Proof: By Euclid

For let the equilateral pentagon $ABCDE$ have been inscribed in the circle $ABCDE$ which has a rational diameter.
I say that the side of pentagon [$ABCDE$] is that irrational (straight line) called minor.
For let the center of the circle, point $F$, have been found [Prop. 3.1].
And let $AF$ and $FB$ have been joined.
And let them have been drawn across to points $G$ and $H$ (respectively).
And let $AC$ have been joined.
And let $FK$ made (equal) to the fourth part of $AF$.
And $AF$ (is) rational.
$FK$ (is) thus also rational.
And $BF$ is also rational.
Thus, the whole of $BK$ is rational.
And since circumference $ACG$ is equal to circumference $ADG$, of which $ABC$ is equal to $AED$, the remainder $CG$ is thus equal to the remainder $GD$.
And if we join $AD$ then the angles at $L$ are inferred (to be) right angles, and $CD$ (is inferred to be) double $CL$ [Prop. 1.4].
So, for the same (reasons), the (angles) at $M$ are also right angles, and $AC$ (is) double $CM$.
Therefore, since angle $ALC$ (is) equal to $AMF$, and (angle) $LAC$ (is) common to the two triangles $ACL$ and $AMF$, the remaining (angle) $ACL$ is thus equal to the remaining (angle) $MFA$ [Prop. 1.32].
Thus, triangle $ACL$ is equiangular to triangle $AMF$.
Thus, proportionally, as $LC$ (is) to $CA$, so $MF$ (is) to $FA$ [Prop. 6.4].
And (we can take) the doubles of the leading (magnitudes).
Thus, as double $LC$ (is) to $CA$, so double $MF$ (is) to $FA$.
And as double $MF$ (is) to $FA$, so $MF$ (is) to half of $FA$.
And, thus, as double $LC$ (is) to $CA$, so $MF$ (is) to half of $FA$.
And (we can take) the halves of the following (magnitudes).
Thus, as double $LC$ (is) to half of $CA$, so $MF$ (is) to the fourth of $FA$.
And $DC$ is double $LC$, and $CM$ half of $CA$, and $FK$ the fourth part of $FA$.
Thus, as $DC$ is to $CM$, so $MF$ (is) to $FK$.
Via composition, as the sum of $DCM$ (i.e., $DC$ and $CM$) (is) to $CM$, so $MK$ (is) to $KF$ [Prop. 5.18].
And, thus, as the (square) on the sum of $DCM$ (is) to the (square) on $CM$, so the (square) on $MK$ (is) to the (square) on $KF$.
And since the greater piece of a (straight line) subtending two sides of a pentagon, such as $AC$, (which is) cut in extreme and mean ratio is equal to the side of the pentagon [Prop. 13.8] - that is to say, to $DC$ - -and the square on the greater piece added to half of the whole is five times the (square) on half of the whole [Prop. 13.1], and $CM$ (is) half of the whole, $AC$, thus the (square) on $DCM$, (taken) as one, is five times the (square) on $CM$.
And the (square) on $DCM$, (taken) as one, (is) to the (square) on $CM$, so the (square) on $MK$ was shown (to be) to the (square) on $KF$.
Thus, the (square) on $MK$ (is) five times the (square) on $KF$.
And the square on $KF$ (is) rational.
For the diameter (is) rational.
Thus, the (square) on $MK$ (is) also rational.
Thus, $MK$ is rational [in square only].
And since $BF$ is four times $FK$, $BK$ is thus five times $KF$.
Thus, the (square) on $BK$ (is) twenty-five times the (square) on $KF$.
And the (square) on $MK$ (is) five times the square on $KF$.
Thus, the (square) on $BK$ (is) five times the (square) on $KM$.
Thus, the (square) on $BK$ does not have to the (square) on $KM$ the ratio which a square number (has) to a square number.
Thus, $BK$ is incommensurable in length with $KM$ [Prop. 10.9].
And each of them is a rational (straight line).
Thus, $BK$ and $KM$ are rational (straight lines which are) commensurable in square only.
And if from a rational (straight line) a rational (straight line) is subtracted, which is commensurable in square only with the whole, then the remainder is that irrational (straight line called) an apotome [Prop. 10.73].
Thus, $MB$ is an apotome, and $MK$ its attachment.
So, I say that (it is) also a fourth (apotome) .
So, let the (square) on $N$ be (made) equal to that (magnitude) by which the (square) on $BK$ is greater than the (square) on $KM$.
Thus, the square on $BK$ is greater than the (square) on $KM$ by the (square) on $N$.
And since $KF$ is commensurable (in length) with $FB$ then, via composition, $KB$ is also commensurable (in length) with $FB$ [Prop. 10.15].
But, $BF$ is commensurable (in length) with $BH$.
Thus, $BK$ is also commensurable (in length) with $BH$ [Prop. 10.12].
And since the (square) on $BK$ is five times the (square) on $KM$, the (square) on $BK$ thus has to the (square) on $KM$ the ratio which 5 (has) to one.
Thus, via convertion, the (square) on $BK$ has to the (square) on $N$ the ratio which 5 (has) to 4 [Prop. 5.19 corr.] 2, which is not (that) of a square (number) to a square (number) .
$BK$ is thus incommensurable (in length) with $N$ [Prop. 10.9].
Thus, the square on $BK$ is greater than the (square) on $KM$ by the (square) on (some straight line which is) incommensurable (in length) with ($BK$).
Therefore, since the square on the whole, $BK$, is greater than the (square) on the attachment, $KM$, by the (square) on (some straight line which is) incommensurable (in length) with ($BK$), and the whole, $BK$, is commensurable (in length) with the (previously) laid down rational (straight line) $BH$, $MB$ is thus a fourth apotome [Def. 10.14] .
And the rectangle contained by a rational (straight line) and a fourth apotome is irrational, and its square root is that irrational (straight line) called minor [Prop. 10.94].
And the square on $AB$ is the rectangle contained by $HBM$, on account of joining $AH$, (so that) triangle $ABH$ becomes equiangular with triangle $ABM$ [Prop. 6.8], and (proportionally) as $HB$ is to $BA$, so $AB$ (is) to $BM$.
Thus, the side $AB$ of the pentagon is that irrational (straight line) called minor.
(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"

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)