PROPOSITION 15.
In a given circle to inscribe an equilateral and equiangular hexagon.
Let
ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle
ABCDEF.
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.
I say that the hexagon
ABCDEF is equilateral and equiangular.
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.
Therefore in the given circle an equilateral and equiangular hexagon has been inscribed. Q. E. F.
PORISM.
From this it is manifest that the side of the hexagon is equal to the radius of the circle.
And, in like manner as in the case of the pentagon, if through the points of division on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon in conformity with what was explained in the case of the pentagon.
And further by means similar to those explained in the case of the pentagon we can both inscribe a circle in a given hexagon and circumscribe one about it. Q. E. F.
