VMO

a) Since $D,E$ are the tangency points of $(I)$ with $BC$, $CA$ respectively, we have $DE\perp CI$. Moreover, $FL\perp CI$ so $DE\parallel FL$. Similarly, we have $DF\parallel EK$.

Clearly, we have $DK=DL=EF$ or $D$ lies on the perpendicular bisector of $KL$. Furthermore, $I$ also lies on the perpendicular bisector of $KL$, therefore $D,I$ and $J$ are collinear.



b) Let $T$ be the midpoint of $BC$, $G$ be the intersection of $AI$ with $BC$ and $U$ be the reflection of $T$ over $G$. We will show that the perpendicular bisector of $PQ$ always passes through $U$. Let $X,Y$ be the intersections of $EF$ with $BI$ and $CI$, respectively. We have $$\begin{gathered} \angle XIC=180^{\circ }-\angle BIC=90^{\circ }-\frac{\angle BAC}{2} \\ =\angle AEI-\angle EFI=\angle AEF=\angle XEC. \end{gathered}$$ Hence, $C,I,E$ and $X$ are concyclic, which implies $\angle BXC=90^{\circ }$. Similarly, we obtain $\angle BYC=90^{\circ }$. Since $\angle BXC=\angle BYC=90^{\circ }$, the four points $B,C,X$ and $Y$ lie on a circle with center $T$ and diameter $BC$. It follows that $T$ lies on the perpendicular bisector of the segment $XY$.

It is easy to see that the pairs of points $M,E$ and $N,F$ are symmetric with respect to $I$, so $EF\parallel MN$ or $XY\parallel PQ$. From there, we have $△IEX=△IMP$, which implies that $X$ and $P$ are symmetric with respect to $I$. Similarly, we also have $Y$ and $Q$ are symmetric with respect to $I$. Hence two perpendicular bisectors of $XY$ and $PQ$ are symmetric with respect to $AI$. Thus, the perpendicular bisector of $PQ$ passes through $U$.

Since $B,C$ are fixed, then $T$ is fixed. Note that $\frac{GB}{GC}=\frac{AB}{AC}$ remains unchanged so $G$ is fixed. Hence $U$ is fixed. So the perpendicular bisector of $PQ$ always passes through the fixed point $U$. $\square$