Invert in the circle of center $D$ and radius $DB=DC=DI=DI_a$ (by Incenter-Excenter Lemma). $X$ belongs to the circle $(DI)$, which gets mapped to the line $\ell$ through $I$ perpendicular to $DI$, and similarly $Y$ belongs to $(DI_a)$ which gets mapped to the line $m$ trough $I_a$ perpendicular to $DI_a$. So $X^*=\ell\cap BC$ and $Y^*=m\cap BC$. $Z$ gets mapped to the intersection of $(X^*X^*D)$ (i.e. the circle trough $X^*$ and $D$ tangent to $BC$) and $(Y^*Y^*D)$. It is well known that this is the $D$-Humpty point of $DX^*Y^*$, so $W=DZ^*\cap BC$ is the midpoint of $Y^*X^*$. Since $W$ is the midpoint of $X^*Z^*$, it follows that the projection of $W$ on $AD$ must be the midpoint of $I$ and $I_a$, which are the projections of $X^*$ and $Y^*$ respectively. In other words, this projection is $D$, which implies $AD\perp W$. The inverse of $W$ must belong to $(ABC)$, and must satisfy $AD\perp DW^*$, so this point must be the antipode $A'$ of $A$. Therefore, $Z^*,Z,D,A',W$ are all aligned as wanted.

Let $D'$ be the antipode of $D$ wrt $\Gamma$. Note that $X,I,D'$ are collinear, $I_A, Y, D'$ are collinear, $A'D' \parallel AD$, and $D$ is the midpoint of $II_A$ by incenter-excenter lemma. Hence $-1=(I,I_A;D,\infty_{AD}) = (D'I, D'I_A; D'D, D'A') = (X,Y;D,A')$. This means that quadrilateral $XDYA'$ is harmonic and we are done.

Same as this

Another projective overkill Let $N$ the midpoint of the arc $BAC$ on $\Gamma$ then by the alge conditions u have $X,I,N$ colinear and $Y,I_A,N$ colinear, now clearly $A,A'$ and $D,N$ are diametricaly oposite on $\Gamma$ so $AD \parallel NA'$ and now let $P_{\infty}$ the point of infinity along $AD,NA'$, and also by I-E lemma $D$ is the midpoint of $II_A$ so now projecting cross ratios u have $$-1=(I, I_A; D, P_{\infty}) \overset{N}{=} (X, Y;D, A') \implies Z,D,A' \; \text{colinear}$$Thus we are done

khina wrote: Same as this I wouldn't say that it's the same. There is some work to do before concluding that $X$ is the point where the mixtilinear circle touches the circumcircle (similarly for $Y$). I was not aware of the Taiwanese problem though.

Note that $XI$ and $I_AY$ intersect on the midpoint of major arc $BC$, let it be $M$. We need to prove that quadrilateral $XDYA'$ is harmonic. Note that $\dfrac{XD}{DY} \cdot \dfrac{A'Y}{A'X}=\dfrac{XD/DM}{DY/DM} \cdot \dfrac{A'Y/AA'}{A'X/AA'}=\dfrac{\sin \angle IMD}{\sin \angle I_AMD} \cdot \dfrac{\cos \angle AMI_A}{\cos \angle AMI}=\dfrac{ID}{DI_A} \cdot \dfrac{MI_A}{MI} \cdot \dfrac{MI}{MI_A}=1,$ as desired. We freely used the Ratio Lemma. Moreover, we used that $DI=DI_A$, as $D$ is the center of $(BICI_A)$.