Proof: By Euclid

fig02e

For if (double $AC$ is) not (greater than $BC$), if possible, let $BC$ be double $CA$.
Thus, the (square) on $BC$ (is) four times the (square) on $CA$.
Thus, the (sum of) the (squares) on $BC$ and $CA$ (is) five times the (square) on $CA$.
And the (square) on $BA$ was assumed (to be) five times the (square) on $CA$.
Thus, the (square) on $BA$ is equal to the (sum of) the (squares) on $BC$ and $CA$.
The very thing (is) impossible [Prop. 2.4].
Thus, $CB$ is not double $AC$.
So, similarly, we can show that a (straight line) less than $CB$ is not double $AC$ either.
For (in this case) the absurdity is much [greater].
Thus, double $AC$ is greater than $CB$.
(Which is) the very thing it was required to show.
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