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.

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.*

Q.E.D.