Proof: By Euclid
(related to Lemma: Lem. 10.041: Side of Sum of Medial Areas is Irrational)
 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 $DB$.
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes