Kitap 4
15
Therefore in the given circle an equilateral and equiangular hexagon has been inscribed. Q. E. F.
15
Porism. From this it is manifest that the side of the hexagon is equal to the radius of the circle.
15
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.
15
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.
PROPOSITION 16.
16
In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular.
16
Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular.
16
In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments.
16
Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD.
