Let the incircle of $\triangle ABC$ hit $BC$ at $D$, then we use external Pitot theorem twice to get that $ABXC$ contains an excircle whose center is in fact $T$, and also that the X-excircle of $\triangle XBC$ is tangent to $BC$ also at $D$, let $I$ incenter of $\triangle ABC$, let $I_X$ the X-excenter of $\triangle XBC$. Now first note from tangencies that $BZ=BF=BD$ and $CY=CE=CD$ so in fact from perpendicular bisectors symetry we get that $TZ=TF=TE=TY$ therefore $(ZFEY)$ is cyclic with center $T$. so from here just notice if you let $TB \cap (BFZ)=B'$ and $TC \cap (CEY)=C'$ then from symetry previously seen and tangencies we can easly conclude that $IF \cap I_XZ=B'$ and $IE \cap I_XY=C'$ and then angle chase because $\angle ECY=\angle EC'I_X=\angle FB'Z$ which means that $\angle BB'F=\angle XCT$ but also we have $\angle IBC'=\angle II_XC'=\angle BCX$ and adding those up gives $\angle BB'C'=\angle BCT$ therefore $BB'C'C$ is cyclic and it means by PoP that $T$ lies on the radical axis of $(BFZ), (CEY)$ as desired thus we are done .

solved before mathluis but i had to sleep different solution Let $M', M$ be the midpoints of minor and major arcs respectively. Let $N$ be the midpoint of $BC$, and $D$ the foot of $I$ onto $BC$. Redefine $X$ as the intersection of $ID$ with the circle centered at $M$ with radius $MB$. Let $I_X$ be the $X$-excenter of $\triangle XBC$. Let $P$ and $Q$ be the intersections $(ECY)$, $(BFZ)$ with $I_XY$ and $I_XZ$ respectively. Note now that $I,F,Q$ and $I,E,P$ are collinear because of right angles. Further, by Fact 5, $I_XBI'C$ is cyclic, where $I'$ is the incentre of $\triangle BCX$. Also by right angles, $I_XZXY$ is cyclic. Note by opposite right angles and collinearities, $I_XPIQ$ is cyclic. Lemma 1: $\angle ABQ=\angle CPY$ Proof: Superficially we get $D$ is the $X$-extouch point of $\triangle XBC$ being the foot. Thus, it follows that $ZB=BD$, and $DC=CY$. However, $BD=BF$ and $CD=CE$. Hence, along with $B,Q$ being antipodes by right angles in $(QZBF)$, $BF=BZ$ implies that $BQ$ bisects $\angle FBZ$. Analogously, $CP$ bisects $\angle YCE$. Now, angle chasing, and using $(ABXC)$ is cyclic, \[\angle ABQ=90^\circ-\frac{\angle XBA}{2}=\frac{\angle ACX}{2}=90^\circ-\angle YCP=\angle CPY\] Now, angle chasing we have \begin{align*} \angle QPC&=180^\circ-\angle I_XPQ-\angle CPY &\text{Because collinearity }I_X, P, Y\\ &=180^\circ-\angle I_XIQ-\angle ABQ & \text{By cyclics and Lemma }1\\ &=\angle QID-\angle ABQ &\text{From }I_X, P, Y \text{ collinear}\\ &=360^\circ-90^\circ-\angle BQI-\angle ABQ-\angle DBQ & \text{Due to quadrilateral angle sum in }IDBQ\\ &=180^\circ-\angle DBQ & \text{As }\angle QFB \text{ is right} \end{align*}Thus $BCPQ$ is cyclic. Hence radical axis theorem on $(BQ), (CP), (BCPQ)$ implies that $BQ, CP$ and the radical axis of $(BFZ)$ and $(CYP)$ concur. Lemma 2: $\triangle ABC \sim\triangle I_XPQ$. Proof: Note that \[\angle PQI_X=\angle PII_X=\angle IPC+\angle PCA+\angle ACB+90^\circ-180^\circ=\angle ACB\]By noting $I,E,P$ collinear and perpendicular to $AC$, and from quadrilateral angle sum in $IDCP$. Analogously, $\angle I_XPQ=\angle BCA$, thus we are done. Lemma 3: $(BQI)$ and $(CPI)$ concur on $AI$. Proof: Suppose the intersection point is $G$. Note that it suffices to show that $M'$ lies on the radical axis of the two circles, because $A,I,M'$ collinear by Fact 5. Let $O_B$ and $O_C$ be the centres of $(BQI)$ and $(CPI)$ respectively. We just need $M'O_B^2-O_BI^2=M'O_C^2-O_CI^2$ by Power of a Point, but by perpendicularity criterion, this is the same as $M'I\perp O_BO_C$. Firstly, from angle at center theorem, \[\angle IO_BQ=2\angle IBQ=\angle DBZ=180^\circ-\angle QI_XI\]Hence $O_B$ lies on $(IQI_XP)$. Similarly $O_C$ also lies on this circle. From $M'B=M'I=M'C$ by Fact 5, $M'O_B$ bisects $\angle IM'B$. Thus, \[\angle IM'O_B=\frac{\angle IM'B}{2}=\frac{\angle ACB}{2}\]by Fact 5 with angle at center theorem. Now, we also have \begin{align*} \angle M'O_BO_C&=\angle M'O_BI+\angle IO_BO_C & \text{Splitting the angle}\\ &=\angle BQO_B+\angle IQO_C & \text{Angle at center theorem}\\ &=\frac{\angle ZQP}{2} &\text{Doubling our angles and using Fact 5}\\ &=90^\circ- \frac{\angle ACB}{2} &\text{By similarity} \end{align*}Hence $O_BO_C\perp IM'$ as desired. Now, from this, we may use radical axis theorem on $(BQI), (ICP), (BQPC)$ so $BQ, CP, AI$ concur. From before, this concurrency point also lies on the radical axis of $(BQ)$ and $(CP)$. Finally, by Trig Ceva, along with $AX$ bisecting $\angle QI_XP$ and $AI$ bisecting $\angle BAC$, in triangles $\triangle I_XPQ$, and $\triangle ABC$, we get \[\frac{\sin \angle CBQ }{\sin\angle ABQ}\times \frac{\sin \angle PCA}{\sin\angle PCB}=1\]But then as $BQPC$ cyclic, $\angle PCB=180^\circ-\angle BQP$, and $\angle CBQ=180^\circ-\angle QPC$. In conjunction with Lemma 1, this implies that \[\frac{\sin \angle ZQB}{\sin\angle BQP}\times \frac{\sin \angle QPC}{\sin \angle CPY}=1\]Thus $BQ, CP$ and $I_XX$ concur. This implies that $BQ, CP, I_XX, AI$ and the radical axis of $(BQ)$, $(CP)$ concur. It only remains to prove $X$ is the same point as that in the problem. Indeed, if $X_C$ and $X_B$ respectively lie on $AC$ and $AB$ such $CX_C=CX$ and $BX_B=BX$, we only need to show that $AX_B=AX_C$. This however is because by Sine Law, \[\frac{AX}{AX_B}=\frac{\sin\angle XX_BA}{\sin\angle AXX_B}\]Now, we check that $\angle AXX_B=\angle BQF+\angle ACB=\angle PCB$ by Lemma 1, and also $\angle XX_BA=\angle PCA$. The analogous result $AX_C$ means we only need \[\frac{\sin\angle PCB}{\sin\angle PCA}=\frac{\sin\angle CBQ}{\sin\angle ABQ}\]But this is equivalent to Trig Ceva on $\triangle ABC$. Thus, we are done.