Proof: By Euclid
(related to Lemma: Lem. 13.02: Converse of Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"