1. In a convex quadrilateral $ABCD$,
$$
\angle A-\angle B=\angle B-\angle C=\angle C-\angle D>0 \text {. }
$$

If one of the interior angles is $84^{\circ}$, then the other three interior angles are $\qquad$
Let $\angle A-\angle B=\angle B-\angle C=\angle C-\angle D=x>0$, then $\angle C=\angle D+x, \angle B=\angle D+2 x$, $\angle A=\angle D+3 x$.
Also, $\angle A+\angle B+\angle C+\angle D=360^{\circ}$, so $4 \angle D+6 x=360^{\circ}$.

Therefore, $\angle D+\frac{3}{2} x=90^{\circ}>84^{\circ}$.
Clearly, $\angle D+\frac{3}{2} x<\angle B<\angle A$.
When $\angle D=84^{\circ}$, we get $x=4^{\circ}$.
Then $\angle C=88^{\circ}, \angle B=92^{\circ}, \angle A=96^{\circ}$.
When $\angle C=84^{\circ}$, we get $x=12^{\circ}$.
Then $\angle D=72^{\circ}, \angle B=96^{\circ}, \angle A=108^{\circ}$.