Let $I$ be its incenter. Let $M_A,M_B,M_C$ be midpoints of minor arcs $BC,CA,AB$ respectively and $N$ be the midpoint of major arc $BC$. It is known that $Y,I,N$ are collinear. Note that $-1 = (N,M_A;B,C) \overset{I}{=} (Y,A;M_B,M_C)$. Therefore, the tangents to $\Omega$ at $A,Y$ and the line $M_BM_C$ are concurrent at a point called $R$. Moreover, by Pascal's theorem on the hexagon $ABM_BM_CCA$, we get that $E,F,R$ are collinear. Obviously, it suffices to show that $(AYZ)$ is tangent to $\Omega$ which is not hard to see since $\angle DXE = 90^\circ$ implies that $D,X,A$, intersection of $EF,BC$ are concyclic, and by angle chasing, $\angle RAZ = \angle AYZ$.

Another completion after the first paragraph. It's well known that $(AI;ZD)=-1$. However, $\angle ZXD=90^\circ$, so $XZ$ bisects $\angle AXI$, implying that $AXIR$ is concyclic (because $R$ lies on the perpendicular bisector of $AI$, which is $M_BM_C$). Then, shooting lemma gives $RX\cdot RZ = RA^2$.

Let $I$ and $I_A$ be the incenter and the $A$ exincenter of the triangle $ABC$, respectively. Let $M$ and $N$ be the midpoints of the major and the minor arcs $BC$, respectively. For any point $U$, define $U'$ as its image under the inversion at $A$ with radius $\sqrt{AB\cdot AC}$ followed by reflection across $AD$. It's well known that $D'=N$, $I'=I_A$, $M'$ is the intersection of $BC$ and the external angle bisector of $\angle BAC$, which is also on $EF$, and $Y'$ is the tangency point of the $A$-exincircle with $BC$. Let $P=A'$ be the point at infinity on $AD$. Note that $Z'$ is a point on $AD$ satisfying $(D',Z';I',P)=(D,Z;I,A)\overset{F}{=}(FD\cap AC,E,C,A)=-1$, then $Z'$ is the reflection of $N$ across $I_A$. Finally, $X'$ is a point on $(AE'F')$ and on the circle with diameter $NZ'$ ($A$, $D$, and $Z$ are colinear, so the circle with diameter $DZ$ goes to the circle with diameter $NZ'$). Note that $Z',M\in (AE'F')$, so $\angle Z'X'M=\angle Z'AM=\frac{\pi}{2}=\angle Z'X'M$. Therefore $X'\in MN=$ perpendicular bisector of $BC$ and $BC\parallel X'Z'$. Moreover, if $Q=X'Z'\cap Y'I_A$, $I_A$ midpoint of $NZ'$ implies $Q$ midpoint of $X'Z'$, so the triangle $Y'X'Z'$ is isosceles in $Y'$ and $\angle BY'X'=\angle Y'X'Z'=\angle Y'Z'X'$ and $(X'Y'Z')$ is tangent to $BC$.

Let $T$ be midpoint of arc $BAC$ and let $BI$ and $CI$ meet $\Omega$ at $L$ and $K$. Let $T'$ be midpoint of arc $BC$. Note that $(BC ; TT') = (LK ; YA) = -1$ so tangents at $A$ and $Y$ to $\Omega$ concur on $KL$ on point $S$. Note that by applying Pascal on $BAACKL$ we would have $S,E,F$ are collinear. Note that if we prove $AS$ is tangent to $AXZ$ then by applying Radical Axis Theorem on $AXZ$ and $XZY$ and $\Omega$ we would have $SY$ is tangent to $XZY$. We want to prove $\angle AXZ = \angle ACT' = \angle ADB$. Let $EF$ and $BC$ meet at $P$. we want to prove $AXDP$ is cyclic which is true since $AP$ is external angle bisector and then $\angle PAD = \angle 90 = \angle PXD$.

See teacher buratino's generalization and my solution: https://artofproblemsolving.com/community/c374081h3074408_four_concyclic_points