Problem: Let $ABCD$ be a cyclic quadrilateral,
and let diagonals $AC$ and $BD$ intersect at $X$.
Let $C_1,D_1$ and $M$ be the midpoints of segments $CX$, $DX$ and $CD$, respectively.
Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$
at different points $E$ and $F$, respectively.
Prove that line $XY$ is tangent to the circle through $E$, $F$ and $X$.

Solution: We present two approaches.

\paragraph{First approach through isogonality lemma.}
Note $ABC_1D_1$ is cyclic.
By the ``isogonality lemma'' applied to $\triangle YC_1D_1$,
it follows that $YX$ and $YM$ are isogonal
with respect to $\triangle YC_1D_1$.
Then
\[ \angle EXY = \angle XC_1Y + \angle C_1YX
= \angle AD_1X + \angle MYA = \angle YFX \]
and we're done (angles are directed).
\begin{remark*}
Note that the points $C$ and $D$ can be more or less deleted immediately.
The ``isogonality lemma'' people has mentioned has appeared in other places too.
In addition to the Taiwan TST 2014 round 2 problem 6, it also shows itself in
ELMO 2012/5, % http://aops.com/community/c6h486936p2728469
BrMO2 2013/2, % http://www.aops.com/community/c6h518987
SL 2009 G4, SL 2012 G2.
\end{remark*}

\paragraph{Solution with complex numbers, by Sanjana Das.}
Quadrilateral $ABC_1D_1$ is also cyclic ($\dang AC_1D_1 = \dang ACD = \dang ABD_1$).
We want $\dang YXE = \dang XFE$,
or \[\dang(XY, AC) = \dang(BD, MY).\]

Now use complex numbers with unit circle $(ABC_1D_1)$
(using $c$ and $d$ to refer to $C_1$ and $D_1$ for berevity).

We want to check that $\frac{(x - y)(m - y)}{(a - c)(b - d)}$ is real.
First, if $Z = AB \cap C_1D_1$, then $XY \perp OZ$ where $O$ is the
circumcenter of $(ABC_1D_1)$ by Brocard's Theorem, so we want to show
\[ \frac{(ab(c + d) - cd(a + b))\cdot (y - m)}
  {(a - c)(b - d)(ab - cd)} \in i\RR.\]
Now we have $m = c + d - x$, which means
\begin{align*}
    y - m &= x + y - c - d \\
    &= \frac{ac(b +d) - bd(a +c)}{ac - bd} + \frac{ad(b + c) - bc(a+ d)}{ad - bc} - c - d \\
    &= \frac{ab(2acd + 2bcd + c^3 + d^3 - ac^2 - ad^2 - bc^2 - bd^2 - c^2d - cd^2)}{(ac - bd)(ad - bc)} \\
    &= -\frac{ab(a + b - c - d)(c - d)^2}{(ac - bd)(ad - bc)}.
\end{align*}
This means our final expression is
\[-\frac{ab(ab(c + d) - cd(a + b))(a + b - c - d)(c - d)^2}{(ac - bd)(ad - bc)(a - c)(b - d)(ab - cd)},\]
which is visibly the negative of its conjugate and is therefore imaginary.