Solution:

The fact that $DX, EY, FZ$ are concurrent follows from the fact that triangles $DEF$ and $XYZ$ are homothetic; indeed, note that $EF$ and $YZ$ are both perpendicular to the internal angle bisector of $\angle BAC$.

Now, to see that the concurrence point lies on $IO$, note that point $I$ is the orthocenter of triangle $XYZ$, and $O$ is the nine-point center of triangle $XYZ$. Thus line $IO$ is the Euler line of triangle $XYZ$ and thus passes through the circumcenter $S$ of triangle $XYZ$. But $I$ is the circumcenter of triangle $DEF$, hence line $SI$ passes through the concurrency point.