Proof: By Euclid

Let $AB$ be a straight line, and let it have been cut unequally at $C$, and let $AC$ be greater (than $CB$).
I say that (the sum of) the (squares) on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$.

fig059ae

For let $AB$ have been cut in half at $D$.
Therefore, since a straight line has been cut into equal (parts) at $D$, and into unequal (parts) at $C$, the (rectangle contained) by $AC$ and $CB$, plus the (square) on $CD$, is thus equal to the (square) on $AD$ [Prop. 2.5].
Hence, the (rectangle contained) by $AC$ and $CB$ is less than the (square) on $AD$.
Thus, twice the (rectangle contained) by $AC$ and $CB$ is less than double the (square) on $AD$.
But, (the sum of) the (squares) on $AC$ and $CB$ [is] double (the sum of) the (squares) on $AD$ and $DC$ [Prop. 2.9].
Thus, (the sum of) the (squares) on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$.
(Which is) the very thing it was required to show.
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