# Sizes of Pyramids of Same Height with Polygonal Bases are as Bases

## Theorem

In the words of Euclid:

(*The Elements*: Book $\text{XII}$: Proposition $6$)

## Proof

Let there be two pyramids of the same height whose bases are the polygons $ABCDE$ and $FGHKL$ and whose apices are $M$ and $N$.

It is to be demonstrated that the ratio of $ABCDE$ to $FGHKL$ equals the ratio of pyramid $ABCDEM$ to pyramid $FGHKLN$.

Let $AC, AD, FH, FK$ be joined.

We have that $ABCM$ and $ACDM$ are tetrahedra of the same height.

So by Proposition $5$ of Book $\text{XII} $: Sizes of Tetrahedra of Same Height are as Bases:

From Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded:

- $ABCD : \triangle ACD = ABCDM : ACDM$

But by Proposition $5$ of Book $\text{XII} $: Sizes of Tetrahedra of Same Height are as Bases:

Therefore from Proposition $22$ of Book $\text{V} $: Equality of Ratios Ex Aequali:

- $ABCD : \triangle ADE = ABCDM : ADEM$

Again from Proposition $18$ of Book $\text{V} $: Magnitudes Proportional Separated are Proportional Compounded:

- $ABCDE : \triangle ADE = ABCDEM : ADEM$

Similarly it can be shown that:

- $FGHKL : \triangle FGH = FGHKLN : FGHN$

We have that $ADEM$ and $FGHN$ are two tetrahedra of the same height.

Therefore from Proposition $5$ of Book $\text{XII} $: Sizes of Tetrahedra of Same Height are as Bases:

But we have:

- $\triangle ADE : ABCDE = ADEM : BCDEM$

Therefore by Proposition $22$ of Book $\text{V} $: Equality of Ratios Ex Aequali:

- $ABCDE : \triangle FGH = ABCDEM : FGHN$

We also have:

- $\triangle FGH : FGHKL = FGHN : FGHKLN$

Further by Proposition $22$ of Book $\text{V} $: Equality of Ratios Ex Aequali:

- $ABCDE : FGHKL = ABCDEM : FGHKLN$

$\blacksquare$

## Historical Note

This proof is Proposition $6$ of Book $\text{XII}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XII}$. Propositions