VMO

In case $AB=AC$, it is obvious that $DB=DC$ and the bisector of $\angle MTN$ passes through the midpoint of the minor arc $BC$ of $O$ (which is the fixed point that is revealed in part b), hence we only need to consider the case where $AB\ne AC$.

a) Let $R,S$ be the second intersections of $I$ with $AB,AC$ respectively. Since $E,F$ lie on a circle with diameter $BC$, and quadrilateral $EFRS$ is cyclic, we conclude $BC\parallel RS$ by Reim's theorem.



By the assumption, $I$ touches $BC$ at $D$ then $$\frac{B{D}^2}{C{D}^2}=\frac{BF\cdot BR}{CE\cdot CS}=\frac{BF\cdot BR}{CE\cdot CS}=\frac{BF\cdot AB}{CE\cdot AC}=\frac{BF\cdot BE}{CE\cdot CF}=\frac{\cot B}{\cot C}$$ which means $\frac{BD}{CD}=\sqrt{\frac{\cot B}{\cot C}}$.

b) Denoted by $G$ the intersection of $EF$ and $BC$. We have The radical axis of $(BHC)$ and $I$ is $PQ$, The radical axis of $I$ and $(BFCE)$ is $EF$, The radical axis of $(BHC)$ and $(BFCE)$ is $BC$.



Hence $G$ is the radical center of these circles, which means $G$ lies on $PQ$. On the other hand, we have Let the radical axis of $O$ and $(TPQ)$ be $d$, The radical axis of $(TPQ)$ and $(BHC)$ is $PQ$, The radical axis of $(BHC)$ and $O$ is $BC$.

then $d$, $PQ$ and $BC$ concur at $G$ which means $GT$ is the radical axis of $O$ and $(TPQ)$, hence $GT$ touches $(TPQ)$ and $O$. Since $PQMN$ is a cyclic quadrilateral, $$\overline{GM}\cdot \overline{GN}=\overline{GP}\cdot \overline{GQ}=G{T}^2,$$ hence $GT$ touches $(TMN)$.

In conclusion, $(TMN)$ touches $O$ at $T$, we conclude that $TM,TN$ are isogonal with respect to $\angle BTC$ then the bisector of $\angle MTN$ is coincident with the bisector of $\angle BTC$, which passes through $J$, the midpoint of minor arc $BC$ of $O$. $\square$