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数学 / 几何原本 / Proposition XI.29

第11卷命题 29 · 同底等高且侧棱共线之平行六面体相等

Parallelepipedal solids which are on the same base and of the same height, and in which the extremities of the sides which stand up are on the same straight lines, are equal to one another.

同底等高的平行六面体,若其侧棱的端点位于同一直线上,则它们彼此相等。

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第11卷命题 29 · 同底等高且侧棱共线之平行六面体相等 · 3D 示意 本页以“同底等高且侧棱共线之平行六面体相等”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。(已改为 3D 线框/截面示意,点位沿用原命题字母。)
分步证明Step-by-step proof
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  1. Let CM, CN be parallelepipedal solids on the same base AB and of the same height, and let the extremities of their sides which stand up, namely AG, AF, LM, LN, CD, CE, BH, BK, be on the same straight lines FN, DK; I say that the solid CM is equal to the solid CN. For, since each of the figures CH, CK is a parallelogram, CB is equal to each of the straight lines DH, EK, [I. 34] hence DH is also equal to EK.

    设CM、CN为同底AB且等高的平行六面体,其侧棱AG、AF、LM、LN、CD、CE、BH、BK的端点位于直线FN、DK上。

  2. Let EH be subtracted from each; therefore the remainder DE is equal to the remainder HK. Hence the triangle DCE is also equal to the triangle HBK, [I. 8, 4] and the parallelogram DG to the parallelogram HN.

    由于CH、CK均为平行四边形,故CB等于DH和EK,从而DH等于EK。减去公共部分EH,得DE等于HK。

  3. [I. 36] For the same reason the triangle AFG is also equal to the triangle MLN. But the parallelogram CF is equal to the parallelogram BM, and CG to BN, for they are opposite; therefore the prism contained by the two triangles AFG, DCE and the three parallelograms AD, DG, CG is equal to the prism contained by the two triangles MLN, HBK and the three parallelograms BM, HN, BN.

    因此三角形DCE等于三角形HBK,平行四边形DG等于平行四边形HN。同理,三角形AFG等于三角形MLN。

  4. Let there be added to each the solid of which the parallelogram AB is the base and GEHM its opposite; therefore the whole parallelepipedal solid CM is equal to the whole parallelepipedal solid CN.

    又平行四边形CF等于BM,CG等于BN(相对面相等),故由两三角形AFG、DCE及三平行四边形AD、DG、CG构成的棱柱等于由两三角形MLN、HBK及三平行四边形BM、HN、BN构成的棱柱。加上以AB为底、GEHM为对面的立体,则整个平行六面体CM等于CN。