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 AG, AF, LM, LN, CD, CE, BH, and BK, be on the same straight lines FN and DK.
I say that the solid CM equals the solid CN.
Since each of the figures CH and CK is a parallelogram, therefore CB equals each of the straight lines DH and EK. Therefore DH also equals EK.
Subtract EH from each, therefore the remainder DE equals the remainder HK. Therefore the triangle DCE also equals the triangle HBK, and the parallelogram DG equals the parallelogram HN. For the same reason the triangle AFG equals the triangle MLN.
But the parallelogram CF equals the parallelogram BM, and CG equals BN, for they are opposite, therefore the prism contained by the two triangles AFG and DCE and the three parallelograms AD, DG, and CG equals the prism contained by the two triangles MLN and HBK and the three parallelograms BM, HN, and BN.
Add to each the solid of which the parallelogram AB is the base and GEHM its opposite, therefore the whole parallelepipedal solid CM equals the whole parallelepipedal 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 on the same straight lines, equal one another.