Proof: By Euclid
(related to Proposition: Prop. 10.025: Rationality of Rectangle Contained by Medial Straight Lines Commensurable in Square)
- For let the squares $AD$ and $BE$ have been described on (the straight lines) $AB$ and $BC$ (respectively).
- $AD$ and $BE$ are thus each medial.
- And let the rational (straight line) $FG$ be laid out.
- And let the rectangular parallelogram $GH$, equal to $AD$, have been applied to $FG$, producing $FH$ as breadth.
- And let the rectangular parallelogram $MK$, equal to $AC$, have been applied to $HM$, producing $HK$ as breadth.
- And, finally, let $NL$, equal to $BE$, have similarly been applied to $KN$, producing $KL$ as breadth.
- Thus, $FH$, $HK$, and $KL$ are in a straight line.
- Therefore, since $AD$ and $BE$ are each medial, and $AD$ is equal to $GH$, and $BE$ to $NL$, $GH$ and $NL$ (are) thus each also medial.
- And they are applied to the rational (straight line) $FG$.
- $FH$ and $KL$ are thus each rational, and incommensurable in length with $FG$ [Prop. 10.22].
- And since $AD$ is commensurable with $BE$, $GH$ is thus also commensurable with $NL$.
- And as $GH$ is to $NL$, so $FH$ (is) to $KL$ [Prop. 6.1].
- Thus, $FH$ is commensurable in length with $KL$ [Prop. 10.11].
- Thus, $FH$ and $KL$ are rational (straight lines which are) commensurable in length.
- Thus, the (rectangle contained) by $FH$ and $KL$ is rational [Prop. 10.19].
- And since $DB$ is equal to $BA$, and $OB$ to $BC$, thus as $DB$ is to $BC$, so $AB$ (is) to $BO$.
- But, as $DB$ (is) to $BC$, so $DA$ (is) to $AC$ [Prop. 6.1].
- And as $AB$ (is) to $BO$, so $AC$ (is) to $CO$ [Prop. 6.1].
- Thus, as $DA$ is to $AC$, so $AC$ (is) to $CO$.
- And $AD$ is equal to $GH$, and $AC$ to $MK$, and $CO$ to $NL$.
- Thus, as $GH$ is to $MK$, so $MK$ (is) to $NL$.
- Thus, also, as $FH$ is to $HK$, so $HK$ (is) to $KL$ [Prop. 6.1], [Prop. 5.11].
- Thus, the (rectangle contained) by $FH$ and $KL$ is equal to the (square) on $HK$ [Prop. 6.17].
- And the (rectangle contained) by $FH$ and $KL$ (is) rational.
- Thus, the (square) on $HK$ is also rational.
- Thus, $HK$ is rational.
- And if it is commensurable in length with $FG$ then $HN$ is rational [Prop. 10.19].
- And if it is incommensurable in length with $FG$ then $KH$ and $HM$ are rational (straight lines which are) commensurable in square only: thus, $HN$ is medial [Prop. 10.21].
- Thus, $HN$ is either rational or medial.
- And $HN$ (is) equal to $AC$.
- Thus, $AC$ is either rational or medial.
- Thus, the ... by medial straight lines (which are) commensurable in square only, and so on ....
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
