Proof: By Euclid

Let $AB$ be a major (straight line) which has been divided at $C$, so that $AC$ and $CB$ are incommensurable in square, making the sum of the squares on $AC$ and $CB$ rational, and the (rectangle contained) by $AC$ and $CD$ medial [Prop. 10.39].
I say that $AB$ cannot be (so) divided at another point.
For, if possible, let it also have been divided at $D$, such that $AD$ and $DB$ are also incommensurable in square, making the sum of the (squares) on $AD$ and $DB$ rational, and the (rectangle contained) by them medial.
And since, by whatever (amount the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$, twice the (rectangle contained) by $AD$ and $DB$ also differs from twice the (rectangle contained) by $AC$ and $CB$ by this (same amount).
But, (the sum of) the (squares) on $AC$ and $CB$ exceeds (the sum of) the (squares) on $AD$ and $DB$ by a rational (area).
For (they are) both rational (areas).
Thus, twice the (rectangle contained) by $AD$ and $DB$ also exceeds twice the (rectangle contained) by $AC$ and $CB$ by a rational (area), (despite both) being medial (areas).
The very thing is impossible [Prop. 10.26].
Thus, a major (straight line) cannot be divided (into its component terms) at different points.
Thus, it can only be (so) divided at the same (point).
(Which is) the very thing it was required to show.
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