Problem: Let $ABCD$ be a cyclic quadrilateral.
Prove that there exists a point $X$ on segment $\ol{BD}$
such that $\angle BAC=\angle XAD$ and $\angle BCA=\angle XCD$
if and only if there exists a point $Y$ on segment $\ol{AC}$
such that $\angle CBD=\angle YBA$ and $\angle CDB=\angle YDA$.

Solution: Both conditions are equivalent to $ABCD$ being harmonic.

Here is a complex solution. Extend $U$ and $V$ and shown. Thus $u = bd/a$ and $v = bd/c$.

\begin{center}
\begin{asy}
  size(4.5cm);
  pair A = Drawing("A", dir(110), dir(110));
  pair B = Drawing("B", dir(160), dir(160));
  pair D = Drawing("D", dir( 20), dir( 20));
  pair C = Drawing("C", dir(240), dir(240));
  draw(unitcircle);
  draw(A--B--C--D--cycle);
  draw(B--D);
  pair U = -conj(A);
  pair V = -conj(C);
  Drawing("U", U, U);
  Drawing("V", V, V);
  draw(C--U);
  draw(A--V);
  Drawing("X", extension(C,U,A,V), dir(0));
  draw(A--U, dashed);
  draw(C--V, dashed);
  draw(A--C);
\end{asy}
\end{center}

Note $\ol{AV} \cap \ol{CU}$ lies on the
perpendicular bisector of $\ol{BD}$ unconditionally.
Then $X$ exists as described if and only if
the midpoint of $\ol{BD}$ lies on $\ol{AV}$.
In complex numbers this is $a + v = m + av \ol m$, or
\[
  a + \frac{bd}{c} = \frac{b+d}{2} + \frac{abd}{c} \cdot \frac{b+d}{2bd}
  \iff 2(ac+bd) = (b+d)(a+c)
\]
which is symmetric.