# Proof: By Euclid  • And so let the (straight lines) standing up, $AG$, $HK$, $BE$, $LM$, $CO$, $PQ$, $DF$, and $RS$, not be at right angles to the bases $AB$ and $CD$.
• Again, I say that solid $AE$ (is) equal to solid $CF$.
• For let $KN$, $ET$, $GU$, $MV$, $QW$, $FX$, $OY$, and $SI$ have been drawn from points $K$, $E$, $G$, $M$, $Q$, $F$, $O$, and $S$ (respectively) perpendicular to the reference plane (i.e., the plane of the bases $AB$ and $CD$), and let them have met the plane atpoints $N$, $T$, $U$, $V$, $W$, $X$, $Y$, and $I$ (respectively).
• And let $NT$, $NU$, $UV$, $TV$, $WX$, $WY$, $YI$, and $IX$ have been joined.
• So solid $KV$ is equal to solid $QI$.
• For they are on the equal bases $KM$ and $QS$, and (have) the same height, and the (straight lines) standing up in them are at right angles to their bases (see first part of proposition).
• But, solid $KV$ is equal to solid $AE$, and $QI$ to $CF$.
• For they are on the same base, and (have) the same height, and the (straight lines) standing up in them are not on the same straight lines [Prop. 11.30].
• Thus, solid $AE$ is also equal to solid $CF$.
• Thus, parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another.
• (Which is) the very thing it was required to show.

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