We start with some general observations. Set $\alpha = \angle A / 2$, $\beta = \angle B / 2$, $\gamma = \angle C / 2$. Then obviously $\alpha + \beta + \gamma = 90^{\circ}$. Since $\angle UIC = 90^{\circ}$, we obtain $\angle IUC = \alpha + \beta$. Therefore $\angle BIV = \angle IUC - \angle IBC = \alpha = \angle BAI = \angle BYV$, which implies that the points $B$, $Y$, $I$, and $V$ lie on a common circle (see Figure 1).

Assume now that the points $I$, $X$ and $Y$ are collinear. We prove that $\angle YIA = 90^{\circ}$.
Let the line $XU$ intersect $AB$ at $N$. Since the lines $AI$, $UX$, and $VY$ are parallel, we get
$$\frac{NX}{AI} = \frac{YN}{YA} = \frac{VU}{VI} = \frac{XU}{AI},$$
implying $NX = XU$. Moreover, $\angle BIU = \alpha = \angle BNU$. This implies that the quadrilateral $BUIN$ is cyclic, and since $BI$ is the angle bisector of $\angle UBN$, we infer that $NI = UI$. Thus in the isosceles triangle $NIU$, the point $X$ is the midpoint of the base $NU$. This gives $\angle IXN = 90^{\circ}$, i.e., $\angle YIA = 90^{\circ}$.

Figure 1

Let $S$ be the midpoint of the segment $VC$. Let moreover $T$ be the intersection point of the lines $AX$ and $SI$, and set $x = \angle BAV = \angle BCV$. Since $\angle CIA = 90^{\circ} + \beta$ and $SI = SC$, we obtain
$$\angle TIA = 180^{\circ} - \angle AIS = 90^{\circ} - \beta - \angle CIS = 90^{\circ} - \beta - \gamma - x = \alpha - x = \angle TAI,$$
which implies that $TI = TA$. Therefore, since $\angle XIA = 90^{\circ}$, the point $T$ is the midpoint of $AX$, i.e., $T = W$.

To complete our solution, it remains to show that the intersection point of the lines $IS$ and $BC$ coincide with the midpoint of the segment $BC$. But since $S$ is the midpoint of the segment $VC$, it suffices to show that the lines $BV$ and $IS$ are parallel.

Since the quadrilateral $BYIV$ is cyclic, $\angle VBI = \angle VYI = \angle YIA = 90^{\circ}$. This implies that $BV$ is the external angle bisector of the angle $ABC$, which yields $\angle VAC = \angle VCA$. Therefore $2\alpha - x = 2\gamma + x$, which gives $\alpha = \gamma + x$. Hence $\angle SCI = \alpha$, so $\angle VSI = 2\alpha$.

On the other hand, $\angle BVC = 180^{\circ} - \angle BAC = 180^{\circ} - 2\alpha$, which implies that the lines $BV$ and $IS$ are parallel. This completes the solution.

As in Solution 1, we first prove that the points $B$, $Y$, $I$, $V$ lie on a common circle and $\angle YIA = 90^{\circ}$. The remaining part of the solution is based on the following lemma, which holds true for any triangle $ABC$, not necessarily with the property that $I$, $X$, $Y$ are collinear.

Lemma. Let $ABC$ be the triangle inscribed in a circle $\Gamma$ and let $I$ be its incentre. Assume that the line passing through $I$ and perpendicular to the line $AI$ intersects the side $AB$ at the point $Y$. Let the circumcircle of the triangle $BYI$ intersect the circle $\Gamma$ for the second time at $V$, and let the excircle of the triangle $ABC$ opposite to the vertex $A$ be tangent to the side $BC$ at $E$. Then
$$\angle BAV = \angle CAE.$$
Proof. Let $\rho$ be the composition of the inversion with centre $A$ and radius $\sqrt{AB \cdot AC}$, and the symmetry with respect to $AI$. Clearly, $\rho$ interchanges $B$ and $C$.

Let $J$ be the excentre of the triangle $ABC$ opposite to $A$ (see Figure 2). Then we have $\angle JAC = \angle BAI$ and $\angle JCA = 90^{\circ} + \gamma = \angle BIA$, so the triangles $ACJ$ and $AIB$ are similar, and therefore $AB \cdot AC = AI \cdot AJ$. This means that $\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $AB$ and $\angle AIY = 90^{\circ}$, the point $Y' = \rho(Y)$ lies on $AC$, and $\angle JY'A = 90^{\circ}$. Thus $\rho$ maps the circumcircle $\gamma$ of the triangle $BYI$ to a circle $\gamma'$ with diameter $JC$.

Finally, since $V$ lies on both $\Gamma$ and $\gamma$, the point $V' = \rho(V)$ lies on the line $\rho(\Gamma) = AB$ as well as on $\gamma'$, which in turn means that $V' = E$. This implies the desired result.

Now we turn to the solution of the problem.
Assume that the incircle $\omega_1$ of the triangle $ABC$ is tangent to $BC$ at $D$, and let the excircle $\omega_2$ of the triangle $ABC$ opposite to the vertex $A$ touch the side $BC$ at $E$ (see Figure 3). The homothety with centre $A$ that takes $\omega_2$ to $\omega_1$ takes the point $E$ to some point $F$, and the tangent to $\omega_1$ at $F$ is parallel to $BC$. Therefore $DF$ is a diameter of $\omega_1$. Moreover, $Z$ is the midpoint of $DE$. This implies that the lines $IZ$ and $FE$ are parallel.

Let $K = YI \cap AE$. Since $\angle YIA = 90^{\circ}$, the lemma yields that $I$ is the midpoint of $XK$. This implies that the segments $IW$ and $AK$ are parallel. Therefore, the points $W$, $I$ and $Z$ are collinear.