第4卷命题 11 · 圆内接正五边形作法

Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon. Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10] let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD. Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined.

作等腰三角形FGH，使其底角各为顶角的两倍，并在圆内作三角形ACD与之等角。

Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another. But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another.

作直线CE、DB分别平分角ACD和CDA，并连接AB、BC、DE、EA。

But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral. I say next that it is also equiangular. For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB.

由于角ACD和CDA各为角CAD的两倍且被平分，得五个角DAC、ACE、ECD、CDB、BDA相等，从而五段弧AB、BC、CD、DE、EA相等。

And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27] For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular.

等弧对等弦，故五边相等；又弧ABCD等于弧EDCB，得角BAE等于角AED，同理其余角相等，故五边形等角。