Proof: By Euclid
(related to Proposition: Prop. 10.081: Construction of Second Apotome of Medial is Unique)
- For, if possible, let $BD$ be (so) attached.
- Thus, $AD$ and $DB$ are also medial (straight lines which are) commensurable in square only, containing a medial (area) - (namely, that contained) by $AD$ and $DB$ [Prop. 10.75].
- 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 subtracted (from $EG$), producing $HM$ as breadth.
- The remainder $EL$ is thus equal to the (square) on $AB$ [Prop. 2.7].
- Hence, $AB$ is the square root of $EL$.
- So, again, let $EI$, equal to the (sum of the squares) on $AD$ and $DB$ have been applied to $EF$, producing $EN$ as breadth.
- And $EL$ is also equal to the square on $AB$.
- Thus, the remainder $HI$ is equal to twice the (rectangle contained) by $AD$ and $DB$ [Prop. 2.7].
- And since $AC$ and $CB$ are (both) medial (straight lines), the (sum of the squares) on $AC$ and $CB$ is also medial.
- And it is equal to $EG$.
- Thus, $EG$ is also medial [Prop. 10.15], [Prop. 10.23 corr.] .
- And it is applied to the rational (straight line) $EF$, producing $EM$ as breadth.
- Thus, $EM$ is rational, and incommensurable in length with $EF$ [Prop. 10.22].
- Again, since the (rectangle contained) by $AC$ and $CB$ is medial, twice the (rectangle contained) by $AC$ and $CB$ is also medial [Prop. 10.23 corr.] .
- And it is equal to $HG$.
- Thus, $HG$ is also medial.
- And it is applied to the rational (straight line) $EF$, producing $HM$ as breadth.
- Thus, $HM$ is also rational, and incommensurable in length with $EF$ [Prop. 10.22].
- And since $AC$ and $CB$ are commensurable in square only, $AC$ is thus incommensurable in length with $CB$.
- And as $AC$ (is) to $CB$, so the (square) on $AC$ is to the (rectangle contained) by $AC$ and $CB$ [Prop. 10.21 lem.] .
- Thus, the (square) on $AC$ is incommensurable with the (rectangle contained) by $AC$ and $CB$ [Prop. 10.11].
- But, the (sum of the squares) on $AC$ and $CB$ is commensurable with the (square) on on $AC$, and twice the (rectangle contained) by $AC$ and $CB$ is commensurable with the (rectangle contained) by $AC$ and $CB$ [Prop. 10.6].
- Thus, the (sum of the squares) on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $AC$ and $CB$ [Prop. 10.13].
- And $EG$ is equal to the (sum of the squares) on $AC$ and $CB$.
- And $GH$ is equal to twice the (rectangle contained) by $AC$ and $CB$.
- Thus, $EG$ is incommensurable with $HG$.
- And as $EG$ (is) to $HG$, so $EM$ is to $HM$ [Prop. 6.1].
- Thus, $EM$ is incommensurable in length with $MH$ [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], and $HM$ (is) attached to it.
- So, similarly, we can show that $HN$ (is) also (commensurable in square only with $EN$ and is) attached to ($EH$).
- Thus, different straight lines, which are commensurable in square only with the whole, are attached to an apotome.
- The very thing is impossible [Prop. 10.79].
- Thus, only one medial straight line, which is commensurable in square only with the whole, and contains a medial (area) with the whole, can be attached to a second 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"
