Generalization wrote:

Given a $\triangle {ABC}$ with non-collinear points $D,E,F$ on $BC,CA,AB$ respectively. $X,Y,Z$ are the second intersections of $\odot {DEF}=\omega$ with $BC,CA,AB$ respectively. Let $X'$ be the reflection of $X$ in $EF$ and let $X_2$ be the isogonal conjugate of $X'$ WRT $\triangle {AEF}$ . Analogously, we can construct $Y',Z',Y_2,Z_2$ . Prove that $X_2,Y_2,Z_2$ are collinear.

Solution:
For any two different points $P,Q$ and a line $l$ : Define $\ell_{PQ}$ to be the perpendicular bisector of segment $PQ$ , and define $P_l$ to be the reflection of $P$ in $l$ .
It is easy to see that $F$ is the center of $(YY'_{AB}Y')$ , hence $F\in \ell_{YY'_{AB}}$ , similarly, $E\in \ell_{ZZ'_{AC}}$ . Let $M,N$ be the second intersections of $\ell_{YY'_{AB}},\ell_{ZZ'_{AC}}$ with $\omega$ respectively. By directed angles since
$\measuredangle MFY+\measuredangle NEZ= \frac{1}{2}\measuredangle Y'_{AB}FY+ \frac{1}{2}\measuredangle Z'_{AC}EZ= \measuredangle ZFY'+\measuredangle Y'FD+\measuredangle YEZ'+\measuredangle Z'ED= \measuredangle ZFD+\measuredangle YED=\overarc {ZY}$ we get that $M=N=D'=\ell_{YY'_{AB}}\cap\ell_{Z'Z_{AC}} \in \omega$ . Similarly, $E'=\ell_{XX'_{AC}}\cap\ell_{YY'_{BC}}\in \omega$ and $F'=\ell_{XX'_{AB}}\cap \ell_{ZZ'_{BC}}\in\omega$ .
Now it is well know that $X_2$ is the center of $(XX'_{AB}X'_{AC})$ , hence $X_2=\ell_{XX'_{AB}}\cap\ell_{XX'_{AC}}$ , similarly $Y_2=\ell_{YY'_{AB}}\cap\ell_{YY'_{BC}}$ and $Z_2=\ell_{ZZ'_{BC}}\cap\ell_{ZZ'_{AC}}$ . Hence by Pascal on $D'EE'DFF'$ we get the desired result. $\blacksquare$