Problem: Let $ABCD$ be a quadrilateral with $AC = BD$.
Diagonals $AC$ and $BD$ meet at $P$.
Let $\omega_1$ and $O_1$ denote the circumcircle and circumcenter of triangle $ABP$.
Let $\omega_2$ and $O_2$ denote the circumcircle and circumcenter of triangle $CDP$.
Segment $BC$ meets $\omega_1$ and $\omega_2$ again at $S$ and $T$
(other than $B$ and $C$), respectively.
Let $M$ and $N$ be the midpoints of minor arcs $\arc{SP}$ (not including $B$)
and $\arc{TP}$ (not including $C$).
Prove that $\ol{MN} \parallel \ol{O_1O_2}$.

Solution: Let $Q$ be the second intersection point of $\omega_1$, $\omega_2$.
Suffice to show $\ol{QP} \perp \ol{MN}$.
Now $Q$ is the center of a spiral \emph{congruence}
which sends $\ol{AC} \mapsto \ol{BD}$.
So $\triangle QAB$ and $\triangle QCD$ are similar isosceles triangles.
Now, \[ \dang QPA = \dang QBA = \dang DCQ = \dang DPQ \]
and so $\ol{QP}$ is bisects $\angle BPC$.

\begin{center}
\begin{asy}
size(12cm);
pair Q = dir(80);
pair B = dir(190);
pair C = dir(350);
pair A = shift(Q)*rotate(-40)*shift(-Q)*B;
pair D = shift(Q)*rotate(40)*shift(-Q)*C;
pair P = extension(A, C, B, D);
draw(circumcircle(A, B, P));
draw(circumcircle(C, D, P));
filldraw(A--B--C--D--cycle, opacity(0.1)+lightcyan, lightcyan);
filldraw(Q--A--C--cycle, opacity(0.1)+lightgreen, lightblue);
filldraw(Q--B--D--cycle, opacity(0.1)+lightgreen, lightblue);
pair O_1 = circumcenter(A, B, P);
pair O_2 = circumcenter(C, D, P);
pair S = -B+2*foot(O_1, B, C);
pair T = -C+2*foot(O_2, B, C);
pair M = circumcenter(S, P, incenter(B, S, P));
pair N = circumcenter(T, P, incenter(C, T, P));
draw(M--N, blue);
draw(O_1--O_2, blue);
pair I = incenter(P, B, C);
draw(Q--I, blue);
draw(B--I--C, blue);

dot("$Q$", Q, dir(Q));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$D$", D, dir(D));
dot("$P$", P, dir(60));
dot("$O_1$", O_1, dir(150));
dot("$O_2$", O_2, dir(10));
dot("$S$", S, dir(-90));
dot("$T$", T, dir(-90));
dot("$M$", M, dir(100));
dot("$N$", N, dir(35));
dot("$I$", I, dir(45));

/* TSQ Source:

Q = dir 80
B = dir 190
C = dir 350
A = shift(Q)*rotate(-40)*shift(-Q)*B
D = shift(Q)*rotate(40)*shift(-Q)*C
P = extension A C B D
circumcircle A B P
circumcircle C D P
A--B--C--D--cycle 0.1 lightcyan / lightcyan
Q--A--C--cycle 0.1 lightgreen / lightblue
Q--B--D--cycle 0.1 lightgreen / lightblue
O_1 = circumcenter A B P
O_2 = circumcenter C D P
S = -B+2*foot O_1 B C
T = -C+2*foot O_2 B C
M = circumcenter S P incenter B S P
N = circumcenter T P incenter C T P
M--N blue
O_1--O_2 blue
I = incenter P B C
Q--I blue
B--I--C blue

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


Now, let $I = \ol{BM} \cap \ol{CN} \cap \ol{PQ}$
be the incenter of $\triangle PBC$.
Then $IM \cdot IB = IP \cdot IQ = IN \cdot IC$, so $BMNC$ is cyclic,
meaning $\ol{MN}$ is antiparallel to $\ol{BC}$ through $\angle BIC$.
Since $\ol{QPI}$ passes through the circumcenter of $\triangle BIC$,
it follows now $\ol{QPI} \perp \ol{MN}$ as desired.