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 rational, can be attached to that (straight line) which with a rational (area) makes a medial whole. * Let $AB$ be a (straight line) which with a rational (area) makes a medial whole, and let $BC$ be (so) attached to $AB$. * Thus, $AC$ and $CB$ are (straight lines which are) incommensurable in square, fulfilling the (other) proscribed (conditions) [Prop. 10.77]. * I say that another (straight line) fulfilling the same (conditions) cannot be attached to $AB$.
In other words, \[\sqrt{\frac{\sqrt{1+\alpha^2}+\alpha}{2\,(1+\alpha^2)}} - \sqrt{\frac{\sqrt{1+\alpha^2}-\alpha}{2\,(1+\alpha^2)}}=\sqrt{\frac{\sqrt{1+\beta^2}+\beta}{2\,(1+\beta^2)}} - \sqrt{\frac{\sqrt{1+\beta^2}-\beta}{2\,(1+\beta^2)}}\] has only one solution: i.e., \[\beta=\alpha,\]
where \(\alpha,\beta\) denote positive rational numbers.
This proposition corresponds to [Prop. 10.46], with minus signs instead of plus signs.
Proofs: 1
Propositions: 1