Solution:

Suppose the reflection of $AC$ across $BI$ intersects $BC$ at $E$. Define $F$ similarly for the reflection of $AB$ across $CI$. Also suppose $CI$ intersects $AB$ at $M$ and $BI$ intersects $AC$ at $N$. Since $CA$ and $CF = BC$ are reflections across $CI$, and so are $MA$ and $MF = XM$, we have that $A$ and $F$ are reflections across $CI$. Similarly $A$ and $E$ are reflections across $BI$.

Thus $\angle XFC = \angle BAC = \angle XEB$ if $\angle BAC$ is acute (and $\angle XFC = \angle XEB = \pi - \angle BAC$, when $\angle BAC$ is obtuse), so $XF = XE$. Moreover we also find that $IF = IA = IE$ by the aforementioned reflection properties, so thus $XI$ is the perpendicular bisector of $EF$ and is hence perpendicular to $BC$.