Book 4
14
Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed.
14
Let it be circumscribed, and let it be ABCDE.
14
Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed. Q. E. F.
PROPOSITION 15.
15
In a given circle to inscribe an equilateral and equiangular hexagon.
15
Let ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF.
15
Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described; let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined.
15
I say that the hexagon ABCDEF is equilateral and equiangular.
15
For, since the point G is the centre of the circle ABCDEF, GE is equal to GD.
15
Again, since the point D is the centre of the circle GCH, DE is equal to DG.
15
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]
15
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.
15
Similarly, the angle DGC can also be proved to be onethird of two right angles.
15
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.
15
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]
15
Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another.
