Let $I$ be the incenter of triangle $ABC$.Prove that the circumcenter of triangle $BIC$ lies just in the middle of arc $BC$ of circumcircle of $ABC$.

Question

Let $I$ be the incenter of triangle $ABC$.Prove that the circumcenter of triangle $BIC$ lies just in the middle of arc $BC$ of circumcircle of $ABC$.

Since $MB=MC$ so $M$ lies on orthogonal bisector of $BC$.Now it remains to prove $MO=BO$ or $MO=CO$ (O is circumcenter of $ABC$) but how?

Answer 1

Suppose the ray $AI$ intersects the circumcircle at $M′\not=A$. It's obvious that $M′B=M′C$ since $\angle M'AB=\angle M'AC $. Now we prove $M'I=MB$, so that $M'I=M'B=M'C$ and $M'$ is the circumcenter of $\triangle ABC$.
\begin{align} \angle IBM'&=\color{red}{\angle IBC}+\color{blue}{\angle CBM'}\\ &=\color{red}{\frac 1 2\angle B}+\color{blue}{\angle CAM'}\\ &=\frac{\color{red}{\angle B}+\color{blue}{\angle A}}2\\ &=\frac 1 2(180°-\angle C)\\&=90°-\frac 1 2\angle C \end{align} \begin{align} \angle IM'B=\angle AM'B=\angle ACB=\angle C \end{align} So $\angle M'IB=180°-\angle IBM'-\angle IM'B=90°-\frac{\angle C}2=\angle IBM'$, hence $\triangle M'IB$ is isosceles and $M'I=M'B$. Done.

Answer 2

Clearly $A,I$ and $M$ are collinear, because $$\angle BAI =\frac{1}{2} \angle BAC= \angle BAM$$ Then $\angle BIM$ is a external angle in the triangle $\triangle BAI$, and his value is $\frac{B}{2}+\frac{A}{2}$.

By other hand $\angle IBM$ is the sum of $\angle IBC$ and $\angle CBM$ then his value is also $\frac{B}{2}+\frac{A}{2}$.

Finally $\triangle BIM $ is isosceles in $M$. Analogy $\triangle CIM$ is isosceles in $M$, and we conclude that $MB=MI=MC$.