(related to Proposition: Prop. 12.09: Tetrahedra are Equal iff Bases are Reciprocally Proportional to Heights)

- For let there be (two) equal pyramids having the triangular bases $ABC$ and $DEF$, and apexes the points $G$ and $H$ (respectively).
- I say that the bases of the pyramids $ABCG$ and $DEFH$ are reciprocally proportional to their heights, and (so) that as base $ABC$ is to base $DEF$, so the height of pyramid $DEFH$ (is) to the height of pyramid $ABCG$.

- For let the parallelepiped solids $BGML$ and $EHQP$ have been completed.
- And since pyramid $ABCG$ is equal to pyramid $DEFH$, and solid $BGML$ is six times pyramid $ABCG$ (see previous proposition), and solid $EHQP$ (is) six times pyramid $DEFH$, solid $BGML$ is thus equal to solid $EHQP$.
- And the bases of equal parallelepiped solids are reciprocally proportional to their heights [Prop. 11.34].
- Thus, as base $BM$ is to base $EQ$, so the height of solid $EHQP$ (is) to the height of solid $BGML$.
- But, as base $BM$ (is) to base $EQ$, so triangle $ABC$ (is) to triangle $DEF$ [Prop. 1.34].
- And, thus, as triangle $ABC$ (is) to triangle $DEF$, so the height of solid $EHQP$ (is) to the height of solid $BGML$ [Prop. 5.11].
- But, the height of solid $EHQP$ is the same as the height of pyramid $DEFH$, and the height of solid $BGML$ is the same as the height of pyramid $ABCG$.
- Thus, as base $ABC$ is to base $DEF$, so the height of pyramid $DEFH$ (is) to the height of pyramid $ABCG$.
- Thus, the bases of pyramids $ABCG$ and $DEFH$ are reciprocally proportional to their heights.
- And so, let the bases of pyramids $ABCG$ and $DEFH$ be reciprocally proportional to their heights, and (thus) let base $ABC$ be to base $DEF$, as the height of pyramid $DEFH$ (is) to the height of pyramid $ABCG$.
- I say that pyramid $ABCG$ is equal to pyramid $DEFH$.
- For, with the same construction, since as base $ABC$ is to base $DEF$, so the height of pyramid $DEFH$ (is) to the height of pyramid $ABCG$, but as base $ABC$ (is) to base $DEF$, so parallelogram $BM$ (is) to parallelogram $EQ$ [Prop. 1.34], thus as parallelogram $BM$ (is) to parallelogram $EQ$, so the height of pyramid $DEFH$ (is) also to the height of pyramid $ABCG$ [Prop. 5.11].
- But, the height of pyramid $DEFH$ is the same as the height of parallelepiped $EHQP$, and the height of pyramid $ABCG$ is the same as the height of parallelepiped $BGML$.
- Thus, as base $BM$ is to base $EQ$, so the height of parallelepiped $EHQP$ (is) to the height of parallelepiped $BGML$.
- And those parallelepiped solids whose bases are reciprocally proportional to their heights are equal [Prop. 11.34].
- Thus, the parallelepipedal solid $BGML$ is equal to the parallelepipedal solid $EHQP$.
- And pyramid $ABCG$ is a sixth part of $BGML$, and pyramid $DEFH$ a sixth part of parallelepiped $EHQP$.
- Thus, pyramid $ABCG$ is equal to pyramid $DEFH$.
- Thus, the bases of equal pyramids which also have triangular bases are reciprocally proportional to their heights.
- And those pyramids having triangular bases whose bases are reciprocally proportional to their heights are equal.
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"