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