Proof: By Euclid

Let the parallelepiped solids $CM$ and $CN$ be on the same base, $AB$, and (have) the same height, and let the (ends of the straight lines) standing up in them, $AF$, $AG$, $LM$, $LN$, $CD$, $CE$, $BH$, and $BK$, not be on the same straight lines.
I say that the solid $CM$ is equal to the solid $CN$.

fig30e

For let $NK$ and $DH$ have been produced, and let them have joined one another at $R$.
And, further, let $FM$ and $GE$ have been produced to $P$ and $Q$ (respectively).
And let $AO$, $LP$, $CQ$, and $BR$ have been joined.
So, solid $CM$, whose base (is) parallelogram $ACBL$, and opposite (face) $FDHM$, is equal to solid $CP$, whose base (is) parallelogram $ACBL$, and opposite (face) $OQRP$.
For they are on the same base, $ACBL$, and (have) the same height, and the (ends of the straight lines) standing up in them, $AF$, $AO$, $LM$, $LP$, $CD$, $CQ$, $BH$, and $BR$, are on the same straight lines, $FP$ and $DR$ [Prop. 11.29].
But, solid $CP$, whose base is parallelogram $ACBL$, and opposite (face) $OQRP$, is equal to solid $CN$, whose base (is) parallelogram $ACBL$, and opposite (face) $GEKN$.
For, again, they are on the same base, $ACBL$, and (have) the same height, and the (ends of the straight lines) standing up in them, $AG$, $AO$, $CE$, $CQ$, $LN$, $LP$, $BK$, and $BR$, are on the same straight lines, $GQ$ and $NR$ [Prop. 11.29].
Hence, solid $CM$ is also equal to solid $CN$.
Thus, parallelepiped solids (which are) on the same base, and (have) the same height, and in which the (ends of the straight lines) standing up are not on the same straight lines, are equal to one another.
(Which is) the very thing it was required to show.
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