Problem: Trapezoid $ABCD$, with $\ol{AB} \parallel \ol{CD}$,
is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$.
Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively.
Let the line through $G$ parallel to $\ol{AB}$ intersect
$\ol{BD}$ and $\ol{BC}$ at points $R$ and $S$, respectively.
Prove that quadrilateral $PQRS$ is cyclic
if and only if $\ol{BG}$ bisects $\angle CBD$.

Solution: Perform an inversion around $B$ with arbitrary radius,
and denote the inverse of a point $Z$ with $Z^\ast$.

\begin{center}
  \begin{asy}
    unitsize(2cm);
    pair A = Drawing("A", dir(130), dir(130));
    pair B = Drawing("B", dir(50), dir(50));
    pair C = Drawing("C", dir(-20), dir(-20));
    pair D = Drawing("D", dir(200), dir(200));
    pair Q = Drawing("Q", dir(270), dir(270));
    pair G = Drawing("G", 0.3*Q+0.7*B, 1.414*dir(290));
    pair R = Drawing("R", extension(B,D,G,G-(1,0)), dir(135));
    pair S = Drawing("S", extension(B,C,G,G+(1,0)), dir(45));
    pair P = Drawing("P", 2*foot(origin,A,G)-A, dir(G-A));
    draw(unitcircle);
    draw(D--A--B--C--D--B--Q--P--A);
    draw(R--S);
    draw(circumcircle(P,Q,R), dashed);
  \end{asy}
  \quad
  \begin{asy}
    unitsize(2cm);
    pair B = Drawing("B", dir(130), dir(110));
    pair R = Drawing("R^\ast", dir(210), dir(210));
    pair S = Drawing("S^\ast", dir(330), dir(330));
    pair G = Drawing("G^\ast", dir(270), dir(290));
    pair D = Drawing("D^\ast", .6*R+.4*B, 1.414*dir(230));
    pair C = Drawing("C^\ast", .6*S+.4*B, 1.414*dir(-90));
    pair Q = Drawing("Q^\ast", extension(B,G,C,D), dir(45));
    pair A = Drawing("A^\ast", extension(B,2/(B+conj(B)),C,D), dir(180));
    pair P = Drawing("P^\ast", 2*foot(circumcenter(A,B,G),C,D)-A,dir(-25));
    draw(unitcircle);
    draw(circumcircle(B,C,D));
    draw(A--P);
    draw(R--B--S--cycle);
    draw(G--B);
    draw(circumcircle(A,B,G));
    draw(A--B);
    /*
    pair X = Drawing("X", 2*foot(origin,P,G)-G, dir(P-G));
    draw(G--X, dashed);
    */
  \end{asy}
\end{center}

After inversion, we obtain a cyclic quadrilateral
$BS^\ast G^\ast R^\ast$ and points $C^\ast$, $D^\ast$
on $\ol{BS^\ast}$, $\ol{BR^\ast}$,
such that $(BC^\ast D^\ast)$ is tangent to $(BS^\ast G^\ast R^\ast)$ ---
in other words, so that $\ol{C^\ast D^\ast}$
is parallel to $\ol{S^\ast R^\ast}$.
Point $A^\ast$ lies on line $\ol{C^\ast D^\ast}$
so that $\ol{A^\ast B}$ is tangent to $(BS^\ast G^\ast R^\ast)$.
Points $P^\ast$ and $Q^\ast$ are the
intersections of $(A^\ast BG^\ast)$ and $\ol{BG^\ast}$
with line $C^\ast D^\ast$.

Observe that $P^\ast Q^\ast R^\ast S^\ast$ is a trapezoid,
so it is cyclic if and only if it isosceles.

Let $X$ be the second intersection of line $G^\ast P^\ast$
with $(B S^\ast R^\ast)$.  Because
\[ \dang Q^\ast P^\ast G^\ast
  = \dang A^\ast B G^\ast = \dang BXG^\ast \]
we find that $BXS^\ast R^\ast$ is an isosceles trapezoid.

If $G^\ast$ is indeed the midpoint of the arc
then everything is clear by symmetry now.
Conversely, if $P^\ast R^\ast = Q^\ast S^\ast$ then
that means $P^\ast Q^\ast R^\ast S^\ast$ is a cyclic trapezoid,
and hence that the perpendicular bisectors of
$\ol{P^\ast Q^\ast}$ and $\ol{R^\ast S^\ast}$ are the same.
Hence $B$, $X$, $P^\ast$, $Q^\ast$ are symmetric around this line.
This forces $G^\ast$ to be the midpoint of arc $R^\ast S^\ast$ as desired.