Problem (2012CGMOp5): As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$.
AlphaGeometry formalization: a b c = triangle; i = incenter a b c; d = foot i a b; e = foot i a c; o = circumcenter b c i ? eqangle o d d b c e e o
HAGeo formalization: A B C = triangle; I = incenter A B C; AB = line A B; D = foot I AB; AC = line A C; E = foot I AC; O = circumcenter B C I; Prove: equal_angle O D B C E O
Difficulty score: 1.4