Easy for this position, though still a nice problem.

Let $\omega$ touch $AB,AC$ at $K,L$ respectively. $KL\cap BC=N$ Let $P$ be the miquel point of $CBKL$. WLOG $AB<AC$. Claim: $A,Z,K,L,I$ are cyclic. Proof: $\angle AKI=\angle ILA=\angle IZA=90$ hence all of them lie on the circle with diameter $AI$. Claim: $N,P,Z$ are collinear. Proof: \[\angle ZPK+\angle KPN=\angle ZPK+\angle B=\angle ZAK+\angle B=\angle A +\angle C +\angle B=180\] Claim: $P,X,Y,Z$ are cyclic. Proof: \[NP.NZ=NK.NL=NX.NY\] Claim: $\angle PXZ=\angle PNB$ Proof: Take the inversion centered at $Z$ with radius $ZX=ZY$. $Z,X,Y,P$ are cyclic thus $P\leftrightarrow N \iff ZP.ZN=ZX^2\iff \angle PXZ=\angle ZNB=\angle PNB$ Claim: $(ABCP)$ and $(XYZP)$ are tangent at $P$. Proof: Let $SP$ be tangent to $(ABCP)$. \begin{align*} \angle SPZ &= \angle SPA+\angle APZ \\ &= \angle PBA+\frac{\angle B-\angle C}{2} \\ &= \angle PBK+\frac{\angle B-\angle C}{2} \\ &=\angle PNK+\frac{\angle B-\angle C}{2} \\ &= \angle PNB \\ &=\angle PXZ \end{align*}This gives that $SP$ is also tangent to $(XYZP)$ as desired.$\blacksquare$

Solution by @n_bug: We'll use Casey's theorem. We must check that Ptolemy's relation holds for the side lengths of $ABC$ and the lengths of the tangents from $A$, $B$, and $C$ to the circumcircle of $XYZ$. One of these tangents is $AZ$. Let $\omega$ touch $AB$ and $AC$ at $D$ and $E$; then, by power-of-a-point, the other two tangents equal $BD$ and $CE$. So we really want Ptolemy's relation for the side lengths of $ABC$ and the lengths of $AZ$, $BD$, and $CE$. Let $I$ move linearly along the interior angle bisector through $A$. (While $ABC$ remains fixed.) Then $D$, $E$, and $Z$ will also move linearly along their respective lines, since they are the projections of $I$ onto them. Hence, the signed lengths of $AZ$, $BD$, and $CE$ will be linear functions of $|AI|$. It follows that Ptolemy's expression will be a linear function of $|AI|$ as well. We'll check now that both the constant term and the slope vanish. The constant term is the value at zero, when $A \equiv I$. Then also $A \equiv D \equiv E \equiv Z$, so Ptolemy's expression does vanish. For the slope, let $|AI|$ grow without bound. Then the ratios $AZ : BD : CE$ approach $AZ : AD : AE$. To see that the slope vanishes, we must verify that these satisfy Ptolemy's relation with the side lengths of $ABC$. But that's clear because $ABC$ and $ZDE$ are similar, and the points $A$, $D$, $E$, and $Z$ are concyclic.

The point of tangency is the Miquel point $S$ of $BCEF$, where $E,F$ are the tangency points of $\omega$ and $AC$, $AB$ respectively. Note that $A,S,E,F,I,Z$ are concyclic with diameter $AI$. Let $T = EF \cap BC$. By Miquel point properties we have $T,B,F,S$ concyclic, so \[ \measuredangle TSF = \measuredangle TBF = \measuredangle ZAF = \measuredangle ZSF \]implies $T,S,Z$ collinear. Now $TX \cdot TY = TF \cdot TE = TS \cdot TZ$ implies $X,Y,Z,S$ concyclic. Since obviously $ZX=ZY$, we have $AZ$ is tangent to $(XYZS)$. Let $AI$ meet $(ABC)$ again at $K$, and $AZ$ meet $(ABC)$ again at $A'$. Then \begin{align*} \measuredangle KSZ &= \measuredangle KSA + \measuredangle ASZ = \measuredangle KA'A + (90^\circ - \measuredangle ZAI) \\ &= \measuredangle KA'A + 90^\circ - \measuredangle A'AK = 90^\circ \end{align*}so $SZ$ passes through the midpoint $J$ of arc $BAC$. This implies there is a circle tangent to $(ABC)$ and $AA'$ while passing through $S$ and $Z$ - of course, it must be precisely $(XYZS)$, hence proved.

Suppose the circle through $X$, $Y$ tangent to $(ABC)$ meets $(ABC)$ at $K$. Define $E$, $F$ as the touch points from $\omega$ to $AB$, $AC$. Claim 1: $K$ lies on $(AEFZ)$, or $K$ is the Miquel point of $BCEF$. The tangency gives us the angle condition $\angle BKX = \angle CKY$. Ratio Lemma and Power of a Point then says \[\frac{BK}{CK} = \sqrt{\frac{BK \sin \angle BKX}{CK \sin \angle CKX} \cdot \frac{BK \sin \angle BKY}{CK \sin \angle CKY}} = \sqrt{\frac{BX}{CX} \cdot \frac{BY}{CY}} = \frac{BF}{CE},\] which gives the desired spiral similarity. ${\color{blue} \Box}$ Claim 2: $Z$ lies on $(KXY)$, as desired. Since $\triangle XYZ$ is isosceles, it suffices to show $KZ$ is an external angle bisector of $\angle XKY$, which is the same as the external angle bisector of $\angle BKC$. This is equivalent to \[\angle ZKC = \frac{\angle B + \angle C}{2} \iff \angle AKZ = \frac{|\angle B - \angle C|}{2} \iff \angle AIZ = \frac{|\angle B - \angle C|}{2},\] which is true. $\blacksquare$

Let the perpendicular bisector of $BC$ intersect the circumcircle of $ABC$ at $M$ and $N$ such that $N$ and $A$ lie on the same side of line $BC$. Let $D$ be the antipode of $Z$ on the circumcircle of $XYZ$. Let $w$ touch $AB$ and $AC$ at $E$ and $F$. Let $T$ be the Miquel point of complete quadrilateral $BEFC$. Let $EF$ and $BC$ intersect at $P$. Claim: $N$, $Z$, and $T$ are collinear $$\angle ATZ=\angle AIZ=\angle AMN=\angle ATN$$Claim: $N$, $T$, and $P$ are collinear $$\angle CTP=\angle CFP=\angle AFE=\angle NBC=180^{\circ}-\angle NTC$$Claim: $XYZT$ is concyclic $$PX\cdot PY=PE\cdot PF=PT\cdot PZ$$Claim: The circumcircles of $ABC$ and $XYZT$ are tangent at $T$ Notice that $MT\perp NT$ and $DT\perp ZT$ so $M$, $D$, and $T$ are collinear. Since $ZD||NM$ the result follows.

i didnt see the easy solution notationally, we use $AA$ for the tangent from $A$ to $(ABC)$. $\sqrt{bc}$ invert. xonks Note that by symmetry $(XYZ)$ is tangent to $AZ$, because $IX=IY$ say, combined with perpendicularity of $IZ$ to $BC$. This is crucial in allowing us to safely invert, because everyone knows how hard centres are to deal with inverting. We actually only show that there is a circle tangent to the parallel to $BC$ through $A$, $(ABC)$ passing through $X,Y$. We first of all note that a circle $\omega'$ intersects $(ABC)$ at $X',Y'$. Suppose $\omega'$ is tangent respectively to $AB,AC$ at $P,Q$. It suffices to show a circle is tangent to $BC$, the tangent line at $A$ to $(ABC)$, and passes through $X', Y'$. We reconstruct $PQ\cap AA$ as $S$, and $BC\cap PQ$ as $T$. We claim that $SX'Y'T$ is the desired circle. Claim: $(SX'Y')$ is tangent to $AA$ Let $X'Y'$ intersect $AA$ at $U$. It suffices to show that $U$ is the midpoint of $AS$, as then we would have $US^2=UA^2=UX'\times UY'$ which would suffice. Thus, consider a half dilation at $A$. We want the image of $PQ$ in this dilation to pass through $U$. Apply Radical axis theorem on $(A)$, $(ABC)$, $\omega'$. This gives that $AA, X'Y'$, and the radical axis of $(A)$ and $\omega'$ concur at $U$. However, it is well-known that the radical axis of point-circle is the half dilation of the polar of the point wrt circle. In this case, the polar of $A$ is $PQ$, so indeed the image of $PQ$ in a half dilation would pass through $U$, as desired Let $X'Y'$ intersect $BC$ at $V$. Let $K$ be the intersection of $X'Y'$ and $PQ$. By Desargues Involution Theorem on $AABC$ with line $PQ$ and conic $(ABC)$, there exists a point $K'$ such \[K'X'\times K'Y'=\textrm{Pow}_{(ABC)}K'=K'T\times K'S=K'P\times K'Q\]Yet $K$ uniquely satisfies the first and last equivalences, so $K=K'$, thus $SX'Y'T$ is cyclic by Power of a Point. It only remains to show that this is tangent to $BC$ as well. By ice cream cone theorem, we just want, if $L=AA\cap BC$, that $LS=LT$. This is the same as $\angle LST=\angle STL$, but this follows because we have $\triangle APS\sim\triangle CQT$, because $\angle SAP=\angle C=\angle QCT$, and from $APQ$ being isosceles by ice cream theorem. yayyayayaya