The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$, and $AB$ at the points $D, E$, and $F$, respectively. Let the points $X, Y$, and $Z$ of the circle $\omega$ be diametrically opposite to the points $D, E$, and $F$, respectively. Line $AX, BY$ and $CZ$ intersect the sides $BC, CA$ and $AB$ at the points $D', E'$ and $F'$, respectively. On the segments $AD', BE'$ and $CF'$ noted the points $X', Y'$ and $Z'$, respectively, so that $D'X'= AX$, $E'Y' = BY$, $F'Z' = CZ$. Prove that the points $X', Y'$ and $Z'$ coincide.