Proof: By Euclid

Let $A$ and $B$ be square numbers, and let $C$ and $D$ be their sides (respectively).
And let $A$ measure $B$.
I say that $C$ also measures $D$.

fig14e

For let $C$ make $E$ (by) multiplying $D$.
Thus, $A$, $E$, $B$ are in continued proportion in the ratio of $C$ to $D$ [Prop. 8.11].
And since $A$, $E$, $B$ are in continued proportion, and $A$ measures $B$, $A$ thus also measures $E$ [Prop. 8.7].
And as $A$ is to $E$, so $C$ (is) to $D$.
Thus, $C$ also measures $D$ [Def. 7.20] .
So, again, let $C$ measure $D$.
I say that $A$ also measures $B$.
For similarly, with the same construction, we can show that $A$, $E$, $B$ are in continued proportion in the ratio of $C$ to $D$.
And since as $C$ is to $D$, so $A$ (is) to $E$, and $C$ measures $D$, $A$ thus also measures $E$ [Def. 7.20] .
And $A$, $E$, $B$ are in continued proportion.
Thus, $A$ also measures $B$.
Thus, if a square (number) measures a(nother) square (number) then the side (of the former) will also measure the side (of the latter).
And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number) .
(Which is) the very thing it was required to show.
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