Proof: By Euclid

For let the straight line $BC$, which is incommensurable in square with $AB$, and fulfils the (other) prescribed (conditions), have been subtracted from the straight line $AB$ [Prop. 10.34].
I say that the remainder $AC$ is the aforementioned irrational (straight line).

fig073e

For since the sum of the squares on $AB$ and $BC$ is medial, and twice the (rectangle contained) by $AB$ and $BC$ rational, the (sum of the squares) on $AB$ and $BC$ is thus incommensurable with twice the (rectangle contained) by $AB$ and $BC$.
Thus, the remaining (square) on $AC$ is also incommensurable with twice the (rectangle contained) by $AB$ and $BC$ [Prop. 2.7], [Prop. 10.16].
And twice the (rectangle contained) by $AB$ and $BC$ is rational.
Thus, the (square) on $AC$ is irrational.
Thus, $AC$ is an irrational (straight line) [Def. 10.4] .
And let it be called that which makes with a rational (area) a medial whole.
(Which is) the very thing it was required to show.
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