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数学 / 几何原本 / Proposition XI.24

第11卷命题 24 · 平行平面所夹立体对面相等且为平行四边形

If a solid be contained by parallel planes, the opposite planes in it are equal and parallelogrammic.

如果一个立体被平行平面所围成,则其相对的面相等且为平行四边形。

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分步证明Step-by-step proof
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  1. For let the solid CDHG be contained by the parallel planes AC, GF, AH, DF, BF, AE; I say that the opposite planes in it are equal and parallelogrammic. For, since the two parallel planes BG, CE are cut by the plane AC, their common sections are parallel. [XI. 16] Therefore AB is parallel to DC.

    设立体CDHG由平行平面AC、GF、AH、DF、BF、AE围成。

  2. Again, since the two parallel planes BF, AE are cut by the plane AC, their common sections are parallel. [XI. 16] Therefore BC is parallel to AD. But AB was also proved parallel to DC; therefore AC is a parallelogram.

    由于两平行平面BG、CE被平面AC所截,其交线平行,故AB平行于DC;同理,BC平行于AD,因此AC是平行四边形。

  3. Similarly we can prove that each of the planes DF, FG, GB, BF, AE is a parallelogram. Let AH, DF be joined. Then, since AB is parallel to DC, and BH to CF, the two straight lines AB, BH which meet one another are parallel to the two straight lines DC, CF which meet one another, not in the same plane; therefore they will contain equal angles; [XI. 10] therefore the angle ABH is equal to the angle DCF.

    类似可证DF、FG、GB、BF、AE均为平行四边形。连接AH、DF,由AB平行DC、BH平行CF,得角ABH等于角DCF;又AB等于DC、BH等于CF,故三角形ABH全等于三角形DCF。

  4. And, since the two sides AB, BH are equal to the two sides DC, CF, [I. 34] and the angle ABH is equal to the angle DCF, therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DCF. [I. 4] And the parallelogram BG is double of the triangle ABH, and the parallelogram CE double of the triangle DCF; [I. 34] therefore the parallelogram BG is equal to the parallelogram CE. Similarly we can prove that AC is also equal to GF, and AE to BF.

    平行四边形BG是三角形ABH的二倍,CE是三角形DCF的二倍,故BG等于CE;同理可证AC等于GF,AE等于BF。