Example 1 In a convex quadrilateral $ABCD$, $\angle A - \angle B = \angle B - \angle C = \angle C - \angle D > 0$, and one of the four interior angles is $84^{\circ}$. Find the measures of the other angles.
Let $\angle A-\angle B=\angle B-\angle C=\angle C-\angle D=\alpha$. Then
$$
\begin{array}{l}
\angle C=\angle D+\alpha, \angle B=\angle D+2 \alpha, \\
\angle A=\angle D+3 \alpha . \\
\text { Also } \angle A+\angle B+\angle C+\angle D=360^{\circ} \text {, then } \\
\angle D+(\angle D+\alpha)+(\angle D+2 \alpha)+ \\
(\angle D+3 \alpha)=360^{\circ} .
\end{array}
$$

Also $\angle A+\angle B+\angle C+\angle D=360^{\circ}$, then $\angle D+(\angle D+\alpha)+(\angle D+2 \alpha)+$ $(\angle D+3 \alpha)=360^{\circ}$.
Thus $\angle D+\frac{3}{2} \alpha=90^{\circ}>84^{\circ}$.
From this, we know there are only two cases.
(1) $\angle D=84^{\circ}$, in this case, $\alpha=4^{\circ}$. Hence $\angle C=88^{\circ}, \angle B=92^{\circ}, \angle A=96^{\circ}$.
(2) $\angle C=84^{\circ}$, in this case, $\alpha=12^{\circ}$. Hence $\angle D=72^{\circ}, \angle B=96^{\circ}, \angle A=108^{\circ}$.