Let CM and CN be parallelepipedal solids on the same base AB and of the same height, and let the ends of their edges which stand up, namely AF, AG, LM, LN, CD, CE, BH, and BK, not be on the same straight lines.
I say that the solid CM equals the solid CN.
Produce NK and DH to meet one another at R, and produce FM and GE to P and Q. Join AO, LP, CQ, and BR.
Then the solid CM, of which the parallelogram ACBL is the base and FDHM its opposite, equals 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 ends of their edges which stand up, namely AF, AO, LM, LP, CD, CQ, BH, and BR, are on the same straight lines FP and DR.
But the solid CP, of which the parallelogram ACBL is the base and OQRP its opposite, equals 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 ends of their edges which stand up, namely AG, AO, CE, CQ, LN, LP, BK, and BR, are on the same straight lines GQ and NR.
Hence the solid CM also equals the solid CN.
Therefore, parallelepipedal solids which are on the same base and of the same height, and in which the ends of their edges which stand up are not on the same straight lines, equal one another.