We use sine bash. Note that \begin{align*} \frac{BX}{XC}=\frac{BA\cdot \sin{\angle FAG}}{AC\cdot \sin{\angle GAE}}=\frac{BA\cdot FG}{AC\cdot GE}=\frac{BA\cdot \sin{\angle DPF}}{AC\cdot \sin{\angle DPE}} =\frac{BA\cdot \sin{\angle PFD}}{AC\cdot \sin{\angle PED}}=\frac{BA\cdot \sin{\frac{\angle A-\angle B}{2}}}{AC\cdot \sin{\frac{\angle A-\angle C}{2}}}. \end{align*}Similarly, we get other ratios. Thus, we have \begin{align*} \frac{BX}{XC}\cdot \frac{CY}{YA}\cdot \frac{AZ}{ZB}=\frac{BA\cdot \sin{\frac{\angle A-\angle B}{2}}}{AC\cdot \sin{\frac{\angle A-\angle C}{2}}}\cdot \frac{CB\cdot \sin{\frac{\angle B-\angle C}{2}}}{BA\cdot \sin{\frac{\angle B-\angle A}{2}}}\cdot \frac{AC\cdot \sin{\frac{\angle C-\angle A}{2}}}{CB\cdot \sin{\frac{\angle C-\angle B}{2}}}=-1. \end{align*}We are done by Menelaus.

Let $H_A$ be the orthocenter of $BIC$ and define $H_B,H_C$ similarly. Let $M_A,M_B,M_C$ be the midpoints of $BC,AC,AB$. $-1=(A,I;\overline{AI}\cap \overline{EF},P)\overset{D}{=} (\overline{AD}\cap\overline{EF},\overline{DI}\cap\overline{EF};\overline{AI}\cap\overline{EF},\overline{AX}\cap\overline{EF})\overset{A}{=}(D,\overline{DI}\cap\overline{EF};I,\overline{AX}\cap\overline{DI})$. By Iran Lemma, $(D,\overline{DI}\cap\overline{EF};I,H_A)=-1$, so $A-H_A-X$. Let $P_A=\overline{EF}\cap\overline{M_BM_C}$. By Repeated applications of Iran Lemma, $H_A$ is the pole of $\overline{M_BM_C}$, so $\overline{AH_A}$ is the pole of $P_A$ (not by Iran Lemma, but this can be generalized). $\overline{DP_A}$ passes through the Feuerbach Point $Fe$ by First Fontene's Theorem, so $X$ is the pole of $\overline{DFe}$. $\overline{XYZ}$ is thus the polar of $Fe$, as desired.

Similar solution to above but I try to describe it in a complicated, poisonous way. But I am a newbie to these things so let me know if I have any mistakes. Solution: Let $H_A$ be the orthocenter of $BIC$, $Fe$ be the Feuerbach point, midline of $A$ means the line passing the mid point of $AB,AC$. Lemma 1: $(D,ID\cap EF;I,H_A)=-1$ Proof: It's well-known that the gergonne point lies on Feuerbach hyperbola so $EF$ is the polar of $D$ w.r.t the Feuerbach hyperbola. And the conclusion immediately follows by the definition. Corollary: The polar of $H_A$ w.r.t incircle is the midline of $A$. Proof: Just perform polar transformation w.r.t the incircle. Same as above, by cross ratio, $(A,I;AI\cap EF,P)=-1$, then projecting this cross ratio from $D$ to $EF$ then from $A$ to $DI$ we conclude that $A,H_A,X$ are collinear. By applying Fontene first on $(A,B,C,I)$, we obtain that $DFe,EF,$ midline of $A$ are concurrent. Perform a polar transformation w.r.t. the incircle we see that pole of $DFe,A,H_A$ are collinear. It's trivial that the pole of $DFe\in BC$ so actually that pole is $X$. Hence $X\in FeFe$, analogously we conclude that $X,Y,Z$ are on the polar of $Fe$ w.r.t. the in circle, indeed collinear.$\quad\blacksquare$