Apparently there is a nice theorem related to the anticenter of a cyclic quadrilateral that is not mentioned in the wikipedia:
Anticenter of the cyclic quadrilateral, the intersection point of its diagonals and two intersection points of the lines that contain its opposite sides are always concyclic.
illustration Points E, F, G, H belong to a circle. Can you prove it?
Here is an "ugly" but straightforward solution with complex numbers. $WLOG$, assume $(ABC)$ to be the unit circle. Let $T$ be the anticenter and $X$ be the intersection of the diagonals of the quadrilateral $ABCD$. Lowercase letters denote the affixes.
Since the anticenter is symmetric to the circumcenter in the centroid: $$t=\frac{a+b+c+d}{2}$$
The intersection of $AB$ and $CD$ is given by:
$$f=\frac{(\bar{a} b-a \bar{b})(c-d)-(a-b)(\bar{c} d-c \bar{d})}{(\bar{a}-\bar{b})(c-d)-(a-b)(\bar{c}-\bar{d})}$$
Since $|a|=|b|=|c|=|d|=1$, this simplifies to:
$$f=\frac{a b(c+d)-c d(a+b)}{a b-c d}$$
Similarly:
$$e=\frac{a d(c+b)-c b(a+d)}{a d-c b} \; \; \text{and} \; \; x=\frac{a c(b+d)-b d(a+c)}{a c-b d}$$
It is not hard to show that $T, F, E, X$ are concyclic iff:
$$\frac{f-t}{e-t} \div \frac{f-x}{e-x} \; \in \; \mathbb{R} $$
Therefore, we need to show that:
$$\huge{ \frac{\frac{\frac{a b(c+d)-c d(a+b)}{a b-c d}-\frac{a+b+c+d}{2}}{\frac{a d(c+b)-c b(a+d)}{a d-c b}-\frac{a+b+c+d}{2}}}{\frac{{\frac{a b(c+d)-c d(a+b)}{a b-c d}-\frac{a c(b+d)-b d(a+c)}{a c-b d}}}{{\frac{a d(c+b)-c b(a+d)}{a d-c b}-\frac{a c(b+d)-b d(a+c)}{a c-b d}}}}\; \in \; \mathbb{R}}$$
To show this, we can substitute $\frac{1}{a}$ for $a$, $\frac{1}{b}$ for $b$, $\frac{1}{c}$ for $c$, $\frac{1}{d}$ for $d$ (recall $z\bar{z}=1$ for the unit circle), then show that the new expression is equal to the original one (since $z$ is a real number iff $z + \bar{z}=0$). Using Mathematica shows that the two expressions are equivalent. $ \; \; \blacksquare$
Here is what I checked in Mathematica.
More detailed explanations of the formulas I used can be found in Evan Chen's this article.