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数学 / 几何原本 / Proposition III.20

第3卷命题 20 · 圆心角是圆周角的两倍

elem.3.20

在一个圆中,当圆心角和圆周角以同一段弧为底时,圆心角是圆周角的两倍。

A B C D E F G
fig-1

本页以“圆心角是圆周角的两倍”整体图解辅助阅读;点、线、角、圆索引已按命题文字和证明步骤校订,可与证明和问答联动。

线

正文图形由校订坐标生成;点、线、角、圆可与证明和问答联动。

分步证明Step-by-step proof
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  1. Let ABC be a circle, let the angle BEC be an angle at its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double of the angle BAC. For let AE be joined and drawn through to F.

    设圆ABC,圆心角BEC和圆周角BAC以同一弧BC为底。连接AE并延长至F。

  2. Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5] therefore the angles EAB, EBA are double of the angle EAB. But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angle EAB.

    因为EA等于EB,所以角EAB等于角EBA,因此角EAB与角EBA之和是角EAB的两倍。

  3. For the same reason the angle FEC is also double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAC.

    又因为角BEF等于角EAB与角EBA之和,所以角BEF是角EAB的两倍。同理,角FEC是角EAC的两倍。

  4. Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G. Similarly then we can prove that the angle GEC is double of the angle EDC, of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC.

    因此整个角BEC是整个角BAC的两倍。对于另一圆周角BDC,类似可证角BEC是角BDC的两倍。