We use moving points. Fix $ABCD$ (hence $T,I$ are fixed) and move $W$ with constant velocity (not just linearly) on $\overline{AB}$, so $X,Y,Z$ move linearly as well. Thus $\deg \overline{WY},\deg \overline{XZ} \leq 2$. Furthermore note that these lines coincide when $W$ is at infinity (here $X,Y,Z$ are also at infinity), so $\deg L \leq 3$. We now turn to investigating $G$. It is clear that $O_A$ moves with constant velocity along $\overline{AI}$, and similar statements hold for the three other vertices. Furthermore, when $W$ is the $\overline{AB}$-tangency point, we have $O_A=O_B=O_C=O_D=I$, hence as $W$ varies all the $O_AO_BO_CO_D$ are related through a homothety at $A$. Thus $G$ actually varies linearly (in fact, with constant velocity, although we will not need this). Thus the statement that $\overline{IL}$ and $\overline{TG}$ concur on $\ell_\infty$ has degree $(0+3)+(0+1)+0=4$, so it suffices to check $5$ cases. When $W$ is the $\overline{AB}$-tangency point, $X,Y,Z$ are also their respective tangency points, and $I=G$. Furthermore, by Brianchon we have $T=L$, which finishes. Our $4$ other cases come from when two of $W,X,Y,Z$ coincide with a vertex of $ABCD$; we will only deal with the case of $W=Z=A$, since the others follow symmetrically. Note that in this case, we automatically have $L=A$. We will use moving points again by varying the $\overline{CD}$-touchpoint on the incircle with degree $2$, in which case both $C$ and $D$ have degree $1$. To see this, note that the midpoint of the chord naturally formed by the tangents from $D$ moves with degree $2$ along a circle through $I$, and hence the intersection of the line through $I$ and this midpoint with a fixed line has degree $1$. With this established, it will be more natural to think about varying $D$ linearly along the tangent through $\overline{A}$. Observe that $X$ is fixed, $O_A=A$ is fixed, and $O_B$ is fixed (since $X$ is fixed). $O_C$ may be found by intersecting the fixed perpendicular to $\overline{BC}$ at $X$ with $\overline{CI}$, hence $\deg O_C=1$, and $O_D$ may be found by intersecting the perpendicular to $\overline{AD}$ at $A$ through $\overline{DI}$, hence $\deg O_D=1$. Now note that when we select $D=A$, $\overline{AC}$ and $\overline{BD}$ coincide. Furthermore $\overline{O_AO_C}$ and $\overline{O_BO_D}$ do as well, since $O_C=O_B$ and $O_D=A=O_A$. Hence we actually have $\deg T,\deg G \leq 1$. Additionally, when $D$ is chosen such that $B,D,I$ are collinear, everything is symmetric about $\overline{BD}$ (in particular, $X=C=O_C$), and thus both $T$ and $G$ are $\overline{AC} \cap \overline{BD}$. Thus $\deg \overline{GT} \leq 1$, so it suffices to check $(0+0)+1+0+1$ cases. If $D$ is chosen such that $A,C,I$ are collinear, everything is now symmetric about $\overline{AC}$, so $T,G$ both lie on $\overline{AI}$ as desired (if $T$ and $G$ coincide for some reason, so that $\overline{TG}$ is not well-defined, we can actually reduce $\deg \overline{GT}=0$, so it doesn't matter). For the other case, let $P=\overline{AD} \cap \overline{BC}$, and pick $D$ such that $\overline{AB}$ and $\overline{CD}$ are symmetric about $\overline{PI}$. In this case $T$ is at infinity, so it will suffice to prove that $\overline{O_AO_C} \parallel \overline{O_BO_D}$. I claim that both lines are parallel to $\overline{PI}$. Observe that by symmetry $\overline{O_AO_D}$ and $\overline{O_BO_C}$ are symmetric about some line perpendicular to $\overline{PI}$, so it suffices to show $O_AO_D=O_BO_C$. This is equivalent to $$DA\tan \frac{\angle D}{2}=CX\tan \frac{\angle C}{2}-BX\tan \frac{\angle B}{2} \iff (AB+DA)\tan \frac{\angle B}{2}=(AB-DA)\tan \frac{\angle A}{2}.$$Observe that $\tan \tfrac{\angle A}{2}$ is equal to the radius of the inscribed circle divided by the length of the tangent from $A$. We may now view $\triangle PAB$ as the reference triangle, relabeling it $\triangle ABC$ and defining $s,a,b,c$ as usual. The desired relation is then $$\frac{a+(c-b)}{s-b}=\frac{a-(c-b)}{s-c},$$which is obvious. This finishes the problem. $\blacksquare$

mine! you lost the game

Intended solution is basically MarkBcc168's, however some other notes: Let $\omega_a$ be the circle at $A$ through $W,Z$ and similarly for the others. Then apply Monge to groups of 3 circles, which means that $\overline{WX}\cap\overline{YZ}$ lies on $\overline{AC}$ as well. (This is necessary to establish the orthocentric system $TGUV$ mentioned in post #2.) I wanted to construct some "nice" points to hide the radical axis of $\omega_b$ and $\omega_d$, but didn't get anywhere and was forced to settle for the radical centers of 3 of the 4 circles.

Solved during before IMO 2023

Not a hard problem since even I can solve