elem.4.16
在给定圆内作一个内接十五边形，使其等边且等角。
圆 ABCD 圆心 D，半径 R；AC 为内接等边三角形的边（中心角 120°），AB 为内接正五边形的边（中心角 72°），故弧 BC 占周长的 2/15；E 为弧 BC 中点，弧 BE 与 EC 各占 1/15。连接 BE、EC 等十五段即得正十五边形。
正文图形由校订坐标生成；点、线、角、圆可与证明和问答联动。
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. 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.
在圆ABCD内作等边三角形的一边AC和等边五边形的一边AB。
Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD. If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it.
圆ABCD被分成十五等份，弧ABC占三分之一圆，含五份；弧AB占五分之一圆，含三份；故余弧BC含两份。
Q. E.
将弧BC平分于E，则弧BE和EC各为圆的十五分之一。
F. 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 a fifteen-angled figure which is equilateral and equiangular.
连接BE、EC，并在圆内依次作等于它们的弦，即得等边等角的内接十五边形。