Solution:

Consider the midpoint $M$ of arc $EF$ on the incircle of $\triangle ABC$. Angles $AFM$ and $MFE$ are equal since they intercept equal arcs $FM$ and $ME$, and so $M$ is on the bisector of $\angle AFE$. Similarly, $M$ is on the bisector of $\angle FEA$, and therefore $M$ coincides with $X$. Moreover, angles $FDX$ and $XDE$ are equal since they intercept equal arcs $FX$ and $XE$, and so $DX$ is the angle bisector of $\angle D$ in $\triangle DEF$. Similarly, $EY$ and $FZ$ are the other two angle bisectors in $\triangle DEF$. But the three angle bisectors in a triangle always meet!