# Proof: By Euclid

• Let there be two pyramids with the same height, having the triangular bases $ABC$ and $DEF$, (with) apexes the points $G$ and $H$ (respectively).
• And let each of them have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms [Prop. 12.3].
• I say that as base $ABC$ is to base $DEF$, so (the sum of) all the prisms in pyramid $ABCG$ (is) to (the sum of) all the equal number of prisms in pyramid $DEFH$. • For since $BO$ is equal to $OC$, and $AL$ to $LC$, $LO$ is thus parallel to $AB$, and triangle $ABC$ similar to triangle $LOC$ [Prop. 12.3].
• So, for the same (reasons), triangle $DEF$ is also similar to triangle $RVF$.
• And since $BC$ is double $CO$, and $EF$ (double) $FV$, thus as $BC$ (is) to $CO$, so $EF$ (is) to $FV$.
• And the similar, and similarly laid out, rectilinear (figures) $ABC$ and $LOC$ have been described on $BC$ and $CO$ (respectively), and the similar, and similarly laid out, [rectilinear] (figures) $DEF$ and $RVF$ on $EF$ and $FV$ (respectively).
• Thus, as triangle $ABC$ is to triangle $LOC$, so triangle $DEF$ (is) to triangle $RVF$ [Prop. 6.22].
• Thus, alternately, as triangle $ABC$ is to [triangle] $DEF$, so [triangle] $LOC$ (is) to triangle $RVF$ [Prop. 5.16].
• But, as triangle $LOC$ (is) to triangle $RVF$, so the prism whose base [is] [triangle]bookofproofs$6432$LOC$, and opposite (plane)$PMN$, (is) to the prism whose base (is) triangle$RVF$, and opposite (plane)$STU$(see lemma). • And, thus, as triangle$ABC$(is) to triangle$DEF$, so the prism whose base (is) triangle$LOC$, and opposite (plane)$PMN$, (is) to the prism whose base (is) triangle$RVF$, and opposite (plane)$STU$. • And as the aforementioned prisms (are) to one another, so the prism whose base (is) parallelogram$KBOL$, and opposite (side) straight line$PM$, (is) to the prism whose base (is) parallelogram$QEVR$, and opposite (side) straight line$ST$[Prop. 11.39], [Prop. 12.3]. • Thus, also, (is) the (sum of the) two prisms - that whose base (is) parallelogram$KBOL$, and opposite (side)$PM$, and that whose base (is)$LOC$, and opposite (plane)$PMN$- to (the sum of) the (two) prisms - that whose base (is)$QEVR$, and opposite (side) straight line$ST$, and that whose base (is) triangle$RVF$, and opposite (plane)$STU$[Prop. 5.12]. • And, thus, as base$ABC$(is) to base$DEF$, so the (sum of the first) aforementioned two prisms (is) to the (sum of the second) aforementioned two prisms. And, similarly, if pyramids$PMNG$and$STUH$are divided into two prisms, and two pyramids, as base$PMN$(is) to base$STU$, so (the sum of) the two prisms in pyramid$PMNG$will be to (the sum of) the two prisms in pyramid$STUH$. • But, as base$PMN$(is) to base$STU$, so base$ABC$(is) to base$DEF$. • For the triangles$PMN$and$STU$(are) equal to$LOC$and$RVF$, respectively. • And, thus, as base$ABC$(is) to base$DEF$, so (the sum of) the four prisms (is) to (the sum of) the four prisms [Prop. 5.12]. • So, similarly, even if we divide the pyramids left behind into two pyramids and into two prisms, as base$ABC$(is) to base$DEF$, so (the sum of) all the prisms in pyramid$ABCG$will be to (the sum of) all the equal number of prisms in pyramid$DEFH\$.
• (Which is) the very thing it was required to show.

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