Commentary on Proposition X.72¶
Back to Proposition X.72.
We have to classify, according to the different possible relations between $k, \lambda$, the straight line
$$\sqrt{\sqrt k.\rho^2 + \sqrt\lambda.\rho^2},$$
where $\sqrt k.\rho^2$ and $\sqrt\lambda.\rho^2$ are incommensurable.
Suppose that
$$\begin{aligned} \sigma u &= \sqrt k.\rho^2, \\ \sigma v &= \sqrt\lambda.\rho^2. \end{aligned}$$
It is immaterial whether $\sqrt k.\rho^2$ or $\sqrt\lambda.\rho^2$ is the greater. Suppose, e.g., that the former is.
Now, $\sqrt k.\rho^2, \sqrt\lambda.\rho^2$ being both medial areas, and $\sigma$ rational, $u, v$ are both rational and $\smile \sigma \dotfill (1).$
Again, by hypothesis,
$$\sigma u \smile \sigma v,$$
$$\text{or } u \smile v \dotfill (2).$$
Hence [(1), (2)] $(u + v)$ is a binomial straight line.
Next, $\sqrt{u^2 - v^2}$ is either commensurable or incommensurable in length with $u$.
($\alpha$) Suppose $\sqrt{u^2 - v^2} \frown u$. In this case $(u + v)$ is a third binomial straight line, and therefore [X. 56]
$$\sqrt{\sigma(u + v)} \text{ is a second bimedial straight line.}$$
($\beta$) If $\sqrt{u^2 - v^2} \smile u$, $(u + v)$ is a sixth binomial straight line, and therefore [X. 59]
$$\sqrt{\sigma(u + v)} \text{ is a side of the sum of two medial areas.}$$
The explanation after X. 72 is for the purpose of showing that all the irrational straight lines treated hitherto are different from one another, viz. the medial, the six irrational straight lines beginning with the binomial, and the six consisting of the first, second, third, fourth, fifth and sixth binomials.