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.
Let BC be bisected at E; [III. 30]
therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD.