Proof: By Euclid

Let $A$ and $B$ be two numbers, and let $A$ make the square (number) $C$ (by) multiplying $B$.
I say that $A$ and $B$ are similar plane numbers.
For let $A$ make $D$ (by) multiplying itself.
Thus, $D$ is square.
And since $A$ has made $D$ (by) multiplying itself, and has made $C$ (by) multiplying $B$, thus as $A$ is to $B$, so $D$ (is) to $C$ [Prop. 7.17].
And since $D$ is square, and $C$ (is) also, $D$ and $C$ are thus similar plane numbers.
Thus, one (number) falls (between) $D$ and $C$ in mean proportion [Prop. 8.18].
And as $D$ is to $C$, so $A$ (is) to $B$.
Thus, one (number) also falls (between) $A$ and $B$ in mean proportion [Prop. 8.8].
And if one (number) falls (between) two numbers in mean proportion then [the] [numbers]bookofproofs$2315 are similar plane (numbers) [Prop. 8.20].
Thus, $A$ and $B$ are similar plane (numbers) .
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!