| To inscribe an equilateral and equiangular fifteen-angled figure in a given circle. | ||
| Let ABCD be the given circle.
It is required to inscribe in the circle ABCD a fifteen-angled figure which shall be both equilateral and equiangular. | ||
| Inscribe a side AC of an equilateral triangle and a side AB of an equilateral pentagon in in the circle ABCD. 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. | IV.2
IV.11 | |
| Bisect BC at E. Therefore each of the circumferences BE and EC is a fifteenth of the circle ABCD. | III.30 | |
| If therefore we join BE and EC and continually fit into the circle ABCD straight lines equal to them, a fifteen-angled figure which is both equilateral and equiangular will be inscribed in it. | IV.1 | |
| Q.E.F. | ||
And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteen-angled figure and circumscribe one about it.
Book IV Introduction - Proposition IV.15 - Book V Introduction.