Problem: Let $ABCD$ be a quadrilateral inscribed in a circle with center $O$ and $E$ be
the intersection of segments $AC$ and $BD$. Let $\omega_1$ be the circumcircle
of $ADE$ and $\omega_2$ be the circumcircle of $BCE$. The tangent to
$\omega_1$ at $A$ and the tangent to $\omega_2$ at $C$ meet at $P$. The
tangent to $\omega_1$ at $D$ and the tangent to $\omega_2$ at $B$ meet at $Q$.
Show that $OP=OQ$.

Solution: \paragraph{Solution 1.}
Let $R = \ol{AD} \cap \ol{BC}$ (possibly at infinity,
but we'll see it's an Euclidean point later).
\begin{center}
  \begin{asy}
    size(12cm);
    pair A = dir(190), B = dir(350), C = dir(80), D = dir(120), O = (0, 0);
    pair E = extension(A, C, B, D), R = extension(A, D, B, C);
    pair P = reflect(O, (A+C)/2) * R;
    pair Q = reflect(O, (B+D)/2) * R;

    draw(A--C^^B--D, gray(0.5));
    draw(A--R^^C--P^^B--Q, heavycyan);
    draw(B--R^^D--Q^^A--P, heavymagenta);
    draw(O--P^^O--Q^^O--R, brown);
    filldraw(circumcircle(A, D, E), yellow+opacity(0.1), gray(0.5));
    filldraw(circumcircle(B, C, E), yellow+opacity(0.1), gray(0.5));
    draw(unitcircle);

    dot("$A$", A, dir(200));
    dot("$B$", B, dir(340));
    dot("$C$", C, dir(C));
    dot("$D$", D, dir(D));
    dot("$E$", E, 1.5*dir(190));
    dot("$O$", O, dir(270));
    dot("$P$", P, dir(P));
    dot("$Q$", Q, dir(Q));
    dot("$R$", R, dir(R));
  \end{asy}
\end{center}

\begin{claim*}
  $ACRP$ is an isosceles trapezoid with $\ol{AC} \parallel \ol{PR}$.
  Consequently, $R$ is an Euclidean point, and $OP = OR$.
\end{claim*}
\begin{proof}
  It is equivalent to show that $\triangle PAC \stackrel{-}{\cong} \triangle RCA$,
  or without using $R$, that
  \[
    \dang PAC = \dang ACR = \dang ACB, \quad \text{and} \quad
    \dang PCA = \dang CAR = \dang CAD.
  \]
  Indeed, we have
  \[ \dang PAC = \dang ADE = \dang ADB = \dang ACB \]
  and likewise $\dang PCA = \dang CAD$, as requested.
\end{proof}
Similarly $OQ = OR$, so we're done.

\paragraph{Solution 2.}
\begin{center}
  \begin{asy}
    unitsize(70);
    pair A = dir(130);
    pair B = dir(83);
    pair C = dir(-40);
    pair D = dir(-140);
    pair O = 0;
    pair E = extension(A,C,B,D);
    pair O1 = circumcenter(A,E,D);
    pair O2 = circumcenter(B,E,C);
    pair P = extension(A, (O1-A)*dir(90)+A, C, (O2-C)*dir(90)+C);
    pair Q = extension(D, (O1-D)*dir(90)+D, B, (O2-B)*dir(90)+B);
    pair Ap = 2*foot(O,A,P)-A;
    pair Bp = 2*foot(O,B,Q)-B;

    draw(unitcircle, lightblue);
    draw(A--C, green);
    draw(B--D, green);
    draw(circumcircle(A,E,D), dotted);
    draw(circumcircle(B,E,C), dotted);
    draw(A--P--C, red);
    draw(B--Q--D, red);
    draw(Bp--O--D, dashed+red);
    draw(Ap--O--C, dashed+red);

    dot("$A$", A, dir(A));
    dot("$B$", B, dir(B));
    dot("$C$", C, dir(C));
    dot("$D$", D, dir(-100));
    dot("$O$", O, 1.5*dir(-90));
    dot("$E$", E, dir(180));
    dot("$P$", P, dir(-90));
    dot("$Q$", Q, dir(-90));
    dot("$A'$", Ap, 1.5*dir(55));
    dot("$B'$", Bp, dir(Bp));
  \end{asy}
\end{center}

Let $PA$ intersect $(ABCD)$ again at $A'$ and let $QB$ intersect $(ABCD)$ again at $B'$.
The problem follows from the following claim.
\begin{claim*}
  $OA'PC\cong ODQB'$.
\end{claim*}
\begin{proof}
  We use directed angles mod $\pi$. Note that
  \[\dang ODQ = \dang ODA + \dang ADQ = \dang DAO + \dang PAD = \dang PAO = \dang OA'P,\]
  and similarly $\dang OCP = \dang OB'Q$. Furthermore,
  \[\dang A'AD = \dang AED = \dang CEB = \dang CBB',\]
  so $A'D = CB'$, i.e. $A'CDB'$ is an isosceles trapezoid.
  This means $\dang A'OC = \dang DOB'$.
  Combined with $OA'=OB'=OD=DC$, these three angle equalities imply the congruency.
\end{proof}

\paragraph{Solution 3 using complex numbers (Noah Walsh).}
Use complex numbers with $(ABCD)$ as the unit circle.
Because $PA$ is tangent to $(ADE)$, we have $\angle PAC = \angle ADB = \frac{1}{2}\angle AOB$,
and therefore
\begin{align*}
  \frac{\frac{p-a}{c-a}}{\frac{\ol{p} - a^{-1}}{c^{-1} - a^{-1}}} &= \frac{a}{b}\\
  \implies b\frac{p-a}{c-a} &= a\frac{ac\ol{p} - c}{a-c}\\
  \implies bp + a^2c\ol{p} &= ab + ac.\\
\end{align*}
The condition that $PC$ is tangent to $(BCE)$ can be expressed
by swapping $a$ with $c$ and $b$ with $d$, so we get
\begin{align*}
  bp + a^2c\ol{p} &= ab + ac\\
  dp + ac^2\ol{p} &= cd + ac \\
  \implies p(ad-bc) &= acd + a^2c - abc - ac^2\\
  \implies p &= \frac{ac(a+d-b-c)}{ad-bc}.
\end{align*}
Similarly,
\[ q = \frac{bd(a+d-b-c)}{ad-bc}. \]
It is now obvious that $p$ and $q$ have the same magnitude.

\begin{remark*}
  One possible trick one might use to compute $p$ is to compute the coordinates
  of the points $A', C' \in (ABCD)$ such that $AA'$ and $CC'$ are tangent
  to $(ADE)$ and $(BCE)$ respectively, and then use the chords intersection formula.

  If one does this, one gets $a' = ac/b$ and $c' = ac/d$.
  At this point, one can simply recognize that $A'$ and $C'$
  are the reflection of $B$ and $D$ over the perpendicular bisector of $AC$,
  reflect everything over said perpendicular bisector,
  and suddenly realize that one has solved the problem synthetically.
\end{remark*}

\paragraph{Solution 4 by linearity (Noah Walsh).}
If $AE = DE$ the problem is true by symmetry, so we assume $AE \neq DE$.

Fix $A$, $D$, and $E$, and move $B$ linearly along line $DE$.
By e.g.\ Reim's theorem, $C$ moves linearly as well.

How do we compute $O$? It lies on the perpendicular bisector of $AD$, which is fixed.
It also lies on the perpendicular bisector of $BC$,
which is parallel to a fixed line, and passes through the midpoint of $BC$, which moves linearly.
Therefore, $O$ moves linearly as well.

Point $P$ is the intersection of the tangent to $(ADE)$ at $A$, which is fixed,
and the tangent to $(BCE)$ at $C$, which moves linearly, and therefore is linear.
Similarly, $Q$ moves linearly.
Therefore, to verify that
$\ray{OP} \times \ray{OP} = \ray{OQ} \times \ray{OQ}$,
it suffices to check three cases.

The two cases where $EA = EB$ are trivial by symmetry.
We can also check that
$\lim_{B \to \infty_{DE}} \frac{\ray{OP}\times\ray{OP}}{\ray{OQ}\times\ray{OQ}} = 1$.

Indeed, obviously
$\lim_{B \to \infty_{DE}} \frac{|OP|}{|AP|} = \lim_{B \to \infty_{DE}} \frac{|OQ|}{|DQ|} = 1$.
In addition,
\begin{align*}
  \frac{dAP}{dDQ}
  &= \frac{dAP}{dAC} \times \frac{dAC}{dBD} \times \frac{dBD}{dBQ} \\
  &= \frac{\sin \angle ACP}{\sin \angle APC}
    \times \frac{dEC}{dEB} \times \frac{\sin \angle BQD}{\sin \angle BDQ} \\
  &= \frac{\sin \angle ACP}{\sin (\angle ACP + \angle CAP)}
    \times \frac{\sin \angle EBC}{\sin \angle ECB}
    \times \frac{\sin (\angle BDQ + \angle DBQ)}{\sin \angle BDQ} \\
  &= \frac{\sin\angle BCE}{\sin(\angle BCE + \angle ADE)}
    \times \frac{\sin\angle DAE}{\sin\angle ADE}
    \times \frac{\sin(\angle DAE + \angle ECB)}{\sin\angle EAD} \\
  &=\frac{\sin(\angle DAE + \angle ECB)}{\sin(\angle BCE + \angle ADE)} \\
  &= 1.
\end{align*}
It follows that $\lim_{B \to \infty_{DE}}\frac{|AP|}{|DQ|} = 1$.
Therefore,
\begin{align*}
  \lim_{B \to \infty_{DE}}\frac{|OP|}{|OQ|}
  &= \lim_{B \to \infty_{DE}}\frac{|OP|}{|AP|}
    \times \lim_{B \to \infty_{DE}}\frac{|AP|}{|DQ|}
    \times \lim_{B \to \infty_{DE}} \frac{|DQ|}{|OQ|}\\
  &= 1,
\end{align*}
which completes the $B \to \infty_{DE}$ case, and we are done.