$ABCD$ is a cyclic quadrilateral whose diagonals intersect at point $P$. Let the centres of the circumcircles of $ABCD$, $\triangle ABP$, $\triangle BCP$, $\triangle CDP$, $\triangle DAP$ be $O$, $O$$1$, $O$$2$, $O$$3$, $O$$4$ in the given order. Prove that $OP$, $O$$1$$O$$3$, $O$$2$$O$$4$ intersect at one point.
By connecting $O$$1$, $O$$2$, $O$$3$, $O$$4$, we can see that $O$$1$$O$$2$ is perpendicular to common cord $BP$. Similarly, $O$$3$$O$$4$ is perpendicular to common cord $DP$.
Since $D$, $P$, and $B$ are on the same line; $O$$1$$O$$2$ is parallel to $O$$3$$O$$4$. Similarly, $O$$1$$O$$4$ is parallel to $O$$2$$O$$3$. So, $O$$1$$O$$2$$O$$3$$O$$4$ is a parallelogram
Let us consider the inversion $*$ with center in $P$ and power $PA\cdot PC=PB\cdot PD$. (The image of a point $X$ is thus denoted by $X^*$.)
Then we have by the choice of the center and the power $A^*=C$, $B^*=D$, $C^*=A$, $D^*=B$.
The circle $(O_1)=(PAB)$ is transformed in the line $A^*B^*=CD$, and $PO_1\perp CD$.
Since $O,O_3$ are both on the perpedicular bisector of $CD$, we have $OO_3\perp CD$.
This gives $PO_1\| OO_3$.
Similarly, $PO_3\| OO_1$.
So $PO_1OO_3$ is a parallelogram, its diagonals intersect in the midpoint $Q$, say, of $OP$.
So $O_1O_3$ passes through $Q$. Similarly, $O_2O_4$ passes through $Q$.
$\square$