If an area is contained by a rational (straight line) and a third binomial (straight line) then the square root of the area is the irrational (straight line which is) called second bimedial. * For let the area $ABCD$ be contained by the rational (straight line) $AB$ and by the third binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, of which $AE$ is the greater. * I say that the square root of area $AC$ is the irrational (straight line which is) called second bimedial.
If the rational straight line has unit length then this proposition states that the square root of a third binomial straight line is a second bimedial straight line: i.e., a third binomial straight line has a length \[\sqrt{\alpha}\,\left(1+\sqrt{1-\beta^{\,2}}\right),\] whose square root can be written \[\rho\,\left(\alpha^{1/4}+\frac{\sqrt \delta}{\alpha^{1/4}}\right),\] where \[\rho=\sqrt{\frac {1+\beta}2}\quad\text{ and }\quad\delta=\alpha\,\frac{1-\beta}{1+\beta}.\]
This is the length of a second bimedial straight line (see [Prop. 10.38]), since $\rho, \delta,\alpha,\beta$ are all positive rational numbers.
Proofs: 1
