Proof: By Euclid

For let the rational (area) $AC$ have been applied to the rational (straight line) $AB$, producing the (straight line) $BC$ as breadth.
I say that $BC$ is rational, and commensurable in length with $BA$.

fig020e

For let the square $AD$ have been described on $AB$.
$AD$ is thus rational [Def. 10.4] .
And $AC$ (is) also rational.
$DA$ is thus commensurable with $AC$.
And as $DA$ is to $AC$, so $DB$ (is) to $BC$ [Prop. 6.1].
Thus, $DB$ is also commensurable (in length) with $BC$ [Prop. 10.11].
And $DB$ (is) equal to $BA$.
Thus, $AB$ (is) also commensurable (in length) with $BC$.
And $AB$ is rational.
Thus, $BC$ is also rational, and commensurable in length with $AB$ [Def. 10.3] .
Thus, if a rational (area) is applied to a rational (straight line), and so on ....
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