# Proof: By Euclid

• Let there be pyramids of the same height whose bases (are) the triangles $ABC$ and $DEF$, and apexes the points $G$ and $H$ (respectively).
• I say that as base $ABC$ is to base $DEF$, so pyramid $ABCG$ (is) to pyramid $DEFH$.

• For if base $ABC$ is not to base $DEF$, as pyramid $ABCG$ (is) to pyramid $DEFH$, then base $ABC$ will be to base $DEF$, as pyramid $ABCG$ (is) to some solid either less than, or greater than, pyramid $DEFH$.
• Let it, first of all, be (in this ratio) to (some) lesser (solid), $W$.
• And let pyramid $DEFH$ have been divided into two pyramids equal to one another, and similar to the whole, and into two equal prisms.
• So, the (sum of the) two prisms is greater than half of the whole pyramid [Prop. 12.3].
• And, again, let the pyramids generated by the division have been similarly divided, and let this be done continually until some pyramids are left from pyramid $DEFH$ which (when added together) are less than the excess by which pyramid $DEFH$ exceeds the solid $W$ [Prop. 10.1].
• Let them have been left, and, for the sake of argument, let them be $DQRS$ and $STUH$.
• Thus, the (sum of the) remaining prisms within pyramid $DEFH$ is greater than solid $W$.
• Let pyramid $ABCG$ also have been divided similarly, and a similar number of times, as pyramid $DEFH$.
• Thus, as base $ABC$ is to base $DEF$, so the (sum of the) prisms within pyramid $ABCG$ (is) to the (sum of the) prisms within pyramid $DEFH$ [Prop. 12.4].
• But, also, as base $ABC$ (is) to base $DEF$, so pyramid $ABCG$ (is) to solid $W$.
• And, thus, as pyramid $ABCG$ (is) to solid $W$, so the (sum of the) prisms within pyramid $ABCG$ (is) to the (sum of the) prisms within pyramid $DEFH$ [Prop. 5.11].
• Thus, alternately, as pyramid $ABCG$ (is) to the (sum of the) prisms within it, so solid $W$ (is) to the (sum of the) prisms within pyramid $DEFH$ [Prop. 5.16].
• And pyramid $ABCG$ (is) greater than the (sum of the) prisms within it.
• Thus, solid $W$ (is) also greater than the (sum of the) prisms within pyramid $DEFH$ [Prop. 5.14].
• But, (it is) also less.
• This very thing is impossible.
• Thus, as base $ABC$ is to base $DEF$, so pyramid $ABCG$ (is) not to some solid less than pyramid $DEFH$.
• So, similarly, we can show that base $DEF$ is not to base $ABC$, as pyramid $DEFH$ (is) to some solid less than pyramid $ABCG$ either.
• So, I say that neither is base $ABC$ to base $DEF$, as pyramid $ABCG$ (is) to some solid greater than pyramid $DEFH$.
• For, if possible, let it be (in this ratio) to some greater (solid), $W$.
• Thus, inversely, as base $DEF$ (is) to base $ABC$, so solid $W$ (is) to pyramid $ABCG$ [Prop. 5.7 corr.] .
• And as solid $W$ (is) to pyramid $ABCG$, so pyramid $DEFH$ (is) to some (solid) less than pyramid $ABCG$, as shown before [Prop. 12.2 lem.] .
• And, thus, as base $DEF$ (is) to base $ABC$, so pyramid $DEFH$ (is) to some (solid) less than pyramid $ABCG$ [Prop. 5.11].
• The very thing was shown (to be) absurd.
• Thus, base $ABC$ is not to base $DEF$, as pyramid $ABCG$ (is) to some solid greater than pyramid $DEFH$.
• And, it was shown that neither (is it in this ratio) to a lesser (solid).
• Thus, as base $ABC$ is to base $DEF$, so pyramid $ABCG$ (is) to pyramid $DEFH$.
• (Which is) the very thing it was required to show.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@Fitzpatrick

### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"