For, since the point G is the centre of the circle ABCDEF,
GE is equal to GD.
Again, since the point D is the centre of the circle GCH,
DE is equal to DG.
But GE was proved equal to GD;
therefore GE is also equal to ED;
therefore the triangle EGD is equilateral;
and therefore its three angles EGD, GDE, DEG are equal to one another,
inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5]
And the three angles of the triangle are equal to two right angles; [I. 32]
therefore the angle EGD is one-third of two right angles.
Similarly, the angle DGC can also be proved to be onethird of two right angles.
And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles,
therefore the remaining angle CGB is also one-third of two right angles.
Therefore the angles EGD, DGC, CGB are equal to one another;
so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15]
Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.
But equal angles stand on equal circumferences; [III. 26]
therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.
And equal circumferences are subtended by equal straight lines; [III. 29]
therefore the six straight lines are equal to one another;
therefore the hexagon ABCDEF is equilateral.
I say next that it is also equiangular.
For, since the circumference FA is equal to the circumference ED,
let the circumference ABCD be added to each;
therefore the whole FABCD is equal to the whole EDCBA;
and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA;
therefore the angle AFE is equal to the angle DEF. [III. 27]
Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED;
therefore the hexagon ABCDEF is equiangular.
But it was also proved equilateral;
and it has been inscribed in the circle ABCDEF.
Complete text
Book I