Pyramids which are of the same height and have polygonal bases are to one another as the bases.
同高且底面为多边形的棱锥，其体积之比等于底面面积之比。
Let there be pyramids of the same height of which the polygons ABCDE, FGHKL are the bases and the points M, N the vertices; I say that, as the base ABCDE is to the base FGHKL, so is the pyramid ABCDEM to the pyramid FGHKLN. For let AC, AD, FH, FK be joined. Since then ABCM, ACDM are two pyramids which have triangular bases and equal height, they are to one another as the bases; [XII. 5] therefore, as the base ABC is to the base ACD, so is the pyramid ABCM to the pyramid ACDM.
连接AC、AD、FH、FK，将多边形底面分割成三角形。
And, componendo, as the base ABCD is to the base ACD, so is the pyramid ABCDM to the pyramid ACDM. [V. 18] But also, as the base ACD is to the base ADE, so is the pyramid ACDM to the pyramid ADEM. [XII. 5] Therefore, ex aequali, as the base ABCD is to the base ADE, so is the pyramid ABCDM to the pyramid ADEM.
由XII.5，同高三角形底棱锥体积比等于底面积比，得棱锥ABCM与ACDM之比等于三角形ABC与ACD之比。
[V. 22] And again componendo, as the base ABCDE is to the base ADE, so is the pyramid ABCDEM to the pyramid ADEM. [V. 18] Similarly also it can be proved that, as the base FGHKL is to the base FGH, so is the pyramid FGHKLN to the pyramid FGHN. And, since ADEM, FGHN are two pyramids which have triangular bases and equal height, therefore, as the base ADE is to the base FGH, so is the pyramid ADEM to the pyramid FGHN.
通过合比和等比推理，逐步推导出棱锥ABCDEM与ADEM之比等于多边形ABCDE与三角形ADE之比。
[XII. 5] But, as the base ADE is to the base ABCDE, so was the pyramid ADEM to the pyramid ABCDEM. Therefore also, ex aequali, as the base ABCDE is to the base FGH, so is the pyramid ABCDEM to the pyramid FGHN. [V. 22] But further, as the base FGH is to the base FGHKL, so also was the pyramid FGHN to the pyramid FGHKLN.
同理处理另一棱锥，并利用三角形底棱锥ADEM与FGHN之比等于三角形ADE与FGH之比，最终通过等比得出原命题。