Problem: Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\omega$ and point $P$ lies outside the circle.
The given points are such that
(i) lines $PB$ and $PD$ are tangent to $\omega$,
(ii) $P$, $A$, $C$ are collinear,
and (iii) $\ol{DE} \parallel \ol{AC}$.
Prove that $\ol{BE}$ bisects $\ol{AC}$.

Solution: We present two solutions.

\paragraph{First solution using harmonic bundles.}
Let $M = \ol{BE} \cap \ol{AC}$
and let $\infty$ be the point at infinity
along $\ol{DE} \parallel \ol{AC}$.
\begin{center}
\begin{asy}
size(8cm);
pair B = dir(100);
pair D = dir(210);
pair E = dir(330);
pair P = 2*B*D/(B+D);
pair A = OP(P--(P+8*(E-D)), unitcircle);
pair M = extension(B, E, A, P);
pair C = 2*M-A;

filldraw(unitcircle, opacity(0.1)+lightblue, lightblue);
draw(P--A, lightblue);
draw(B--P--D, lightblue);
draw(D--E, heavygreen);
draw(B--E, heavygreen);
draw(A--E--C, heavygreen);

draw(B--A--D--C--cycle, heavycyan);

dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$P$", P, dir(P));
dot("$A$", A, dir(A));
dot("$M$", M, dir(45));
dot("$C$", C, dir(100));

/* TSQ Source:

B = dir 100
D = dir 210
E = dir 330
P = 2*B*D/(B+D)
A = IP P--(P+8*(E-D)) unitcircle
M = extension B E A P R45
C = 2*M-A R100

unitcircle 0.1 lightblue / lightblue
P--A lightblue
B--P--D lightblue
D--E heavygreen
B--E heavygreen
A--E--C heavygreen

B--A--D--C--cycle heavycyan

*/
\end{asy}
\end{center}
Note that $ABCD$ is harmonic, so
\[ -1 = (AC;BD) \overset{E}{=} (AC;M\infty) \]
implying $M$ is the midpoint of $\ol{AC}$.


\paragraph{Second solution using complex numbers (Cynthia Du).}
Suppose we let $b$, $d$, $e$ be free on unit circle,
so $p = \frac{2bd}{b+d}$.
Then $d/c = a/e$, and $a + c = p + ac \ol p$.
Consequently,
\begin{align*}
  ac &= de \\
  \half(a + c) &= \frac{bd}{b+d} + de \cdot \frac{1}{b+d}
    = \frac{d(b+e)}{b+d}. \\
  \frac{a+c}{2ac} &= \frac{(b+e)}{e(b+d)}.
\end{align*}
From here it's easy to see
\[ \frac{a+c}{2} + \frac{a+c}{2ac} \cdot be = b + e \]
which is what we wanted to prove.