Proof: By Euclid
(related to Proposition: Prop. 10.092: Side of Area Contained by Rational Straight Line and Second Apotome)
 For let $DG$ be an attachment to $AD$.
 Thus, $AG$ and $GD$ are rational (straight lines which are) commensurable in square only [Prop. 10.73], and the attachment $DG$ is commensurable (in length) with the (previously) laid down rational (straight line) $AC$, and the square on the whole, $AG$, is greater than (the square on) the attachment, $GD$, by the (square) on (some straight line) commensurable in length with ($AG$) [Def. 10.12] .
 Therefore, since the square on $AG$ is greater than (the square on) $GD$ by the (square) on (some straight line) commensurable (in length) with ($AG$), thus if (an area) equal to the fourth part of the (square) on $GD$ is applied to $AG$, falling short by a square figure, then it divides ($AG$) into (parts which are) commensurable (in length) [Prop. 10.17].
 Therefore, 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 $FG$.
 Thus, $AF$ is commensurable in length with $FG$.
 $AG$ is thus also commensurable in length with each of $AF$ and $FG$ [Prop. 10.15].
 And $AG$ (is) a rational (straight line), and incommensurable in length with $AC$.
 $AF$ and $FG$ are thus also each rational (straight lines), and incommensurable in length with $AC$ [Prop. 10.13].
 Thus, $AI$ and $FK$ are each medial (areas) [Prop. 10.21].
 Again, since $DE$ is commensurable (in length) with $EG$, thus $DG$ is also commensurable (in length) with each of $DE$ and $EG$ [Prop. 10.15].
 But, $DG$ is commensurable in length with $AC$ [thus, $DE$ and $EG$ are also each rational, and commensurable in length with $AC$].
 Thus, $DH$ and $EK$ are each rational (areas) [Prop. 10.19].
 Therefore, let the square $LM$, equal to $AI$, have been constructed.
 And let $NO$, equal to $FK$, which is about the same angle $LPM$ as $LM$, have been subtracted (from $LM$).
 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 $AI$ and $FK$ are medial (areas), and are equal to the (squares) on $LP$ and $PN$ (respectively), [thus] the (squares) on $LP$ and $PN$ are also medial.
 Thus, $LP$ and $PN$ are also medial (straight lines which are) commensurable in square only.^{1}
 And since the (rectangle contained) by $AF$ and $FG$ is equal to the (square) on $EG$, thus as $AF$ is to $EG$, so $EG$ (is) to $FG$ [Prop. 10.17].
 But, as $AF$ (is) to $EG$, so $AI$ (is) to $EK$.
 And as $EG$ (is) to $FG$, so $EK$ [is] to $FK$ [Prop. 6.1].
 Thus, $EK$ is in mean proportion3 to $AI$ and $FK$ [Prop. 5.11].
 And $MN$ is also in mean proportion3 to the squares $LM$ and $NO$ [Prop. 10.53 lem.] .
 And $AI$ is equal to $LM$, and $FK$ to $NO$.
 Thus, $MN$ is also equal to $EK$.
 But, $DH$ [is] equal to $EK$, and $LO$ equal to $MN$ [Prop. 1.43].
 Thus, the whole (of) $DK$ is equal to the gnomon $UVW$ and $NO$.
 Therefore, since the whole (of) $AK$ is equal to $LM$ and $NO$, of which $DK$ is equal to the gnomon $UVW$ and $NO$, the remainder $AB$ is thus equal to $TS$.
 And $TS$ is the (square) on $LN$.
 Thus, the (square) on $LN$ is equal to the area $AB$.
 $LN$ is thus the square root of area $AB$.
 So, I say that $LN$ is the first apotome of a medial (straight line).
 For since $EK$ is a rational (area), and is equal to $LO$, $LO$  that is to say, the (rectangle contained) by $LP$ and $PN$  is thus a rational (area).
 And $NO$ was shown (to be) a medial (area) .
 Thus, $LO$ is incommensurable with $NO$.
 And as $LO$ (is) to $NO$, so $LP$ is to $PN$ [Prop. 6.1].
 Thus, $LP$ and $PN$ are incommensurable in length [Prop. 10.11].
 $LP$ and $PN$ are thus medial (straight lines which are) commensurable in square only, and which contain a rational (area).
 Thus, $LN$ is the first apotome of a medial (straight line) [Prop. 10.74].
 And it is the square root of area $AB$.
 Thus, the square root of area $AB$ is the first apotome of a medial (straight line).
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes