Commentary on Proposition X.14

Back to Proposition X.14.

[Folio 0042 | p. 38]

Suppose $a$, $b$, $c$, $d$ to be straight lines such that

$$a : b = c : d \tag{1}$$

It follows [vi. 22] that

$$a^2 : b^2 = c^2 : d^2 \tag{2}$$

In order to prove that, convertendo,

$$a^2 : (a^2 - b^2) = c^2 : (c^2 - d^2)$$

Euclid has to use a somewhat roundabout method owing to the absence of a convertendo proposition in his Book v. (which omission Simson supplied by his Prop. E).

It follows from (2) that

$$\{(a^2 - b^2) + b^2\} : b^2 = \{(c^2 - d^2) + d^2\} : d^2,$$

whence, separando,

$$(a^2 - b^2) : b^2 = (c^2 - d^2) : d^2, \tag{v. 17}$$

and, inversely,

$$b^2 : (a^2 - b^2) = d^2 : (c^2 - d^2).$$

From this and (2), ex aequali,

$$a^2 : (a^2 - b^2) = c^2 : (c^2 - d^2). \tag{v. 22}$$

Hence

$$a : \sqrt{a^2 - b^2} = c : \sqrt{c^2 - d^2}. \tag{vi. 22}$$

According therefore as

$$\begin{aligned} a &⌢\text{ or } ⌣ \sqrt{a^2 - b^2}, \\ c &⌢\text{ or } ⌣ \sqrt{c^2 - d^2}. &&\text{[x. 11]} \end{aligned}$$

If $a ⌢ \sqrt{a^2 - b^2}$, we may put $\sqrt{a^2 - b^2} = ka$, where $k$ is of the form $m/n$ and $m$, $n$ are integers. And if $\sqrt{a^2 - b^2} = ka$, it follows in this case that $\sqrt{c^2 - d^2} = kc$.