Proof: By Euclid
(related to Proposition: Prop. 10.054: Root of Area contained by Rational Straight Line and First Binomial)
 For let the area $AC$ be contained by the rational (straight line) $AB$ and by the first binomial (straight line) $AD$.

I say that square root of area $AC$ is the irrational (straight line which is) called binomial.

For since $AD$ is a first binomial (straight line), let it have been divided into its (component) terms at $E$, and let $AE$ be the greater term.
 So, (it is) clear that $AE$ and $ED$ are rational (straight lines which are) commensurable in square only, and that the square on $AE$ is greater than (the square on) $ED$ by the (square) on (some straight line) commensurable (in length) with ($AE$), and that $AE$ is commensurable (in length) with the rational (straight line) $AB$ (first) laid out [Def. 10.5] .
 So, let $ED$ have been cut in half at point $F$.
 And since the square on $AE$ is greater than (the square on) $ED$ by the (square) on (some straight line) commensurable (in length) with ($AE$), thus if a (rectangle) equal to the fourth part of the (square) on the lesser (term)  that is to say, the (square) on $EF$  falling short by a square figure, is applied to the greater (term) $AE$, then it divides it into (terms which are) commensurable (in length) [Prop. 10.17].
 Therefore, let the (rectangle contained) by $AG$ and $GE$, equal to the (square) on $EF$, have been applied to $AE$.
 $AG$ is thus commensurable in length with $EG$.
 And let $GH$, $EK$, and $FL$ have been drawn from (points) $G$, $E$, and $F$ (respectively), parallel to either of $AB$ or $CD$.
 And let the square $SN$, equal to the parallelogram $AH$, have been constructed, and (the square) $NQ$, equal to (the parallelogram) $GK$ [Prop. 2.14].
 And let $MN$ be laid down so as to be straighton to $NO$.
 $RN$ is thus also straighton to $NP$.
 And let the parallelogram $SQ$ have been completed.
 $SQ$ is thus a square [Prop. 10.53 lem.] .
 And since the (rectangle contained) by $AG$ and $GE$ is equal to the (square) on $EF$, thus as $AG$ is to $EF$, so $FE$ (is) to $EG$ [Prop. 6.17].
 And thus as $AH$ (is) to $EL$, (so) $EL$ (is) to $KG$ [Prop. 6.1].
 Thus, $EL$ is in mean proportion3 to $AH$ and $GK$.
 But, $AH$ is equal to $SN$, and $GK$ (is) equal to $NQ$.
 $EL$ is thus in mean proportion3 to $SN$ and $NQ$.
 And $MR$ is also in mean proportion3 to the same  (namely), $SN$ and $NQ$ [Prop. 10.53 lem.] .
 $EL$ is thus equal to $MR$.
 Hence, it is also equal to $PO$ [Prop. 1.43].
 And $AH$ plus $GK$ is equal to $SN$ plus $NQ$.
 Thus, the whole of $AC$ is equal to the whole of $SQ$  that is to say, to the square on $MO$.
 Thus, $MO$ (is) the square root of (area) $AC$.
 I say that $MO$ is a binomial (straight line).
 For since $AG$ is commensurable (in length) with $GE$, $AE$ is also commensurable (in length) with each of $AG$ and $GE$ [Prop. 10.15].
 And $AE$ was also assumed (to be) commensurable (in length) with $AB$.
 Thus, $AG$ and $GE$ are also commensurable (in length) with $AB$ [Prop. 10.12].
 And $AB$ is rational.
 $AG$ and $GE$ are thus each also rational.
 Thus, $AH$ and $GK$ are each rational (areas), and $AH$ is commensurable with $GK$ [Prop. 10.19].
 But, $AH$ is equal to $SN$, and $GK$ to $NQ$.
 $SN$ and $NQ$  that is to say, the (squares) on $MN$ and $NO$ (respectively)  are thus also rational and commensurable.
 And since $AE$ is incommensurable in length with $ED$, but $AE$ is commensurable (in length) with $AG$, and $DE$ (is) commensurable (in length) with $EF$, $AG$ (is) thus also incommensurable (in length) with $EF$ [Prop. 10.13].
 Hence, $AH$ is also incommensurable with $EL$ [Prop. 6.1], [Prop. 10.11].
 But, $AH$ is equal to $SN$, and $EL$ to $MR$.
 Thus, $SN$ is also incommensurable with $MR$.
 But, as $SN$ (is) to $MR$, (so) $PN$ (is) to $NR$ [Prop. 6.1].
 $PN$ is thus incommensurable (in length) with $NR$ [Prop. 10.11].
 And $PN$ (is) equal to $MN$, and $NR$ to $NO$.
 Thus, $MN$ is incommensurable (in length) with $NO$.
 And the (square) on $MN$ is commensurable with the (square) on on $NO$, and each (is) rational.
 $MN$ and $NO$ are thus rational (straight lines which are) commensurable in square only.
 Thus, $MO$ is (both) a binomial (straight line) [Prop. 10.36], and the square root of $AC$.
 (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"