Proof: By Euclid
(related to Proposition: Prop. 10.091: Side of Area Contained by Rational Straight Line and First Apotome)
 For let the area $AB$ have been contained by the rational (straight line) $AC$ and the first apotome $AD$.

I say that the square root of area $AB$ is an apotome.

For since $AD$ is a first apotome, let $DG$ be its attachment.
 Thus, $AG$ and $DG$ are rational (straight lines which are) commensurable in square only [Prop. 10.73].
 And the whole, $AG$, is commensurable (in length) with the (previously) laid down rational (straight line) $AC$, and the square on $AG$ is greater than (the square on) $GD$ by the (square) on (some straight line) commensurable in length with ($AG$) [Def. 10.11] .
 Thus, if (an area) equal to the fourth part of the (square) on $DG$ is applied to $AG$, falling short by a square figure, then it divides ($AG$) into (parts which are) commensurable (in length) [Prop. 10.17].
 Let $DG$ have been cut in half at $E$.
 And let (an area) equal to the (square) on $EG$ have been applied to $AG$, falling short by a square figure.
 And let it be the (rectangle contained) by $AF$ and $FG$.
 $AF$ is thus commensurable (in length) with $FG$.
 And let $EH$, $FI$, and $GK$ have been drawn through points $E$, $F$, and $G$ (respectively), parallel to $AC$.
 And since $AF$ is commensurable in length with $FG$, $AG$ is thus also commensurable in length with each of $AF$ and $FG$ [Prop. 10.15].
 But $AG$ is commensurable (in length) with $AC$.
 Thus, each of $AF$ and $FG$ is also commensurable in length with $AC$ [Prop. 10.12].
 And $AC$ is a rational (straight line).
 Thus, $AF$ and $FG$ (are) each also rational (straight lines).
 Hence, $AI$ and $FK$ are also each rational (areas) [Prop. 10.19].
 And since $DE$ is commensurable in length with $EG$, $DG$ is thus also commensurable in length with each of $DE$ and $EG$ [Prop. 10.15].
 And $DG$ (is) rational, and incommensurable in length with $AC$.
 $DE$ and $EG$ (are) thus each rational, and incommensurable in length with $AC$ [Prop. 10.13].
 Thus, $DH$ and $EK$ are each medial (areas) [Prop. 10.21].
 So let the square $LM$, equal to $AI$, be laid down.
 And let the square $NO$, equal to $FK$, have been subtracted (from $LM$), having with it the common angle $LPM$.
 Thus, the squares $LM$ and $NO$ are about the same diagonal [Prop. 6.26].
 Let $PR$ be their (common) diagonal, and let the (rest of the) figure have been drawn.
 Therefore, since the rectangle contained by $AF$ and $FG$ is equal to the square $EG$, thus as $AF$ is to $EG$, so $EG$ (is) to $FG$ [Prop. 6.17].
 But, as $AF$ (is) to $EG$, so $AI$ (is) to $EK$, and as $EG$ (is) to $FG$, so $EK$ is to $KF$ [Prop. 6.1].
 Thus, $EK$ is in mean proportion3 to $AI$ and $KF$ [Prop. 5.11].
 And $MN$ is also in mean proportion3 to $LM$ and $NO$, as shown before [Prop. 10.53 lem.] .
 And $AI$ is equal to the square $LM$, and $KF$ to $NO$.
 Thus, $MN$ is also equal to $EK$.
 But, $EK$ is equal to $DH$, and $MN$ to $LO$ [Prop. 1.43].
 Thus, $DK$ is equal to the gnomon $UVW$ and $NO$.
 And $AK$ is also equal to (the sum of) the squares $LM$ and $NO$.
 Thus, the remainder $AB$ is equal to $ST$.
 And $ST$ is the square on $LN$.
 Thus, the square on $LN$ is equal to $AB$.
 Thus, $LN$ is the square root of $AB$.
 So, I say that $LN$ is an apotome.
 For since $AI$ and $FK$ are each rational (areas), and are equal to $LM$ and $NO$ (respectively), thus $LM$ and $NO$  that is to say, the (squares) on each of $LP$ and $PN$ (respectively)  are also each rational (areas).
 Thus, $LP$ and $PN$ are also each rational (straight lines).
 Again, since $DH$ is a medial (area) , and is equal to $LO$, $LO$ is thus also a medial (area) .
 Therefore, since $LO$ is medial, and $NO$ rational, $LO$ is thus incommensurable with $NO$.
 And as $LO$ (is) to $NO$, so $LP$ is to $PN$ [Prop. 6.1].
 $LP$ is thus incommensurable in length with $PN$ [Prop. 10.11].
 And they are both rational (straight lines).
 Thus, $LP$ and $PN$ are rational (straight lines which are) commensurable in square only.
 Thus, $LN$ is an apotome [Prop. 10.73].
 And it is the square root of area $AB$.
 Thus, the square root of area $AB$ is an apotome.
 Thus, if an area is contained by a rational (straight line), and so on ....
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"