# Proposition 31

Parallelepipedal solids which are on equal bases and of the same height equal one another.

Let the parallelepipedal solids AE and CF of the same height be on equal bases AB and CD.

I say that the solid AE equals the solid CF.

First, let the sides which stand up, HK, BE, AG, LM, PQ, DF, CO, and RS, be at right angles to the bases AB and CD. Produce the straight line RT in a straight line with CR. Construct the angle TRU equal to the angle ALB at the point R on the straight line RT. Make RT equal to A, and RU equal to LB. Complete the base RW and the solid XU.

Now, since the two sides TR and RU equal the two sides AL and LB, and they contain equal angles, therefore the parallelogram RW equals and is similar to the parallelogram HL. Since again AL equals RT, and LM equals RS, and they contain right angles, therefore the parallelogram RX equals and is similar to the parallelogram AM. For the same reason LE also equals and is similar to SU.

Therefore three parallelograms of the solid AE equal and are similar to three parallelograms of the solid XU. But the former three equal and are similar to the three opposite, and the latter three equal and are similar the three opposite, therefore the whole parallelepipedal solid AE equals the whole parallelepipedal solid XU.

I.31

Draw DR and WU through to meet one another at Y, draw aTb through T parallel to DY, produce PD to a, and complete the solids YX and RI.

XI.29

Then the solid XY, of which the parallelogram RX is the base and Yc its opposite, equals the solid XU, of which the parallelogram RX is the base and UV its opposite, for they are on the same base RX and of the same height, and the ends of their edges which stand up, namely RY, RU, Tb, TW, Se, Sd, Xc, and XV, are on the same straight lines YW and eV. But the solid XU equals AE, therefore the solid XY also equals the solid AE.

I.35

And, since the parallelogram RUWT equals the parallelogram YT, for they are on the same base RT and in the same parallels RT and YW, and RUWT equals CD, since it also equals AB, therefore the parallelogram YT also equals CD.

V.7

But DT is another parallelogram, therefore the base CD is to DT as YT is to DT.

XI.25

And, since the parallelepipedal solid CI is cut by the plane RF which is parallel to opposite planes, therefore the base CD is to the base DT as the solid CF is to the solid RI. For the same reason, since the parallelepipedal solid YI is cut by the plane RX which is parallel to opposite planes, therefore the base YT is to the base TD as the solid YX is to the solid RI.

V.11

But the base CD is to DT as YT is to DT, therefore the solid CF is to the solid RI as the solid YX is to RI.

V.9

Therefore each of the solids CF and YX has to RI the same ratio. Therefore the solid CF equals the solid YX. But YX was proved equal to AE, therefore AE also equals CF.

Next, let the sides standing up, AG, HK, BE, LM, CN, PQ, DF, and RS, not be at right angles to the bases AB and CD.

I say again that the solid AE equals the solid CF.

XI.11

Draw KO, ET, GU, MV, QW, FX, NY and SI from the points K, E, G, M, Q, F, N, and S perpendicular to the plane of reference, and let them meet the plane at the points O, T, U, V, W, X, Y, and I.

Above

Then the solid KV equals the solid QI, for they are on the equal bases KM and QS and of the same height, and their sides which stand up are at right angles to their bases.

XI.30

But the solid KV equals the solid AE, and QI equals CF, for they are on the same base and of the same height, while the ends of their edges which stand up are not on the same straight lines.

Therefore the solid AE also equals the solid CF.

Therefore, parallelepipedal solids which are on equal bases and of the same height equal one another.

Q.E.D.

## Guide

The statement that started with XI.29 has now been generalized two steps. In the next proposition the heights of the two parallelepipeds remain equal, but the bases vary. The present proposition is used not only in the proof of the next, but also in three more of the remaining propositions of Book XI.