Side NoteA crucial fact which shall be used to frame the inverted versions of the problems is that if $\omega$ is a circle with center $O$ and $P$ is a point not on $\omega$ and $O'$ is the image of $O$ under an inversion about $P$ and $O_1$ is the center of the image of the circle under the inversion then $P,O',O_1$ are collinear. LemmaLemma. Let $P$ be a point on the internal bisector of angle $BAC$ and not on the circumcircle of triangle $ABC$ (with center $O$). Let the circumcircles of $APC$ and $APB$ meet the lines $AB$ and $AC$ at points $X$ and $Y$, respectively. Then $PO \perp XY$. Proof 1Proof 1. Let $V,W$ be the feet of perpendiculars from $P$ onto $AB,AC$, respectively. Let $AB=c$, $AC=b$ and $AV=AW=x$ and notice that $PO$ is perpendicular to the radical axis of $(ABC)$ and $(P,PV=PW)$. Thus, by linearity of PoP, it suffices to showing that $$XA\cdot XB-XV^2=YA\cdot YB-YW^2$$which follows as $XA=(2x-b)$, $XB=(b+c-2x)$, $XV=(b-x)$ and $YA=(2x-c)$, $YC=(b+c-2x)$, $YW=(c-x)$ and so the desired condition becomes $$(b+c-2x)\cdot \left((2x-b)-(2x-c)\right)=-(b-x)^2+(c-x)^2$$which is obviously true. Proof 2Proof 2. Let $XY$ meet $(ABC)$ at $S,T$. Let $M,N,F$ be the feet of perpendiculars from $P$ onto $BY,CX$ and $XY$; respectively. Notice that $\angle MFN=\angle MFP+\angle NFP=\angle MYP+\angle NXP=\angle BAP+\angle CAP=\angle BAC$ and by the problem ELMO Shortlist 2010 (or generalized IMO 2009/2) , $F$ is the midpoint of $XY$ or $ST$. As $PX=PC$ and $PY=PB$ and $PB \not= PC$ ($P$ does not lie on $(ABC)$), it follows that $F$ is the midpoint of $ST$. But then $OF \perp ST$ and we get that $PO \perp XY$. Case of cyclicNext we consider the case when $ABCD$ is a cyclic quadrilateral with center $O$. Indeed, It is well-known that triangle $PEF$ is self-polar and so $PO \perp EF$. Thus, it is equivalent to showing that $P$, the circumcenters of $XQY$ (call it $O'$) and $O$ are collinear. Let $QX$ meet $\omega_2$ at $X_1$ and $QY$ meet $\omega_1$ at $Y_1$. Note that $\angle XQP=\angle XDP=\angle ADB=\angle ACB=\angle YCP=\angle YQP$ and so $P$ lies on the internal bisector of $\angle XQY$. Thus, by Lemma 1., $PO' \perp X_1Y_1$ and it is equivalent to prove that $PO \perp X_1Y_1$. Apply inversion at $P$ with arbitrary radius and denote by $T'$ the image of a point $T$ in the plane (except when we use the symbol $O$). The newly obtained problem is as follows: Inverted ProblemLet $A'B'C'D'$ be a cyclic quadrilateral with center $O$ and let the diagonals $A'C'$ and $D'B'$ meet at $P$. The lines through $D'$ and $C'$, parallel to $PC'$ and $PD'$, respectively meet at $Q'$. The circumcircles of $A'PD'$ and $B'PC'$ meet $Q'D'$ and $Q'C'$ at $X'$ and $Y'$, respectively and let the circumcircles of $PX'Q'$ and $PY'Q'$ meet $Q'C'$ and $Q'D'$ at $X_1'$ and $Y_1'$ respectively. Prove that the circles with diameter $OP$ and $PX_1'Y_1'$ are tangent to each other. Solution to inverted problemLet $M,N$ be the midpoints of $A'C'$ and $B'D'$, respectively and note that the points $M,N$ lie on the circle with diameter $OP$. Notice that $$\angle D'PX_1'=\angle PX_1'Q'=\angle PX'Q'=\angle PX'D'=\angle PA'D'=\angle PB'C'=\angle PY_1'Q'=\angle C'PY_1'$$and so lines $PX_1',PY_1'$ are isogonal in $\angle D'PC'$. Note that $PX_1'C'B'$ and $PY_1'D'A'$ are parallelograms and so by vectors, we have $X_1'Y_1' \parallel MN$. Thus, the circles $(PMN)$ and $(PX_1'Y_1')$ are tangent to each other at $P$ and the result holds. General caseNow, we proceed to the general case, when $ABCD$ is an arbitrary convex quadrilateral. Apply inversion at $P$ with arbitrary radius and denote by $T'$ the image of a point $T$ (except when we use the symbol $O$). We have the following equivalent problem: Inverted problem$A'B'C'D$ is a convex quadrilateral and the diagonals $A'C'$ and $B'D'$ meet at $P$. The circumcircles of $PD'C'$ and $PA'B'$ meet again at $E'$ and the circumcircles of $PA'D'$ and $PB'C'$ meet again at $F'$. The parallel from $D'$ to $PC'$ and the parallel from $C'$ to $PD'$ meet at $Q'$. The line $Q'D'$ meets the circumcircle of $PA'D'$ at $X'$ and the line $Q'C'$ meet the circumcircle of $PB'C'$ at $Y'$. If $O$ and $O'$ are the circumcenters of $PE'F'$ and $X'Q'Y'$, then prove that the points $P,O,O'$ are collinear. Solution to inverted problemLet $O_1,O_2,O_3,O_4$ be the circumcenters of triangles $PA'D', PD'C',PC'B',PB'A'$ respectively. Notice that $O_2,O_4$ lie on the perpendicular bisector of $PE'$ and $O_1,O_3$ lie on the perpendicular bisector of $PF'$. Thus, $O=O_1O_3 \cap O_2O_4$ and $O_1O_2O_3O_4$ is a parallelogram. Note that $PA'X'D'$ and $PB'Y'C'$ are isosceles trapezoids. Assume without loss of generality that $\angle PD'C'<90^{\circ}$ (we can do so as $\angle PD'C'+\angle PC'D'=180^{\circ}-\angle D'PC'<180^{\circ}$) and fix the points $P,B',C',D'$ and move point $A'$ with a constant velocity on the line $PC'$. As $A'$ moves with a constant velocity and $A'X'$ has a fixed direction; $X'$ moves with a constant velocity on $Q'D'$. Thus, $O'$ moves with a constant velocity on the perpendicular bisector of $Q'Y'$. Moreover, as $A'$ moves with a constant velocity, the points $O_1$ and $O_4$ also move with some constant velocities. Thus, the locus of $O$ is the line passing through the midpoint of $O_2O_3$, perpendicular to $PD'$ and it moves uniformly on that line. As the paths of $O'$ and $O$ are parallel and they moves uniformly, it suffices to prove that $P,O,O'$ are collinear for two positions of $A'$. Case 1. Let $A'$ be such that $A'B'C'D'$ is a cyclic quadrilateral. From the solution to the cyclic case of the original problem, we see that the conclusion holds here as well. Case 2. Let $A'=C'$. We obtain the following equivalent problem: Point $B'$ lies on line $PD'$ in $PD'C'$ with circumcenter $O_2$ and let $O_3$ be the circumcenter of $PB'C'$ and $O$ be the midpoint of $O_2O_3$. The parallel from $D'$ to $PC'$ and from $C'$ to $PD'$ meet at $Q'$ and let the line $Q'D'$ meet the circumcircle of $PD'C'$ at $X'$ and the circumcircle of $PB'C'$ meet $Q'C'$ at $Y'$. If $O'$ is the circumcenter of $X'Q'Y'$, then prove that $P,O,O'$ are collinear. Fix points $P,D',C'$ and vary $B'$ on $PD'$ uniformly. Note that $O_3$ moves uniformly on the perpendicular bisector of $PC'$ and so $O$ moves uniformly on the same line. As $B'Y'$ has a fixed direction ($PB'Y'C'$ is an isosceles trapezoid), $Y'$ also moves uniformly and so, $O'$ moves uniformly on the perpendicular bisector of $X'Q'$. As the lines traced by $O$ and $O'$ are parallel and the points $O$ and $O'$ move uniformly, it suffices to show that $P,O,O'$ are collinear for two positions of $B'$. Sub-case 1. When $B'=D'$. In this case, $Y'=(PD'C') \cap Q'C'$ and so the circumcenters of $PD'C'$ and $X'Q'Y'$ lie on the perpendicular bisector of $X'Y'$ and as $\angle PD'X'=\angle PC'Y'=180^{\circ}-\angle D'PC'$, $PX'=PY'$ and so, $P,O,O'$ are collinear. Sub-case 2. When $C'B'=C'D'$ and $B' \not=D'$. In this case, the radius of circles $PD'C'$ and $PC'B'$ is equal and so $O$ is the midpoint of $PC'$. Note that $PY'C'D'$ is a parallelogram and so, $C'Y'=PD'=C'Q'=C'X'$, thus $C'$ is the circumcenter of $X'Q'Y'$. Th result holds in this case as well. Moreover, such a point $B'$ necessarily exists as $\angle PD'C' \not=90^{\circ}$. All the arguments put together yield the desired conclusion.

While the presented solution is possibly not the nicest one and may appear to be very difficult to come up with; it is quite natural and no step is particularly tricky. The bulk of the solution comes from elaborating each step very carefully so as to not let the lack of diagrams be a reason for this being unclear (This is pretty much true for all solutions using dynamic geometry and inversion). I also believe that one can do the cyclic step without inverting, but again, that will be quite tricky. I hope the patient readers enjoyed!