Proof: By Euclid
(related to Proposition: 4.15: Side of Hexagon Inscribed in a Circle Equals the Radius of that Circle)
 Let the diameter $AD$ of circle $ABCDEF$ have been drawn,^{1} and let the center $G$ of the circle have been found [Prop. 3.1].
 And let the circle $EGCH$ have been drawn, with center $D$, and radius $DG$.
 And $EG$ and $CG$ being joined, let them have been drawn across (the circle) to points $B$ and $F$ (respectively).
 And let $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$ have been joined.
 I say that the hexagon $ABCDEF$ is equilateral and equiangular.
 For since point $G$ is the center of circle $ABCDEF$, $GE$ is equal to $GD$.
 Again, since point $D$ is the center of circle $GCH$, $DE$ is equal to $DG$.
 But, $GE$ was shown (to be) equal to $GD$.
 Thus, $GE$ is also equal to $ED$.
 Thus, triangle $EGD$ is equilateral.
 Thus, its three angles $EGD$, $GDE$, and $DEG$ are also equal to one another, inasmuch as the angles at the base of isosceles triangles are equal to one another [Prop. 1.5].
 And the three angles of the triangle are equal to two right angles [Prop. 1.32].
 Thus, angle $EGD$ is one third of two right angles.
 So, similarly, $DGC$ can also be shown (to be) one third of two right angles.
 And since the straight line $CG$, standing on $EB$, makes adjacent angles $EGC$ and $CGB$ equal to two right angles [Prop. 1.13], the remaining angle $CGB$ is thus also one third of two right angles.
 Thus, angles $EGD$, $DGC$, and $CGB$ are equal to one another.
 And hence the (angles) opposite to them $BGA$, $AGF$, and $FGE$ are also equal [to $EGD$, $DGC$, and $CGB$ (respectively)] [[Prop. 1.15]]bookofproofs$782.
 Thus, the six angles $EGD$, $DGC$, $CGB$, $BGA$, $AGF$, and $FGE$ are equal to one another.
 And equal angles stand on equal circumferences [Prop. 3.26].
 Thus, the six circumferences $AB$, $BC$, $CD$, $DE$, $EF$, and $FA$ are equal to one another.
 And equal circumferences are subtended by equal straight lines [Prop. 3.29].
 Thus, the six straight lines ($AB$, $BC$, $CD$, $DE$, $EF$, and $FA$) are equal to one another.
 Thus, hexagon $ABCDEF$ is equilateral.
 So, I say that (it is) also equiangular.
 For since circumference $FA$ is equal to circumference $ED$, let circumference $ABCD$ have been added to both.
 Thus, the whole of $FABCD$ is equal to the whole of $EDCBA$.
 And angle $FED$ stands on circumference $FABCD$, and angle $AFE$ on circumference $EDCBA$.
 Thus, angle $AFE$ is equal to $DEF$ [Prop. 3.27].
 Similarly, it can also be shown that the remaining angles of hexagon $ABCDEF$ are individually equal to each of the angles $AFE$ and $FED$.
 Thus, hexagon $ABCDEF$ is equiangular.
 And it was also shown (to be) equilateral.
 And it has been inscribed in circle $ABCDE$.
 Thus, an equilateral and equiangular hexagon has been inscribed in the given circle.
 (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes