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 not on the same straight lines, are equal to one another.
同底等高的平行六面体，若其侧棱端点不在同一直线上，则它们彼此相等。
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 AF, AG, LM, LN, CD, CE, BH, BK, not be on the same straight lines; I say that the solid CM is equal to the solid CN. For let NK, DH be produced and meet one another at R, and further let FM, GE be produced to P, Q; let AO, LP, CQ, BR be joined.
延长NK与DH交于R，延长FM与GE至P、Q，连接AO、LP、CQ、BR。
Then the solid CM, of which the parallelogram ACBL is the base, and FDHM its opposite, is equal to the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite; for they are on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AF, AO, LM, LP, CD, CQ, BH, BR, are on the same straight lines FP, DR.
以ACBL为底、FDHM为对面的立体CM，等于以ACBL为底、OQRP为对面的立体CP，因它们同底等高且侧棱端点AF、AO、LM、LP、CD、CQ、BH、BR分别在直线FP、DR上（据XI.29）。
[XI. 29] But the solid CP, of which the parallelogram ACBL is the base, and OQRP its opposite, is equal to the solid CN, of which the parallelogram ACBL is the base and GEKN its opposite; for they are again on the same base ACBL and of the same height, and the extremities of their sides which stand up, namely AG, AO, CE, CQ, LN, LP, BK, BR, are on the same straight lines GQ, NR.
以ACBL为底、OQRP为对面的立体CP，等于以ACBL为底、GEKN为对面的立体CN，因它们同底等高且侧棱端点AG、AO、CE、CQ、LN、LP、BK、BR分别在直线GQ、NR上（据XI.29）。
Hence the solid CM is also equal to the solid CN.
因此，立体CM等于立体CN。