Problem: Let $ABC$ be a triangle with incenter $I$,
incircle $\gamma$ and circumcircle $\Gamma$.
Let $M$, $N$, $P$ be the midpoints of $\ol{BC}$, $\ol{CA}$, $\ol{AB}$
and let $E$, $F$ be the tangency points of $\gamma$
with $\ol{CA}$ and $\ol{AB}$, respectively.
Let $U$, $V$ be the intersections of line $EF$
with line $MN$ and line $MP$, respectively,
and let $X$ be the midpoint of arc $BAC$ of $\Gamma$.
\begin{enumerate}[(a)]
  \ii Prove that $I$ lies on ray $CV$.
  \ii Prove that line $XI$ bisects $\ol{UV}$.
\end{enumerate}

\end{enumerate}

Solution: The fact that $I = \ol{BU} \cap \ol{CV}$ and is Lemma 1.45 from EGMO.

As for (b), we note:
\begin{claim*}
  Line $IX$ is a symmedian of $\triangle IBC$.
\end{claim*}
\begin{proof}
  Recall that $(BIC)$ has circumcenter
  coinciding with the antipode of $X$ (by ``Fact 5'').
  So this follows from the fact
  that $\ol{XB}$ and $\ol{XC}$ are tangent.
\end{proof}
Since $BVUC$ is cyclic with diagonals intersecting at $I$,
and $IX$ is symmedian of $\triangle IBC$,
it is median of $\triangle IUV$, as needed.

\begin{remark*}
  [Alternate solution to (b) by Gunmay Handa]
  It's well known that $X$ is the midpoint of $\ol{I_b I_c}$
  (by considering the nine-point circle of the excentral triangle).
  However, $\ol{UV} \parallel \ol{I_b I_c}$
  and $I = \ol{I_b U} \cap \ol{I_c V}$, implying the result.
\end{remark*}