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 $\angle U I C = 90^\circ$, we obtain $\angle I U C = \alpha + \beta$. Therefore $\angle B I V = \angle I U C - \angle I B C = \alpha = \angle B A I = \angle B Y V$, 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 $\angle Y I A = 90^\circ$. Let the line $X U$ intersect $A B$ at $N$. Since the lines $A I, U X$, and $V Y$ are parallel, we get
$$\frac{N X}{A I} = \frac{Y N}{Y A} = \frac{V U}{V I} = \frac{X U}{A I}$$
implying $N X = X U$. Moreover, $\angle B I U = \alpha = \angle B N U$. This implies that the quadrilateral $B U I N$ is cyclic, and since $B I$ is the angle bisector of $\angle U B N$, we infer that $N I = U I$. Thus in the isosceles triangle $N I U$, the point $X$ is the midpoint of the base $N U$. This gives $\angle I X N = 90^\circ$, i.e., $\angle Y I A = 90^\circ$. !
Figure 1 Let $S$ be the midpoint of the segment $V C$. Let moreover $T$ be the intersection point of the lines $A X$ and $S I$, and set $x = \angle B A V = \angle B C V$. Since $\angle C I A = 90^\circ + \beta$ and $S I = S C$, we obtain
$$\angle T I A = 180^\circ - \angle A I S = 90^\circ - \beta - \angle C I S = 90^\circ - \beta - \gamma - x = \alpha - x = \angle T A I,$$
which implies that $T I = T A$. Therefore, since $\angle X I A = 90^\circ$, the point $T$ is the midpoint of $A X$, i.e., $T = W$.
To complete our solution, it remains to show that the intersection point of the lines $I S$ and $B C$ coincide with the midpoint of the segment $B C$. But since $S$ is the midpoint of the segment $V C$, it suffices to show that the lines $B V$ and $I S$ are parallel.
Since the quadrilateral $B Y I V$ is cyclic, $\angle V B I = \angle V Y I = \angle Y I A = 90^\circ$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which yields $\angle V A C = \angle V C A$. Therefore $2 \alpha - x = 2 \gamma + x$, which gives $\alpha = \gamma + x$. Hence $\angle S C I = \alpha$, so $\angle V S I = 2 \alpha$.
On the other hand, $\angle B V C = 180^\circ - \angle B A C = 180^\circ - 2 \alpha$, which implies that the lines $B V$ and $I S$ are parallel. This completes the solution.