Commentary on Proposition X.61¶
Back to Proposition X.61.
In this case we have to prove that, $(k^{\frac{1}{4}}\rho + k^{\frac{3}{4}}\rho)$ being a first bimedial straight line, as found in X. 37,
$$\frac{(k^{\frac{1}{4}}\rho + k^{\frac{3}{4}}\rho)^2}{\sigma}$$
is a second binomial straight line.
The form of the proposition and the figure being similar to those of X. 60, I can somewhat abbreviate the reproduction of the proof. Take $x, y, z$ such that
$$\begin{aligned} \sigma x &= k^{\frac{1}{2}}\rho^2,\\ \sigma y &= k^{\frac{3}{2}}\rho^2,\\ \sigma \cdot 2z &= 2k\rho^2. \end{aligned}$$
Then shall $(x + y) + 2z$ be a second binomial.
($\alpha$) $k^{\frac{1}{4}}\rho, k^{\frac{3}{4}}\rho$ are medial straight lines commensurable in square only and containing a rational rectangle. [X. 37] The squares $k^{\frac{1}{2}}\rho^2, k^{\frac{3}{2}}\rho^2$ are medial; thus the sum, or $\sigma (x + y)$, is medial. [X. 23, Por.] Therefore $(x + y)$ is rational and $\smile \sigma.$ And $\sigma \cdot 2z$ is rational; therefore
$$2z\text{ is rational and } \sim \sigma \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$
Therefore $(x + y), 2z$ are rational and $\smile \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2),$ so that
$$(x + y) + 2z\text{ is a binomial.}$$
($\beta$) As before,
$$(x + y) > 2z.$$
Now, $k^{\frac{1}{2}}\rho^2, k^{\frac{3}{2}}\rho^2$ being commensurable,
$$x \sim y.$$
And
$$xy = z^2,$$
while
$$x + y = \frac{k^{\frac{1}{2}}\rho^2 + k^{\frac{3}{2}}\rho^2}{\sigma}.$$
Hence [X. 17]
$$\sqrt{(x + y)^2 - (2z)^2} \sim (x + y) \dots\dots\dots\dots\dots\dots\dots\dots(3).$$
But $2z \sim \sigma,$ by (1). Therefore [(1), (2), (3)] $(x + y) + 2z$ is a second binomial straight line.
Of course
$$(x + y) + 2z = \frac{\rho^2}{\sigma}\{\sqrt{k}(1 + k) + 2k\}.$$