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.

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.*

Q.E.D.