第4卷命题 14 · 作圆外切于等边等角五边形

Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE. Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E.

分别作角BCD和角CDE的平分线CF和DF，两线交于点F。

Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively. Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD.

连接FB、FA、FE，类似可证这些直线也平分其余各角。

[I. 6] Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed.

由角相等及平分关系得FC=FD，同理可证FA、FB、FE均等于FC、FD。

Let it be circumscribed, and let it be ABCDE.

以F为圆心、FA为半径作圆，必过其余各点，即为所求外接圆。