10. In isosceles triangle $\mathrm{ABC}$ with base $\mathrm{AC}$, draw altitudes $\mathrm{AA}^{\prime}, \mathrm{BB}^{\prime}$, and $\mathrm{CC}^{\prime}$. If $\frac{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}{\mathrm{AB}}=\frac{1}{\sqrt{3}}$, find the ratio of the areas of $\triangle A^{\prime} B^{\prime} C^{\prime}$ and $\triangle A B C$.
10. $\triangle \mathrm{ABC} \cos \triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}$ (Fig. 10), and the cosine ratio is $\cos \mathrm{C}$ (refer to Question 9). Thus, $\cos \mathrm{C}=\frac{1}{\sqrt{3}}$.

Let $A B=B C=a, \angle A B C=\beta$. Therefore, $A^{\prime} B^{\prime}=\frac{a}{\sqrt{3}}$. Since $\triangle \mathrm{ABC} \sim \triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}$, we also get $\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}=\beta$. Since $\angle \mathrm{C}^{\prime} \mathrm{A}^{\prime} \mathrm{B}^{\prime}$ and $\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}$ are alternate interior angles of two parallel lines, $\therefore \angle \mathrm{C}^{\prime} \mathrm{A}^{\prime} \mathrm{B}^{\prime}=\beta$. Furthermore, we have
$$
\cos \beta=\cos (\pi-2 C)=1-2 \cos ^{2} C=\frac{1}{3} .
$$

Now, we need to find the areas of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}$:
$$
\begin{array}{l}
s_{A, C}=\frac{1}{2} a^{2} \sin \beta, \\
S_{A^{\prime}}^{\prime} c^{\prime}=A^{\prime} D \cdot B^{\prime} D=\frac{a^{2}}{3} \cos \beta \sin \beta .
\end{array}
$$

Thus, $\frac{S_{A}^{\prime} B_{A}^{\prime} e^{\prime}}{S_{\triangle a c}}=\frac{2}{3} \cos \beta=\frac{2}{9}$.