Answer: $70^{\circ}$ and $160^{\circ}$
Arbitrarily defining $B$ and $C$ determines $I$ and $E$ up to reflections across $B C$. $D$ lies on both the circle determined by $\angle E D I=35^{\circ}$ and the line through $C$ tangent to the circle (and on the opposite side of $B)$; since the intersection of a line and a circle has at most two points, there are only two cases for $A B C D$. The diagram below on the left shows the construction made in this solution, containing both cases. The diagram below on the right shows only the degenerate case.

Reflect $B$ across $E C$ to $B^{\prime}$; then $C B=C B^{\prime}$. Since $B A$ and $B C$ are tangent to the circle centered at $I, I B$ is the angle bisector of $\angle A B C$. Then $\angle I B E=\angle I B^{\prime} E=35^{\circ}$. If $B^{\prime}=D$, then $\angle A D C=$ $\angle E B^{\prime} C=70^{\circ}$. Otherwise, since $\angle I B^{\prime} E=35^{\circ}=\angle I D E$ (given), $E B^{\prime} D I$ is a cyclic quadrilateral. Then $\angle I E D=\angle I B^{\prime} D=35^{\circ}$ and $\angle B C I=\angle E C D=30^{\circ}$, so $\triangle C E D \sim \triangle C B I$.
Since $\angle C I D$ is exterior to $\triangle D I E, \angle C I D=\angle I D E+\angle D E I=70^{\circ}$. Then $\triangle C D I \sim \triangle C E B$. Because $E B^{\prime} D I$ is cyclic, $\angle I D C=\angle I E B^{\prime}=\angle I E B=180^{\circ}-70^{\circ}-30^{\circ}=80^{\circ}$. Then $\angle A D C=2 \angle I D C=160^{\circ}$.
Thus, the two possible measures are $70^{\circ}$ and $160^{\circ}$.
9. $[7]$ Let $A B C D$ be a quadrilateral with an inscribed circle centered at $I$. Let $C I$ intersect $A B$ at $E$. If $\angle I D E=35^{\circ}, \angle A B C=70^{\circ}$, and $\angle B C D=60^{\circ}$, then what are all possible measures of $\angle C D A$ ?