Example 22. Using the lower base of a frustum as the base, and the center of the upper base as the vertex, construct a cone. If the lateral surface of this cone divides the volume of the frustum into two equal parts, find the ratio of the radii of the upper and lower bases of the frustum.
Solve the external figure. Let the radii of the upper and lower bases of the frustum be $\mathrm{r}$ and $\mathrm{R}$, and the height be $\mathrm{h}$, then
$$
\begin{array}{l}
\frac{1}{3} \pi h\left(r^{2}+r R+R^{2}\right) \\
=2 \cdot \frac{1}{3} \pi \hbar^{2} h .
\end{array}
$$

Simplifying and rearranging, we get $R^{2}-r R-r^{2}=0$.
Solving it, $R=\frac{(1+\sqrt{5}) \mathrm{r}}{2}$ and $R=\frac{(1-\sqrt{5}) \mathrm{r}}{2}$ (not applicable, discard).
Therefore, $\frac{r}{R}=\frac{2}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2}$.