Just wondered why no one posted the solution for geometry problem from a contest. Anyway, here it is. With usual notations, let $AB=c,CA=b,AC=b$. We have $(R-r)^2+\frac{c^2}{4}=R^2\Rightarrow (a+b-2c)((a-b)^2+ac+bc)=0\Rightarrow c=\frac{a+b}{2}$. Suppose that internal bisector of $\angle{BAC}$ and $\angle{ACB}$ intersect the opposite side at $F$ and $G$, respectively. Side bashing gives us $\frac{BE}{BG}=\frac{BD}{DC}$, so $DE\parallel CG$ and $\angle{CED}=\angle{ECG}=90^{\circ}$.

My sol: By carnot's formula the sum of distances from $O$ to $BC,AC$ is $2r,$ so $OI$ is perpendicular to $CI$. It is well known that $DE$, the trilinear polar of $I$ is perpendicular to $OI$ (proof: radical axis of the nine point circle and circumcircle of the excentral triangle ), so the required angle is $90^\circ$.