Example 5. (Hungarian Mathematical Olympiad Problem) Given the area of $\triangle ABC$ is $S$, and $\angle C=\alpha$. What kind of triangle will make the side $c$ opposite to $\angle C$ the shortest? (Figure 4)
According to the cosine theorem,
$$
\begin{array}{l} 
c^{2}=a^{2}+b^{2}-2 a b \cos C \\
=(a-b)^{2}+2 a b \\
\cdot(1-\cos a) .
\end{array}
$$

The area $S=\frac{1}{2} a b \sin \alpha$. Therefore,
$$
c^{2}=(a-b)^{2}+4 S \frac{1-\cos \alpha}{\sin \alpha} \text {. }
$$

The second term on the right is a constant. If $c$ is to be minimized, only $a=b$, thus, $c^{2}=4 S \frac{1-\cos \alpha}{\sin \alpha}$.

Therefore, when the triangle is isosceles, $\angle C$ is opposite the side $c$ which is the smallest, at this time $a=b=\sqrt{\frac{2 S}{\sin a}}$.