Consider the statement after inversion at $E$: Given a quadrilateral $ABCD$, let $AC$ and $BD$ intersect at $E$, let $\odot(ABE)$ and $\odot(CDE)$ intersect at another point $F$, let $\odot(BCE)$ and $\odot(DAE)$ intersect at another point $G$, let $W, X, Y, Z$ be the circumcenters of $\odot(ABE), \odot(BCE), \odot(CDE), \odot(DAE)$ respectively, prove that: one of the intersection points of $\odot(WFY)$ and $\odot(XGZ)$ lies on $\odot(EFG)$.

Let $M$ be the midpoint of $WY$, let $S$ be the reflection of $E$ with respect to $M$, and let $H$ be the other intersection point of $EF$ and $\odot(WFY)$.

Claim 1. $E$ is the orthocenter of $\triangle WHY$.
Proof. Note that both $W$ and $Y$ lie on the perpendicular bisector of $EF$,
$$\angle(EY, WY) = \angle EYW = \angle WYF = \angle WHF = \angle(WH, HF)$$
Since $WY \perp HF$, it follows that $WH \perp EY$, so $E$ is the orthocenter of $\triangle WHY$.

Claim 2. $M$ is the circumcenter of $\triangle EFG$.
Proof. Since $WX \perp BE$ and $ZY \perp ED$, we get $WX \parallel YZ$; similarly we can obtain $XY \parallel WZ$, so $WXYZ$ is a parallelogram, hence $M$ is also the midpoint of $XZ$. $WY$ is the perpendicular bisector of $EF$, $XZ$ is the perpendicular bisector of $EG$, and $M$ is the intersection point of $WY$ and $XZ$, so $M$ is the circumcenter of $\triangle EFG$.

Claim 3. $S$ is an intersection point of $\odot(WFY)$ and $\odot(EFG)$.
Proof. Since $M$ is the circumcenter of $\odot(EFG)$, $S$ is the antipode of $E$ on $\odot(EFG)$. Also, since $E$ is $\triangle WHY$ and $M$ is the midpoint of $WY$, we know that $S$ is the antipode of $H$ on $\odot(WHY)$. Hence $S$ lies on $\odot(WFY)$ and $\odot(EFG)$. Similarly, we can obtain that the reflection of $E$ with respect to the midpoint of $XZ$ lies on $\odot(XGZ)$, and since the midpoint of $XZ$ is exactly $M$, $S$ also lies on $\odot(XGZ)$, which completes the proof.