Proof: By Euclid
(related to Proposition: Prop. 10.060: Square on Binomial Straight Line applied to Rational Straight Line)
- For let $DH$, equal to the (square) on $AC$, and $KL$, equal to the (square) on $BC$, have been applied to $DE$.
- Thus, the remaining twice the (rectangle contained) by $AC$ and $CB$ is equal to $MF$ [Prop. 2.4].
- Let $MG$ have been cut in half at $N$, and let $NO$ have been drawn parallel [to each of $ML$ and $GF$].
- $MO$ and $NF$ are thus each equal to once the (rectangle contained) by $ACB$.
- And since $AB$ is a binomial (straight line), having been divided into its (component) terms at $C$, $AC$ and $CB$ are thus rational (straight lines which are) commensurable in square only [Prop. 10.36].
- Thus, the (squares) on $AC$ and $CB$ are rational, and commensurable with one another.
- And hence the sum of the (squares) on $AC$ and $CB$ (is rational) [Prop. 10.15], and is equal to $DL$.
- Thus, $DL$ is rational.
- And it is applied to the rational (straight line) $DE$.
- $DM$ is thus rational, and commensurable in length with $DE$ [Prop. 10.20].
- Again, since $AC$ and $CB$ are rational (straight lines which are) commensurable in square only, twice the (rectangle contained) by $AC$ and $CB$ - that is to say, $MF$ - is thus medial [Prop. 10.21].
- And it is applied to the rational (straight line) $ML$.
- $MG$ is thus also rational, and incommensurable in length with $ML$ - that is to say, with $DE$ [Prop. 10.22].
- And $MD$ is also rational, and commensurable in length with $DE$.
- Thus, $DM$ is incommensurable in length with $MG$ [Prop. 10.13].
- And they are rational.
- $DM$ and $MG$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $DG$ is a binomial (straight line) [Prop. 10.36].
- So, we must show that (it is) also a first (binomial straight line).
- Since the (rectangle contained) by $ACB$ is in mean proportion3 to the squares on $AC$ and $CB$ [Prop. 10.53 lem.] , $MO$ is thus also in mean proportion3 to $DH$ and $KL$.
- Thus, as $DH$ is to $MO$, so $MO$ (is) to $KL$ - that is to say, as $DK$ (is) to $MN$, (so) $MN$ (is) to $MK$ [Prop. 6.1].
- Thus, the (rectangle contained) by $DK$ and $KM$ is equal to the (square) on $MN$ [Prop. 6.17].
- And since the (square) on $AC$ is commensurable with the (square) on on $CB$, $DH$ is also commensurable with $KL$.
- Hence, $DK$ is also commensurable with $KM$ [Prop. 6.1], [Prop. 10.11].
- And since (the sum of) the squares on $AC$ and $CB$ is greater than twice the (rectangle contained) by $AC$ and $CB$ [Prop. 10.59 lem.] , $DL$ (is) thus also greater than $MF$.
- Hence, $DM$ is also greater than $MG$ [Prop. 6.1], [Prop. 5.14].
- And the (rectangle contained) by $DK$ and $KM$ is equal to the (square) on $MN$ - that is to say, to one quarter the (square) on $MG$.
- And $DK$ (is) commensurable (in length) with $KM$.
- And if there are two unequal straight lines, and a (rectangle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) commensurable (in length) , then the square on the greater is larger than (the square on) the lesser by the (square) on (some straight line) commensurable (in length) with the greater [Prop. 10.17].
- Thus, the square on $DM$ is greater than (the square on) $MG$ by the (square) on (some straight line) commensurable (in length) with ($DM$).
- And $DM$ and $MG$ are rational.
- And $DM$, which is the greater term, is commensurable in length with the (previously) laid down rational (straight line) $DE$.
- Thus, $DG$ is a first binomial (straight line) [Def. 10.5] .
- (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"
