We start with the following useful lemma:
Lemma. Let $ABC$ be a triangle and $I_A$, $I_B$ and $I_C$ be its excenters. Then the midpoints of the sides of the triangle $I_A I_B I_C$ lie on the circumcircle of $ABC$.
Proof. First notice that $AI_A$ and $I_B I_C$ are the internal and external bisectors of $\angle BAC$, respectively, so $AI_A$ is the altitude relative to $I_B I_C$ in triangle $I_A I_B I_C$. So the circle that goes through the feet of the altitudes $A$, $B$ and $C$ is the nine point circle of $I_A I_B I_C$, and thus the midpoints of the sides of that triangle lie on it.

Since $AE = AF$ and $IE = IF$, $AI$ is the perpendicular bisector of $EF$, so $AI \perp EF$. We also have $AI \perp I_B I_C$. So $EF \parallel I_B I_C$. Since $YZ \parallel I_B I_C$ as well, we conclude that $EF \parallel YZ$. Analogously, $DE \parallel XY$ and $DF \parallel XZ$, so triangles $DEF$ and $XYZ$ are homothetic. Let $\phi$ be the homothety that maps $XYZ$ to $DEF$. Since $X$, $Y$, and $Z$ lie on the circumcircle of $ABC$ (by the lemma) and $D$, $E$ and $F$ lie on the incircle of $ABC$, $\phi$ maps the circumcircle to the incircle, and thus $O$ to $I$. Hence $DX$, $EY$, $FZ$ and $OI$ pass through the center of the homothety $\phi$ and we are done.