In a given circle to inscribe an equilateral and equiangular hexagon.
Proof
Let G be the centre and AD a diameter of the given circle. With
centre D and radius DG describe a circle meeting the given circle
at C and E (Postulate 3). Join GC, GE. The triangle GDC
has GD=DC=CG (radii of equal circles), so it is equilateral,
and ∠DGC=60∘ (I.32). Stepping this 60∘ chord
around the circle six times produces the regular hexagon.