Proof: By Euclid

Let the two rational (straight lines) $AB$ and $BC$, (which are) commensurable in square only, be laid out such that the square on the greater, $AB$, is larger than (the square on) the lesser, $BC$, by the (square) on (some straight line which is) incommensurable (in length) with ($AB$) [Prop. 10.30].
And let $BC$ have been cut in half at $D$.
And let a parallelogram equal to the (square) on either of $BD$ or $DC$, (and) falling short by a square figure, have been applied to $AB$ [Prop. 6.28], and let it be the (rectangle contained) by $AEB$.
And let the semicircle $AFB$ have been drawn on $AB$.
And let $EF$ have been drawn at right angles to $AB$.
And let $AF$ and $FB$ have been joined.¹

fig033e

And since $AB$ and $BC$ are [two] unequal straight lines, and the square on $AB$ is greater than (the square on) $BC$ by the (square) on (some straight line which is) incommensurable (in length) with ($AB$).
And a parallelogram, equal to one quarter of the (square) on $BC$ - that is to say, (equal) to the (square) on half of it - (and) falling short by a square figure, has been applied to $AB$, and makes the (rectangle contained) by $AEB$.
$AE$ is thus incommensurable (in length) with $EB$ [Prop. 10.18].
And as $AE$ is to $EB$, so the (rectangle contained) by $BA$ and $AE$ (is) to the (rectangle contained) by $AB$ and $BE$.
And the (rectangle contained) by $BA$ and $AE$ (is) equal to the (square) on $AF$, and the (rectangle contained) by $AB$ and $BE$ to the (square) on $BF$ [Prop. 10.32 lem.] .
The (square) on $AF$ is thus incommensurable with the (square) on on $FB$ [Prop. 10.11].
Thus, $AF$ and $FB$ are incommensurable in square.
And since $AB$ is rational, the (square) on $AB$ is also rational.
Hence, the sum of the (squares) on $AF$ and $FB$ is also rational [Prop. 1.47].
And, again, since the (rectangle contained) by $AE$ and $EB$ is equal to the (square) on $EF$, and the (rectangle contained) by $AE$ and $EB$ was assumed (to be) equal to the (square) on $BD$, $FE$ is thus equal to $BD$.
Thus, $BC$ is double $FE$.
And hence the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $EF$ [Prop. 10.6].
And the (rectangle contained) by $AB$ and $BC$ (is) medial [Prop. 10.21].
Thus, the (rectangle contained) by $AB$ and $EF$ (is) also medial [Prop. 10.23 corr.] .
And the (rectangle contained) by $AB$ and $EF$ (is) equal to the (rectangle contained) by $AF$ and $FB$ [Prop. 10.32 lem.] .
Thus, the (rectangle contained) by $AF$ and $FB$ (is) also medial.
And the sum of the squares on them was also shown (to be) rational.
Thus, the two straight lines, $AF$ and $FB$, (which are) incommensurable in square, have been found, making the sum of the squares on them rational, and the (rectangle contained) by them medial.
(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"

Footnotes

$AF$ and $FB$ have lengths $\sqrt{[1+k/(1+k^2)^{1/2}]/2}$ and $\sqrt{[1-k/(1+k^2)^{1/2}]/2}$ times that of $AB$, respectively, where $k$ is defined in the footnote to [Prop. 10.30] (translator's note) ↩

◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)

Footnotes