Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$ which is tangent to the sides $AB, BC, CD, DA$ at points $P, Q, R, S$ respectively. We want to show that $AC, BD, PR, QS$ meet at one point. Let $PR ∩ QS = X$. Let $CX$ meet the circle again at points $G$ and $H$.
Describe an inversion which maps $P → R, Q → S$. What happens to $G$ and $H$?
I might be getting confused here but we must have the centre of inversion on the lines $PR$ and $QS$, which gives us point $X$. But $X$ is clearly impossible to be the centre of inversion. Was there some flaw in my logic? Any help is appreciated.
Geogebra screenshot
In an inversion of a given power $K$ and center $X$ we have two cases depending on the sign of $K$. In case of a positive $K$, we transform a point $P$ into $P^*$ with $XP\cdot XP^*=K$, and $P,P^*$ are on the same ray from $X$. It is useful to think of $XP$ and $XP^*$ as signed distances for some or any direction chosen on the line $XP$.
In this sense, if $K<0$, the points $P,P^*$ are on the line $XP=XP^*$, but on different / opposite rays.
In our case, since the points $P,Q,R,S$ are concyclic, we have - using only positive lengths $$ |XP|\cdot |XR| = |XQ|\cdot |XS|:=K\ . $$ Using also signs, we have: $$ XP\cdot XR=XQ\cdot XS=-K\ . $$ We use this $-K$ as power for an inversion. Then $P^*=R$, $R^*=P$ and $Q^*=S$, $S^*=Q$. The circle $\odot(PQRS)$ is thus invariated as a set of points.
In the notations of the $OP$, we have $G^*=H$ and $H^*=G$, since the points $G,H,X$ are colinear, and $G,H$ are on the invariated circle on the one and the other side of the point $X$.
To illustrate further properties of this inversion (denoted by $*$), let us consider further lines and circles and their transforms:
Let $\Xi$ the point of intersection of the circles $\odot(XPQ)$ and $\odot(XRS)$. Then the circle $\odot(XPQ)$ through the center $X$ of inversion is transformed in the line $X^* P^*Q^*=\infty RS$. Similarly, the circle $\odot(XRS)$ is transformed in the line $PQ$. The second intersection of the two circles, is a point which is transformed in a point on the intersection of the lines $RS$ and $PQ$, which is not $X^*=\infty$, so this point is $$ \Xi ^*=PQ\cap RS\ . $$
It may be a fun experiment to apply the Pascal theorem on $QQSRRP$. Then the three intersections $QQ\cap RR=C$, $QS\cap RP=X$, $SR\cap PQ=\Xi^*$ are colinear. This shows $C,X,\Xi^*$ colinear.
With a cyclic permutation of variables so that $ABCD$ becomes $CDAB$ we apply Pascal's theorem on $SSQPPR$. So the three intersections $SS\cap PP=A$, $SQ\cap PR=X$, $QP\cap RS=\Xi^*$ are colinear. This shows $A,X,\Xi^*$ colinear.
In this moment we already have five points on a line, $X,\Xi,\Xi^*,A,C$. So $AC$ passes through $X$. In a similar manner, $BC$ also passes through $X$. This would be one way to conclude.
One can show that $\odot(X\Xi O\Upsilon)$ is a circle with diameter $XO$.
An other approach is to consider the transforms $A^*,B^*,C^*,D^*$ also in the picture:
We consider the point $A=SS\cap PP$. (Here $SS$ and $PP$ are the tangents in $S,P$ of the given circle $\omega=\odot(PQRS)$.) Its transform $A^*$ is then on the transformed curves. $SS$ goes into a circle through $X$ and $S^*=Q$, which is tangent to $\omega$ in $Q$. But at this point $\omega$ is also tangent to $BQC$. So we obtain the circle $\odot(XA^*Q)$ tangent in $Q$ to $BC$. Similarly, the circle $\odot(XA^*R)$ is tangent in $R$ to $CD$. But the circle through $X$ tangent in $R$ to $CD$ is uniquely determined, its center is on $OR$ and the perpendicular bisector of $XR$. So $B^*$ is also on this circle. (Use cyclic permutation of the letters and the same argument.)
So we obtain the above four circles: $\odot(XA^*RB^*)$, $\odot(XB^*SC^*)$, $\odot(XC^*PD^*)$, $\odot(XD^*QA^*)$.
Denote by $O_P\in OP$, $O_Q\in OQ$, $O_R\in OR$, $O_S\in OS$ the four centers of the four circles, the index is the touching point with the given quadrilateral.
The triangles $\Delta ORP$ and $\Delta O_PXP$ are isosceles in $O,O_P$, and share the same angle $P$. So they are similar, this gives $OP\| XO_R$. In a similar manner $OQ\| XO_S$, $OR\|XO_P$, $OS\|XO_R$. We obtain two parallelograms, $OO_PXO_R$ and $OO_QXO_S$, with intersection of diagonals in their mid point, the middle of $OX$, denoted by $\Pi$. This makes $O_PO_QO_RO_S$ also a parallelogram with intersection of diagonals in their mid point $\Pi$.
The points $X,A^*$ are on the two circles $\odot(O_Q)=\odot(XQA^*)$ and $\odot(O_R)=\odot(XRA^*)$, so the radical axis $XA^*$ is perpendicular on the centers line. We obtain: $$ XA^*\perp O_QO_R\ \|\ O_SO_P\perp XC^*\ . $$ So the lines $XA^*$ and $XC^*$ have the same direction, the direction perpendicular on $O_QO_R\|O_SO_P$, they have also a common point, so they coincide. We obtain the collinearity of $X,A^*C^*$, and thus of $X;A,A^*;C,C^*$.
$\square$