Kitap 4
15
But equal angles stand on equal circumferences; [III. 26] therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.
15
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.
15
I say next that it is also equiangular.
15
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]
15
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.
15
But it was also proved equilateral; and it has been inscribed in the circle ABCDEF.
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.
