Problem: Let $ABCD$ be a convex cyclic quadrilateral with
$E = \ol{AC} \cap \ol{BD}$, $F = \ol{AB} \cap \ol{CD}$,
$G = \ol{DA} \cap \ol{BC}$.
The circumcircle of $\triangle ABE$
intersects line $CB$ at $B$ and $P$,
and the circumcircle of $\triangle ADE$
intersects line $CD$ at $D$ and $Q$.
Assume $C$, $B$, $P$, $G$
and $C$, $Q$, $D$, $F$ are collinear in that order.
Let $M = \ol{FP} \cap \ol{GQ}$.
Prove that $\angle MAC = 90\dg$.

Solution: We present three general routes.
(The second route, using the fact that $\ol{AC}$
is an angle bisector, has many possible variations.)

\paragraph{First solution (Miquel points).}
This is indeed a Miquel point problem,
but the main idea is to focus on the self-intersecting
cyclic quadrilateral $PBQD$ as the key player,
rather than on the given $ABCD$.

Indeed, we will prove that $A$ is its Miquel point;
this follows from the following two claims.
\begin{claim*}
  The self-intersecting quadrilateral $PQDB$ is cyclic.
\end{claim*}
\begin{proof}
  By power of a point from $C$:
  $CQ \cdot CD = CA \cdot CE = CB \cdot CP$.
\end{proof}
\begin{claim*}
  Point $E$ lies on line $PQ$.
\end{claim*}
\begin{proof}
  $\dang AEP = \dang ABP = \dang ABC = \dang ADC
  = \dang ADQ = \dang AEQ$.
\end{proof}

\begin{center}
\begin{asy}
pair P = dir(240);
pair B = dir(300);
pair Q = dir(20);
pair D = dir(50);
pair E = extension(Q, P, B, D);
pair C = extension(D, Q, B, P);
pair H = extension(D, P, B, Q);
pair A = foot(H, E, C);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);

pair G = extension(D, A, B, C);
pair F = extension(B, A, C, D);

filldraw(circumcircle(P, A, E), opacity(0.1)+green, heavygreen);
filldraw(circumcircle(Q, A, E), opacity(0.1)+green, heavygreen);
filldraw(circumcircle(A, B, C), opacity(0.05)+yellow, heavygreen+dotted);
draw(B--F--C, gray);
draw(D--G--C, gray);
draw(P--H--B, blue);
draw(P--B--Q--D--cycle, blue);
draw(A--C, gray);
draw(P--Q, blue);
draw(B--D, blue);

pair M = extension(P, F, G, Q);
draw(F--P, heavycyan);
draw(G--Q, heavycyan);

dot("$P$", P, dir(P));
dot("$B$", B, dir(B));
dot("$Q$", Q, dir(Q));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$C$", C, dir(C));
dot("$H$", H, dir(H));
dot("$A$", A, dir(E-C));
dot("$G$", G, dir(G));
dot("$F$", F, dir(F));
dot("$M$", M, dir(B-M));

/* TSQ Source:

P = dir 240
B = dir 300
Q = dir 20
D = dir 50
E = extension Q P B D
C = extension D Q B P
H = extension D P B Q
A = foot H E C RE-C
unitcircle 0.1 lightcyan / lightblue

G = extension D A B C
F = extension B A C D

circumcircle P A E 0.1 green / heavygreen
circumcircle Q A E 0.1 green / heavygreen
circumcircle A B C 0.05 yellow / heavygreen dotted
B--F--C gray
D--G--C gray
P--H--B blue
P--B--Q--D--cycle blue
A--C gray
P--Q blue
B--D blue

M = extension P F G Q
F--P heavycyan
G--Q heavycyan

*/
\end{asy}
\end{center}

To finish, let $H = \ol{PD} \cap \ol{BQ}$.
By properties of the Miquel point, we have $A$ is the
foot from $H$ to $\ol{CE}$.
But also, points $M$, $A$, $H$ are collinear
by Pappus theorem on $\ol{BPG}$ and $\ol{DQF}$, as desired.

\paragraph{Second solution (projective).}
We start with a synthetic observation.
\begin{claim*}
  The line $\ol{AC}$ bisects $\angle PAD$ and $\angle BAQ$.
\end{claim*}
\begin{proof}
  Angle chase:
  $\dang PAC = \dang PAE = \dang PBE = \dang CBD = \dang CAD$.
\end{proof}

There are three ways to finish from here:
\begin{itemize}
  \ii (Michael Kural) Suppose the external bisector
  of $\angle PAD$ and $\angle BAQ$
  meet lines $BC$ and $DC$ at $X$ and $Y$.
  Then \[ -1 = (GP;XC) = (FD;YC) \]
  which is enough to imply that $\ol{XY}$, $\ol{GQ}$, $\ol{PF}$
  are concurrent (by so-called prism lemma).

  \ii (Daniel Liu) Alternatively,
  apply the dual Desargues involution theorem
  to complete quadrilateral $GQFPCM$, through the point $A$.
  This gives that an involutive pairing of
  \[ (AC,AM) \; (AP,AQ) \; (AG, AF). \]
  This is easier to see if we project it onto the
  line $\ell$ through $C$ perpendicular to $\ol{AC}$:
  let $P'$, $Q'$, $G'$, $F'$ be the images of the last four lines,
  so $C$ is the common midpoint of $\ol{F'Q'}$ and $\ol{G'P'}$.
  Hence the involution swapping $G' \leftrightarrow F'$ and $P' \leftrightarrow Q'$
  coincides with negative inversion through $C$ with power $\sqrt{CP' \cdot CQ'}$
  which implies that $\ol{AM} \cap \ell$ is an infinity point, as desired.

  \ii (Kada Williams) The official solution instead shows the external angle
  bisector by a long trig calculation.
\end{itemize}

\paragraph{Third solution (inversion, Andrew Wu).}
Noting that $CE \cdot CA = CP \cdot CB = CQ \cdot CD$,
we perform an inversion at $C$ swapping these pairs of points.
The point $G$ is mapped to a point $G^\ast$ ray $CB$
for which $QEG^\ast C$ is cyclic, but then
\[ \dang CG^\ast E = \dang CQE = \dang CQP = \dang DBC = \dang CBE \]
and so we conclude $EB = EG^\ast$.
Similarly, $ED = EF^\ast$.

Finally, $M^\ast = (CG^\ast D) \cap (CF^\ast B) \neq C$,
and we wish to show that $\angle EM^\ast C = 90\dg$.

\begin{center}
\begin{asy}
pair D = dir(100);
pair B = dir(220);
pair C = dir(320);
pair E = 0.3*D+0.7*B;
pair K = foot(E, B, C);
pair L = foot(E, D, C);
pair Gs = 2*K-B;
pair Fs = 2*L-D;
pair Ms = (D*B-Fs*Gs)/(D+B-Fs-Gs);

filldraw(circumcircle(Gs, D, C), opacity(0.05)+yellow, orange);
filldraw(circumcircle(Fs, B, C), opacity(0.05)+yellow, orange);

fill(B--C--D--cycle, opacity(0.1)+yellow);
draw(B--C--D, orange);
draw(B--D, red);
draw(Gs--E--K, red);
draw(Fs--E--L, red);
filldraw(circumcircle(E, L, K), opacity(0.1)+red, lightred+dashed);

dot("$D$", D, dir(D));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$E$", E, dir(E));
dot("$K$", K, dir(K));
dot("$L$", L, dir(L));
dot("$G^\ast$", Gs, dir(Gs));
dot("$F^\ast$", Fs, dir(30));
dot("$M^\ast$", Ms, dir(Ms));

/* TSQ Source:

D = dir 100
B = dir 220
C = dir 320
E = 0.3*D+0.7*B
K = foot E B C
L = foot E D C
G* = 2*K-B
F* = 2*L-D R30
M* = (D*B-Fs*Gs)/(D+B-Fs-Gs)

circumcircle Gs D C 0.05 yellow / orange
circumcircle Fs B C 0.05 yellow / orange

!fill(B--C--D--cycle, opacity(0.1)+yellow);
B--C--D orange
B--D red
Gs--E--K red
Fs--E--L red
circumcircle E L K 0.1 red / lightred dashed

*/
\end{asy}
\end{center}

Note that $M^\ast$ is the center of the
spiral similarity sending $\ol{BG^\ast}$ to $\ol{F^\ast D}$.
Hence it also maps the midpoint $K$ of $BG^\ast$
to the midpoint $L$ of $\ol{F^\ast E}$.
Consequently, $M^\ast$ lies on the circumcircle $KLC$ as well.
In other words, $ELCKM^\ast$ is a cyclic pentagon
with circumdiameter $\ol{CE}$, as desired.