Proof: By Euclid
(related to Proposition: Prop. 10.055: Root of Area contained by Rational Straight Line and Second Binomial)
- For since $AD$ is a second binomial (straight line), let it have been divided into its (component) terms at $E$, such that $AE$ is the greater term.
- Thus, $AE$ and $ED$ are rational (straight lines which are) commensurable in square only, and the square on $AE$ is greater than (the square on) $ED$ by the (square) on (some straight line) commensurable (in length) with ($AE$), and the lesser term $ED$ is commensurable in length with $AB$ [Def. 10.6] .
- Let $ED$ have been cut in half at $F$.
- And let the (rectangle contained) by $AGE$, equal to the (square) on $EF$, have been applied to $AE$, falling short by a square figure.
- $AG$ (is) thus commensurable in length with $GE$ [Prop. 10.17].
- And let $GH$, $EK$, and $FL$ have been drawn through (points) $G$, $E$, and $F$ (respectively), parallel to $AB$ and $CD$.
- And let the square $SN$, equal to the parallelogram $AH$, have been constructed, and the square $NQ$, equal to $GK$.
- And let $MN$ be laid down so as to be straight-on to $NO$.
- Thus, $RN$ [is] also straight-on to $NP$.
- And let the square $SQ$ have been completed.
- So, (it is) clear from what has been previously demonstrated [Prop. 10.53 lem.] that $MR$ is in mean proportion3 to $SN$ and $NQ$, and (is) equal to $EL$, and that $MO$ is the square root of the area $AC$.
- So, we must show that $MO$ is a first bimedial (straight line).
Since $AE$ is incommensurable in length with $ED$, and $ED$ (is) commensurable (in length) with $AB$, $AE$ (is) thus incommensurable (in length) with $AB$ [Prop. 10.13].
- And since $AG$ is commensurable (in length) with $EG$, $AE$ is also commensurable (in length) with each of $AG$ and $GE$ [Prop. 10.15].
- But, $AE$ is incommensurable in length with $AB$.
- Thus, $AG$ and $GE$ are also (both) incommensurable (in length) with $AB$ [Prop. 10.13].
- Thus, $BA$, $AG$, and ($BA$, and) $GE$ are (pairs of) rational (straight lines which are) commensurable in square only.
- And, hence, each of $AH$ and $GK$ is a medial (area) [Prop. 10.21].
- Hence, each of $SN$ and $NQ$ is also a medial (area) .
- Thus, $MN$ and $NO$ are medial (straight lines).
- And since $AG$ (is) commensurable in length with $GE$, $AH$ is also commensurable with $GK$ - that is to say, $SN$ with $NQ$ - that is to say, the (square) on $MN$ with the (square) on $NO$ [hence, $MN$ and $NO$ are [commensurable in square][bookofproofs$2082]] [[Prop. 6.1]]bookofproofs$1987, [Prop. 10.11].
- And since $AE$ is incommensurable in length with $ED$, but $AE$ is commensurable (in length) with $AG$, and $ED$ commensurable (in length) with $EF$, $AG$ (is) thus incommensurable (in length) with $EF$ [Prop. 10.13].
- Hence, $AH$ is also incommensurable with $EL$ - that is to say, $SN$ with $MR$ - that is to say, $PN$ with $NR$ - that is to say, $MN$ is incommensurable in length with $NO$ [Prop. 6.1], [Prop. 10.11].
- But $MN$ and $NO$ have also been shown to be medial (straight lines) which are commensurable in square.
- Thus, $MN$ and $NO$ are medial (straight lines which are) commensurable in square only.
- So, I say that they also contain a rational (area).
- For since $DE$ was assumed (to be) commensurable (in length) with each of $AB$ and $EF$, $EF$ (is) thus also commensurable with $EK$ [Prop. 10.12].
- And they (are) each rational.
- Thus, $EL$ - that is to say, $MR$ - (is) rational [Prop. 10.19].
- And $MR$ is the (rectangle contained) by $MNO$.
- And if two medial (straight lines), commensurable in square only, which contain a rational (area), are added together, then the whole is (that) irrational (straight line which is) called first bimedial [Prop. 10.37].
- Thus, $MO$ is a first bimedial (straight line).
- (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BY-SA 4.0! 
- Github:
-
- non-Github:
- @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
