Kitap 11
30
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. [XI. 29]
30
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.
30
Hence the solid CM is also equal to the solid CN.
30
Therefore etc. Q. E. D.
PROPOSITION 31.
31
Parallelepipedal solids which are on equal bases and of the same height are equal to one another.
31
Let the parallelepipedal solids AE, CF, of the same height, be on equal bases AB, CD.
31
I say that the solid AE is equal to the solid CF.
31
First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, RS, be at right angles to the bases AB, CD; let the straight line RT be produced in a straight line with CR; on the straight line RT, and at the point R on it, let the angle TRU be constructed equal to the angle ALB, [I. 23] let RT be made equal to AL, and RU equal to LB, and let the base RW and the solid XU be completed.
