Harvard-MIT Mathematics Tournament

Solution:
Arbitrarily defining $B$ and $C$ determines $I$ and $E$ up to reflections across $BC$. $D$ lies on both the circle determined by $\angle EDI = 35^{\circ}$ 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 $ABCD$. 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 $EC$ to $B'$, then $CB = CB'$. Since $BA$ and $BC$ are tangent to the circle centered at $I$, $IB$ is the angle bisector of $\angle ABC$. Then $\angle IBE = \angle IB'E = 35^{\circ}$. If $B' = D$, then $\angle ADC = \angle EB'C = 70^{\circ}$. Otherwise, since $\angle IB'E = 35^{\circ} = \angle IDE$ (given), $EB'DI$ is a cyclic quadrilateral. Then $\angle IED = \angle IB'D = 35^{\circ}$ and $\angle BCI = \angle ECD = 30^{\circ}$, so $\triangle CED \sim \triangle CBI$.
Since $\angle CID$ is exterior to $\triangle DIE$, $\angle CID = \angle IDE + \angle DEI = 70^{\circ}$. Then $\triangle CDI \sim \triangle CEB$. Because $EB'DI$ is cyclic, $\angle IDC = \angle IEB' = \angle IEB = 180^{\circ} - 70^{\circ} - 30^{\circ} = 80^{\circ}$. Then $\angle ADC = 2 \angle IDC = 160^{\circ}$.

Thus, the two possible measures are $70^{\circ}$ and $160^{\circ}$.