# Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere/Lemma

## Lemma to Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere

In the words of Hypsicles of Alexandria:

If two straight lines be cut in extreme and mean ratio, the segments of both are in one and the same ratio.

## Proof

Let $AB$ be cut at $C$ in extreme and mean ratio where $AC$ is the greater segment.

Let $DE$ be cut at $F$ in extreme and mean ratio where $DF$ is the greater segment.

It is to be demonstrated that:

$AB : AC = DE : DF$

Thus:

 $\displaystyle AB \cdot BC$ $=$ $\displaystyle AC^2$ $\displaystyle DE \cdot EF$ $=$ $\displaystyle DF^2$ $\displaystyle \therefore \ \$ $\displaystyle AB \cdot BC : AC^2$ $=$ $\displaystyle DE \cdot EF : DF^2$ $\displaystyle \therefore \ \$ $\displaystyle 4 \cdot AB \cdot BC : AC^2$ $=$ $\displaystyle 4 \cdot DE \cdot EF : DF^2$ $\displaystyle \therefore \ \$ $\displaystyle \left({4 \cdot AB \cdot BC + AC^2}\right) : AC^2$ $=$ $\displaystyle \left({4 \cdot DE \cdot EF + DE^2}\right) : DF^2$ Proposition $18$ of Book $\text{V}$: Magnitudes Proportional Separated are Proportional Compounded $\displaystyle \therefore \ \$ $\displaystyle \left({AB + BC}\right)^2 : AC^2$ $=$ $\displaystyle \left({DE + EF}\right)^2 : DF^2$ Proposition $5$ of Book $\text{II}$: Difference of Two Squares $\displaystyle \therefore \ \$ $\displaystyle \left({AB + BC}\right) : AC$ $=$ $\displaystyle \left({DE + EF}\right) : DF$ $\displaystyle \therefore \ \$ $\displaystyle \left({AB + BC + AC}\right) : AC$ $=$ $\displaystyle \left({DE + EF + DF}\right) : DF$ Proposition $18$ of Book $\text{V}$: Magnitudes Proportional Separated are Proportional Compounded $\displaystyle \therefore \ \$ $\displaystyle 2 \cdot AB : AC$ $=$ $\displaystyle 2 \cdot DE : DF$ $\displaystyle \therefore \ \$ $\displaystyle AB : AC$ $=$ $\displaystyle DE : DF$

$\blacksquare$

## Historical Note

This theorem is Proposition $8$ of Book $\text{XIV}$ of Euclid's The Elements.