BdMO Online Forum

Let ABC be a triangle. The incircle of ABC with center I is tangent to AB at D & AC at E. Let O denote the circumcenter of triangle BCI. Prove that, angle ODB = angle OEC.

$\bullet$Claim 1 : $A ,I,O$ collinear .
proof : well-known fact.

$\bullet$Claim 2:$\triangle ADO \cong \triangle AEO$
proof:

(i)$AD=AE$
(ii)$AO=AO$
(iii)$\angle DAO = \angle EAO$
$\Longrightarrow \triangle ADO \cong \triangle AEO$ $\blacksquare$

$\star$ From $Claim 2$, we get,
$\angle ODA = \angle OEA \Longrightarrow \angle ODB = \angle OEC $ $\dagger$

It's a well known fact that $A,I,O$ are collinear. $AD=AE$ along with $\angle DAO = \angle EAO $ implies that triangle $DAO$ and $EAO$ are congruent. The rest is trivial.

As both of you didn't show the proof of $A$,$I$,$O$ being colinear, I'm just showing that. Let $AI$ intersect the circumcircle of $\triangle ABC$ at $X$. We know that $X$ is the midpoint of arc $BC$. So, $BX=CX$.
Now we will show that $BX=XI$.
We have, $\angle XBI = \angle XBC + \angle CBI = \angle (A/2) + \angle (B/2) = \angle BIX$. We get that $BX=XI=XC$. So, $X$ is the circumcenter of $\triangle BIC$.