Proof: By Euclid

Let $AB$ be a first bimedial (straight line) which has been divided at $C$, such that $AC$ and $CB$ are medial (straight lines), commensurable in square only, (and) containing a rational (area) [Prop. 10.37].
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 medial (straight lines), commensurable in square only, (and) containing a rational (area).
Since, therefore, by whatever (amount) twice the (rectangle contained) by $AD$ and $DB$ differs from twice the (rectangle contained) by $AC$ and $CB$, (the sum of) the (squares) on $AC$ and $CB$ differs from (the sum of) the (squares) on $AD$ and $DB$ by this (same amount) [Prop. 10.41 lem.] .
And twice the (rectangle contained) by $AD$ and $DB$ differs from twice the (rectangle contained) by $AC$ and $CB$ by a rational (area).
For (they are) both rational (areas).
(The sum of) the (squares) on $AC$ and $CB$ thus differs from (the sum of) the (squares) on $AD$ and $DB$ by a rational (area, despite both) being medial (areas).
The very thing is absurd [Prop. 10.26].
Thus, a first bimedial (straight line) cannot be divided into its (component) terms at different points.
Thus, (it can be so divided) at one point only.
(Which is) the very thing it was required to show.
∎

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