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数学 / 几何原本 / Proposition XI.28

第11卷命题 28 · 平行六面体对角面平分

If a parallelepipedal solid be cut by a plane through the diagonals of the opposite planes, the solid will be bisected by the plane.

如果一个平行六面体被一个通过相对平面的对角线的平面所截,则该平面将平行六面体平分。

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第11卷命题 28 · 平行六面体对角面平分 · 3D 示意 本页以“平行六面体对角面平分”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. For let the parallelepipedal solid AB be cut by the plane CDEF through the diagonals CF, DE of opposite planes; I say that the solid AB will be bisected by the plane CDEF.

    设平行六面体AB被平面CDEF所截,该平面通过对面CF、DE的对角线。

  2. For, since the triangle CGF is equal to the triangle CFB, [I. 34] and ADE to DEH, while the parallelogram CA is also equal to the parallelogram EB, for they are opposite, and GE to CH, therefore the prism contained by the two triangles CGF, ADE and the three parallelograms GE, AC, CE is also equal to the prism contained by the two triangles CFB, DEH and the three parallelograms CH, BE, CE; for they are contained by planes equal both in multitude and in magnitude.

    由于三角形CGF等于三角形CFB,三角形ADE等于三角形DEH,且平行四边形CA等于其对边EB,GE等于CH。

  3. 因此,由两个三角形CGF、ADE和三个平行四边形GE、AC、CE所围成的棱柱,等于由两个三角形CFB、DEH和三个平行四边形CH、BE、CE所围成的棱柱。

  4. 因为它们由数量相等且大小相等的平面所围成,所以整个平行六面体被平面CDEF平分。