and draw the diameter AGD;
from the centre D, at the distance DG dscribe the circle EGCH;
join EG, CG, and produce them to the points B,F; and join AB, BC, CD, DB, BF, FA.
The hexagon ABCDBF shall be equilateral and equiangular.
For, because G is the centre of the circle ABCDBF, GE is equal to GD;
and because D is the centre of the circle EGCH, DB is equal to DG;
therefore GE is equal to DE,[Axiom 1.
and the triangle EGD is equilateral;
therefore the three angles EGD, GDE, DEG are equal to one another. [I. 5. Corollary.
But the three angles of a triangle are together equal to two right angles; [I. 32.
therefore the angle EGD is the third part of two right angles.
In the same manner it may be shewn, that the angle DGC is the third part of two right angles.
And because the straight line GC makes with the straight line EB the adjacent angles EGC, CGB together equal to two right angles, [I. 13.
therefore the remaining angle CGB is the third part of two right angles;
therefore the angles EGD, DGC, CGB are equal to one aother.
And to these are equal the vertical opposite angles BGA,AGF,FGB. [1.15.
therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.
