Proof: By Euclid

Let the straight line $AB$ be laid out, and let the whole (straight line) have been cut into unequal parts at each of the (points) $C$ and $D$.
And let $AC$ be assumed (to be) greater than $DB$.
I say that (the sum of) the (squares) on $AC$ and $CB$ is greater than (the sum of) the (squares) on $AD$ and $DB$.

fig041ae

For let $AB$ have been cut in half at $E$.
And since $AC$ is greater than $DB$, let $DC$ have been subtracted from both.
Thus, the remainder $AD$ is greater than the remainder $CB$.
And $AE$ (is) equal to $EB$.
Thus, $DE$ (is) less than $EC$.
Thus, points $C$ and $D$ are not equally far from the point of bisection.
And since the (rectangle contained) by $AC$ and $CB$, plus the (square) on $EC$, is equal to the (square) on $EB$ [Prop. 2.5], but, moreover, the (rectangle contained) by $AD$ and $DB$, plus the (square) on $DE$, is also equal to the (square) on $EB$ [Prop. 2.5], the (rectangle contained) by $AC$ and $CB$, plus the (square) on $EC$, is thus equal to the (rectangle contained) by $AD$ and $DB$, plus the (square) on $DE$.
And, of these, the (square) on $DE$ is less than the (square) on $EC$.
And, thus, the remaining (rectangle contained) by $AC$ and $CB$ is less than the (rectangle contained) by $AD$ and $DB$.
And, hence, twice the (rectangle contained) by $AC$ and $CB$ is less than twice the (rectangle contained) by $AD$ and $DB$.
And thus the remaining sum of the (squares) on $AC$ and $CB$ is greater than the sum of the (squares) on $AD$ and $DB$.¹
(Which is) the very thing it was required to show.

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Since, $AC^{2}+CB^{2}+2\,AC\,CB=AD^{2}+DB^{2}+2\,AD\,DB = AB^{\,2}$ (translator's note). ↩