Proof: By Euclid

Let $AB$ be the square root of a rational plus a medial (area) which has been divided at $C$, so that $AC$ and $CB$ are incommensurable in square, making the sum of the (squares) on $AC$ and $CB$ medial, and twice the (rectangle contained) by $AC$ and $CB$ rational [Prop. 10.40].
I say that $AB$ cannot be (so) divided at another point.
For, if possible, let it also have been divided at $D$, so that $AD$ and $DB$ are also incommensurable in square, making the sum of the (squares) on $AD$ and $DB$ medial, and twice the (rectangle contained) by $AD$ and $DB$ rational.
Therefore, since by whatever (amount) twice the (rectangle contained) by $AC$ and $CB$ differs from twice the (rectangle contained) by $AD$ and $DB$, (the sum of) the (squares) on $AD$ and $DB$ also differs from (the sum of) the (squares) on $AC$ and $CB$ by this (same amount).
And twice the (rectangle contained) by $AC$ and $CB$ exceeds twice the (rectangle contained) by $AD$ and $DB$ by a rational (area).
(The sum of) the (squares) on $AD$ and $DB$ thus also exceeds (the sum of) the (squares) on $AC$ and $CB$ by a rational (area), (despite both) being medial (areas).
The very thing is impossible [Prop. 10.26].
Thus, the square root of a rational plus a medial (area) 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.
∎

Thank you to the contributors under CC BY-SA 4.0!