Proof: By Euclid

Let $A$ and $B$ be similar plane numbers.
I say that $A$ has to $B$ the ratio which (some) square number (has) to a(nother) square number.

fig26e

For since $A$ and $B$ are similar plane numbers, one number thus falls (between) $A$ and $B$ in mean proportion [Prop. 8.18].
Let it (so) fall, and let it be $C$.
And let the least numbers, $D$, $E$, $F$, having the same ratio as $A$, $C$, $B$ have been taken [Prop. 8.2].
The outermost of them, $D$ and $F$, are thus square [Prop. 8.2 corr.] .
And since as $D$ is to $F$, so $A$ (is) to $B$, and $D$ and $F$ are square, $A$ thus has to $B$ the ratio which (some) square number (has) to a(nother) square number.
(Which is) the very thing it was required to show.
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