The bases of equal parallelepiped solids are reciprocally proportional to their heights. And those parallelepiped solids whose bases are reciprocally proportional to their heights are equal. * Let $AB$ and $CD$ be equal parallelepiped solids. * I say that the bases of the parallelepiped solids $AB$ and $CD$ are reciprocally proportional to their heights, and (so) as base $EH$ is to base $NQ$, so the height of solid $CD$ (is) to the height of solid $AB$.
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Proofs: 1
Proofs: 1
This proposition assumes that (a) if two parallelepipeds are equal, and have equal bases, then their heights are equal, and (b) if the bases of two equal parallelepipeds are unequal, then that solid which has the lesser base has the greater height (translator's note). ↩