Proof: By Euclid

Let $A$ and $B$ be square numbers, and let $C$ be the side of $A$, and $D$ (the side) of $B$.
I say that there exists one number in mean proportion to $A$ and $B$, and that $A$ has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$.

fig11e

For let $C$ make $E$ (by) multiplying $D$.
And since $A$ is square, and $C$ is its side, $C$ has thus made $A$ (by) multiplying itself.
And so, for the same (reasons), $D$ has made $B$ (by) multiplying itself.
Therefore, since $C$ has made $A$, $E$ (by) multiplying $C$, $D$, respectively, thus as $C$ is to $D$, so $A$ (is) to $E$ [Prop. 7.17].
And so, for the same (reasons), as $C$ (is) to $D$, so $E$ (is) to $B$ [Prop. 7.18].
And thus as $A$ (is) to $E$, so $E$ (is) to $B$.
Thus, one number (namely, $E$) is in mean proportion to $A$ and $B$.
So I say that $A$ also has to $B$ a squared ratio with respect to (that) $C$ (has) to $D$.
For since $A$, $E$, $B$ are three in (continued) proportion numbers, $A$ thus has to $B$ a squared ratio with respect to (that) $A$ (has) to $E$ [Def. 5.9] .
And as $A$ (is) to $E$, so $C$ (is) to $D$.
Thus, $A$ has to $B$ a squared ratio with respect to (that) side $C$ (has) to (side) $D$.
(Which is) the very thing it was required to show.
∎

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