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 rational, and twice the (rectangle contained) by them medial, can be attached to a minor (straight line). * Let $AB$ be a minor (straight line), and let $BC$ be (so) attached to $AB$. * Thus, $AC$ and $CB$ are (straight lines which are) incommensurable in square, making the sum of the squares on them rational, and twice the (rectangle contained) by them medial [Prop. 10.76]. * I say that another another straight line fulfilling the same (conditions) cannot be attached to $AB$.
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. 10.45], with minus signs instead of plus signs.
Proofs: 1
Propositions: 1
