus signs.
Proofs: 1
Propositions: 1