Problem 808

Proof (1) As shown in Figure 4.
Since points $E, K$ are symmetric with respect to line $BI$, and points $F, D$ are also symmetric with respect to line $BI$, we have $DK = FE$.
Similarly, $DL = EF$.
Thus, $DK = DL$.
Since side $BC$ is tangent to $\odot I$ at point $D$, then $LK \parallel BC$.
Since $ID \perp BC$, we have $ID \perp LK$.
Therefore, $ID$ passes through the midpoint $J$ of chord $KL$ of $\odot I$, i.e., points $D$, $I$, and $J$ are collinear.
(2) Let lines $BI$, $CI$ intersect $EF$ at points $X$, $Y$ respectively.

By Property 1, points $B$, $C$, $X$, $Y$ are concyclic, denoted as $\odot O$. Then $O$ is the midpoint of $BC$. Hence, $O$ is a fixed point.
Let the perpendicular bisector of $XY$ be $l$. Then line $l$ passes through the fixed point $O$.
Let $AI$ intersect side $BC$ at point $T$.
By the Angle Bisector Theorem, we have
$$\frac{BT}{TC} = \frac{AB}{AC} = k.$$

Thus, $T$ is a fixed point.
Consider the point reflection transformation with $I$ as the center, the corresponding points of $E$, $F$ are $M$, $N$ respectively, and the corresponding lines of $BI$, $CI$ are still $BI$, $CI$. Then the corresponding points of $X$, $Y$ are $P$, $Q$ respectively.

Let the perpendicular bisector of segment $PQ$ be $l'$. Then the corresponding line of line $l$ is $l'$.

Since lines $AI$, $l$ are both perpendicular to $EF$, we have $AI \parallel l \parallel l'$, and line $AI$ is equidistant from lines $l$, $l'$.

Let line $l'$ intersect side $BC$ at point $O'$. Then $T$ is the midpoint of segment $OO'$. Thus, $O'$ is a fixed point, i.e., the perpendicular bisector $l'$ of segment $PQ$ always passes through the fixed point $O'$.