Example 2 Given that the interior angles of quadrilateral $ABCD$ satisfy $A>B>C>D$, use $\alpha$ to denote the minimum of $180^{\circ}-A$, $A-B$, $B-C$, and $C-D$. Find the maximum value of $\alpha$.

保留源文本的换行和格式，直接输出翻译结果。
$$
\begin{array}{l}
\text { Solution: } \alpha=\frac{4 \alpha+3 \alpha+2 \alpha+\alpha}{4+3+2+1} \\
\leqslant \frac{4\left(180^{\circ}-A\right)+3(A-B)+2(B-C)+(C-D)}{10} \\
=\frac{4 \times 180^{\circ}-(A+B+C+D)}{10}=36^{\circ} .
\end{array}
$$

This shows that $\alpha$ will not exceed $36^{\circ}$ at most.
On the other hand, when
$$
180^{\circ}-A=A-B=B-C=C-D=36^{\circ}
$$

we have $A=144^{\circ}, B=108^{\circ}, C=72^{\circ}, D=36^{\circ}$, that is, when the four interior angles of the trapezoid $ABCD$ are $144^{\circ}, 108^{\circ}, 72^{\circ}, 36^{\circ}$ respectively, $\alpha$ can reach its maximum value of $36^{\circ}$ (Figure 1).

Furthermore, for a convex $n(n \geqslant 4)$-sided polygon, each interior angle is less than $180^{\circ}$, and their sum is a constant $(n-2) 180^{\circ}$. Therefore, the dependence on quadrilaterals in Example 2 is non-essential, and the generalization to polygons is immediately possible.