Proof: By Euclid
(related to Proposition: Prop. 10.084: Construction of that which produces Medial Whole with Medial Area is Unique)
- For, if possible, let $BD$ be (so) attached.
- Hence, $AD$ and $DB$ are also (straight lines which are) incommensurable in square, making the squares on $AD$ and $DB$ (added) together medial, and twice the (rectangle contained) by $AD$ and $DB$ medial, and, moreover, the (sum of the squares) on $AD$ and $DB$ incommensurable with twice the (rectangle contained) by $AD$ and $DB$ [Prop. 10.78].
- And let the rational (straight line) $EF$ be laid down.
- And let $EG$, equal to the (sum of the squares) on $AC$ and $CB$, have been applied to $EF$, producing $EM$ as breadth.
- And let $HG$, equal to twice the (rectangle contained) by $AC$ and $CB$, have been applied to $EF$, producing $HM$ as breadth.
- Thus, the remaining (square) on $AB$ is equal to $EL$ [Prop. 2.7].
- Thus, $AB$ is the square root of $EL$.
- Again, let $EI$, equal to the (sum of the squares) on $AD$ and $DB$, have been applied to $EF$, producing $EN$ as breadth.
- And the (square) on $AB$ is also equal to $EL$.
- Thus, the remaining twice the (rectangle contained) by $AD$ and $DB$ [is] equal to $HI$ [Prop. 2.7].
- And since the sum of the (squares) on $AC$ and $CB$ is medial, and is equal to $EG$, $EG$ is thus also medial.
- And it is applied to the rational (straight line) $EF$, producing $EM$ as breadth.
- $EM$ is thus rational, and incommensurable in length with $EF$ [Prop. 10.22].
- Again, since twice the (rectangle contained) by $AC$ and $CB$ is medial, and is equal to $HG$, $HG$ is thus also medial.
- And it is applied to the rational (straight line) $EF$, producing $HM$ as breadth.
- $HM$ is thus rational, and incommensurable in length with $EF$ [Prop. 10.22].
- And since the (sum of the squares) on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $AC$ and $CB$, $EG$ is also incommensurable with $HG$.
- Thus, $EM$ is also incommensurable in length with $MH$ [Prop. 6.1], [Prop. 10.11].
- And they are both rational (straight lines).
- Thus, $EM$ and $MH$ are rational (straight lines which are) commensurable in square only.
- Thus, $EH$ is an apotome [Prop. 10.73], with $HM$ attached to it.
- So, similarly, we can show that $EH$ is again an apotome, with $HN$ attached to it.
- Thus, different rational (straight lines), which are commensurable in square only with the whole, are attached to an apotome.
- The very thing was shown (to be) impossible [Prop. 10.79].
- Thus, another straight line cannot be (so) attached to $AB$.
- Thus, only one straight line, which is incommensurable in square with the whole, and (together) with the whole makes the squares on them (added) together medial, and twice the (rectangle contained) by them medial, and, moreover, the (sum of the) squares on them incommensurable with the (rectangle contained) by them, can be attached to $AB$.
- (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"
