Commentary on Proposition VIII.14¶
Back to Proposition VIII.14.
If $a^2$ measures $b^2$, $a$ measures $b$; and, if $a$ measures $b$, $a^2$ measures $b^2$.
$a^2$, $ab$, $b^2$ are in continued proportion in the ratio of $a$ to $b$.
$$\begin{array}{ll} \text{Therefore, since} & a^2 \text{ measures } b^2\text{,} \\ & a^2 \text{ measures } ab\text{.} & \text{[VIII. 7]} \\ \text{But} & a^2 : ab = a : b\text{.} \\ \text{Therefore} & a \text{ measures } b\text{.} \end{array}$$
Since $a$ measures $b$, $a^2$ measures $ab$.
And $a^2$, $ab$, $b^2$ are continuously proportional.
$$\begin{array}{ll} \text{Thus} & ab \text{ measures } b^2\text{.} \\ \text{And} & a^2 \text{ measures } ab\text{.} \\ \text{Therefore} & a^2 \text{ measures } b^2\text{.} \end{array}$$
It will be seen that Euclid puts the last step shortly, saying that, since $a^2$ measures $ab$, and $a^2$, $ab$, $b^2$ are in continued proportion, $a^2$ measures $b^2$. The same thing happens in VIII. 15, where the series of terms is one more than here.