WLOG $A, B, C$ on the unit circle, and denote a lowercase variable complex number as the coordinate of a point labelled with the corresponding uppercase letter. We claim that $G$, the centroid of $\triangle ABC$, is the fixed point. Let $r = AU / AC$ and $\omega$ be the unitary complex number that rotates $AC \mapsto AU$. Therefore, we have the following: \begin{align*} u &= (c - a) r \omega + a \\ y &= \frac{(b - a)r}{\omega} + a \\ x &= (a-b)r \omega + b \\ w &= \frac{(c - b)r}{\omega} + b \\ z &= (b-c)r \omega + c \\ v &= \frac{(a-c)r}{\omega} + c \end{align*} Observe that $(x + z + u)/3 = (y + w + v)/3 = (a + b + c)/3 = g$, which means $G$ is the centroid of $\triangle XZU$ and $\triangle YWV$ as well. We use the following well-known fact: In $\triangle ABC$, the power of its centroid $G$ w.r.t. its circumcircle is $(AB^2 + BC^2 + CA^2)/9$. So, it suffices for us to show that $XZ^2 + ZU^2 + UX^2 = YW^2 + WV^2 + VY^2$. \begin{align*} \sum_{cyc} XZ^2 = \sum_{cyc} |x-z|^2 &= \sum_{cyc} (x-z)(\overline{x-z}) \\ &= \sum_{cyc} \left( (a+c-2b)r\omega + (b-c) \right) \left( \left( \frac{1}{a}+\frac{1}{c}-\frac{2}{b} \right) \frac{r}{\omega} + \frac{1}{b} - \frac{1}{c} \right) \\ &= \sum_{cyc} r^2 \left( 6+\frac{a}{c}+\frac{c}{a}-2\left(\frac{a}{b}+\frac{b}{a}\right)-2\left(\frac{b}{c}+\frac{c}{b}\right) \right) + r \omega \left( \frac{a}{b} + \frac{c}{b} + \frac{2b}{c} - \frac{a}{c} - 3 \right) + \frac{r}{\omega} \left( \frac{b}{a} + \frac{b}{c} + \frac{2c}{b} - \frac{c}{a} - 3 \right) + \left(2 - \frac{b}{c} - \frac{c}{b} \right) \\ &= r^2 \left( 18 - 3\sum_{sym} \frac{a}{b} \right) + r\omega \left( -9 + 3\sum_{cyc} \frac{a}{b} \right) + \frac{r}{\omega} \left(-9 + 3\sum_{cyc} \frac{b}{a} \right) + \left(6 - \sum_{sym} \frac{a}{b} \right), \\ \end{align*} \begin{align*} \sum_{cyc} YW^2 = \sum_{cyc} |y-w|^2 &= \sum_{cyc} (y-w)(\overline{y-w}) \\ &= \sum_{cyc} \left( (2b-a-c)\frac{r}{\omega} + (a-b) \right) \left( \left( \frac{2}{b}-\frac{1}{a}-\frac{1}{c} \right) r \omega + \frac{1}{a} - \frac{1}{b} \right) \\ &= \sum_{cyc} r^2 \left( 6+\frac{a}{c}+\frac{c}{a}-2\left(\frac{a}{b}+\frac{b}{a}\right)-2\left(\frac{b}{c}+\frac{c}{b}\right) \right) + r \omega \left( \frac{b}{a} + \frac{b}{c} + \frac{2a}{b} - \frac{a}{c} - 3 \right) + \frac{r}{\omega} \left( \frac{a}{b} + \frac{c}{b} + \frac{2b}{a} - \frac{c}{a} - 3 \right) + \left(2 - \frac{b}{a} - \frac{a}{b} \right) \\ &= r^2 \left( 18 - 3\sum_{sym} \frac{a}{b} \right) + r\omega \left( -9 + 3\sum_{cyc} \frac{a}{b} \right) + \frac{r}{\omega} \left(-9 + 3\sum_{cyc} \frac{b}{a} \right) + \left(6 - \sum_{sym} \frac{a}{b} \right), \\ \end{align*} From the computations above, we get $XZ^2 + ZU^2 + UX^2 = YW^2 + WV^2 + VY^2$, which completes the proof. $\square$