Possible triangle center associated with Apollonius circle of excircles

Question

When I was playing around with Geogebra, I personally found a possible triangle center, but I'm not 100 % sure if my personal conjecture is true.

Consider the following configuration:

Let $E_A$, $E_B$, $E_C$ be the excircles of a given triangle $ABC$. Let $S$ be the Apollonius circle internally tangent to the three excircles. Denote the point of tangency of $E_A$ with $S$ by $T_A$. Define $T_B$ and $T_C$ similarly. Draw the circle $S_A$ that is tangent to $AB$ and $CA$ and internally tangent to $S$, other than the three excircles. Denote the point of tangency of $S_A$ with $S$ by $U_A$. Define $S_B$, $S_C$, $U_B$, and $U_C$ cyclically.

It seems that the three lines $T_A U_A$, $T_B U_B$, $T_C U_C$ are concurrent.

Question: Are these lines actually concurrent or not? If it's true and is an already known result, does this theorem or point have a name?

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Edit
A demonstration of this proposition on Geogebra applet is available here.

Let $I_A$, $I_B$, $I_C$, $O_A$, $O_B$, and $O_C$ be the centers of $E_A$, $E_B$, $E_C$, $S_A$, $S_B$, and $S_C,$ respectively. Let $I$ be the incircle of $\triangle{ABC}$. Denote by $B_A$ and $A_B$ the intersections of the ray $\overrightarrow{AB}$ with $S$ and the ray $\overrightarrow{BA}$ with $S$, respectively. $A_B$, $A$, $B$, $B_A$ lie on a line in this order. Define $C_B$, $B_C$, $A_C$, $C_A$ cyclically.

What I know about this configuration is as follows:

1. $T_A U_A$, $T_B U_B$, $T_C U_C$ are concurrent
   $$\iff \frac{\vert \overline{U_C T_B} \vert}{\vert \overline{T_B U_A} \vert} \cdot \frac{\vert \overline{U_A T_C} \vert}{\vert \overline{T_C U_B} \vert} \cdot \frac{ \vert \overline{U_B T_A} \vert}{\vert \overline{T_A U_C} \vert}=1$$ (Ceva's theorem for chords)
2. $BC$ and $B_A C_A$ are parallel (parallel tangent theorem). $S$ is a Tucker circle centered at $X_{13323}$ of the triangle bounded by $B_A C_A$, $C_B A_B$, and $A_C B_C$. See also Darij Grinberg and Paul Yiu, "The Apollonius Circle as a Tucker Circle".
3. $T_B$, $T_C$, and the exsimilicenter of $E_B$ and $E_C$ are collinear. This is a special case of Monge's theorem.
4. $A T_A$, $B T_B$, and $C T_C$ meet at the exsimilicenter of $S$ and $I$, known as Apollonius point $X_{181}$. Similarly, $A U_A$, $B U_B$, $C U_C$ meet at the insimilicenter of $S$ and $I$, which is $X_{1682}$. These are consequences of Monge's theorem as well.
5. $A_B A_C$, $B_A C_A$, and $T_B T_C$ are concurrent.
6. $I_B$, $I_C$, the incenter of $\triangle{A_B B_C C_B}$ and the incenter of $\triangle{A_C B_C C_B}$ are collinear. Similarly, $I_B$, $O_C$, the incenter of $\triangle{C_B A_C C_A}$ are collinear (Sawayama-Thébault's theorem).
7. $A_C$, $U_A$, the point of contact of $S_A$ with $A A_C$, the incenter of $\triangle{A A_B A_C}$, and the incenter of $\triangle{A_C B_A A_B}$ are concyclic (attributed to Yūzaburō Sawayama).
8. If $E_A$ is not tangent to $BC$, then the proposition does not hold generally. Even if we assume that $BC \parallel B_A C_A$, $CA \parallel C_B A_B$, and $AB \parallel A_C B_C$, the proposition is generally not true.

I found another way to construct this point: Let $V_1=T_A A_B \cap T_C C_B$, $V_2=T_C C_B \cap T_B B_C$, $V_3=T_B B_C \cap T_A A_C$, $V_4=T_A A_C \cap T_C C_A$, $V_5=T_C C_A \cap T_B B_A$, and $V_6=T_B B_A \cap T_A A_B$. The three lines $V_1 V_4$, $V_2 V_5$, $V_3 V_6$ meet at this point.

I could not find this center in Kimberling's Encyclopedia of Triangle Centers, at least in the list from $X_1$ to $X_{47627}$.

Answer 1

This claim was algebraically confirmed. César Lozada found the following barycentric coordinates for this point.

Let $a$, $b$, $c$ be the lengths of the sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, respectively. Then, $$X=(f(a,b,c):f(b,c,a):f(c,a,b)),$$ where $$f(a,b,c)=a^2(b^2+c^2+a(b+c))(a^4(b+c)^2-bc(b+c)^2(b^2+c^2)+a^3(b+c)(b^2+bc+c^2)-a(b^2-c^2)(b^3-c^3)-a^2(b^4+c^4-bc(b^2+6bc+c^2)))$$

This triangle center was not listed in the Encyclopedia of Triangle Centers. Now it is $X_{50032}$.

This theorem may be generalized to circular triangles.