Problem: Let $ABCD$ be a cyclic quadrilateral,
and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively.
Let $W$, $X$, $Y$ and $Z$ be the orthocenters of
triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively.
Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.

Solution: The following solution is due to Grace Wang.

We begin with:
\begin{claim*}
  Point $W$ has coordinates $\half(2a+b+d)$.
\end{claim*}
\begin{proof}
  The orthocenter of $\triangle DAB$ is $d+a+b$,
  and $\triangle AHE$ is homothetic to $\triangle DAB$
  through $A$ with ratio $1/2$.
  Hence $w = \half(a+(d+a+b))$ as needed.
\end{proof}

By symmetry, we have
\begin{align*}
  w &= \half(2a+b+d) \\
  x &= \half(2b+c+a) \\
  y &= \half(2c+d+b) \\
  z &= \half(2d+a+c).
\end{align*}
We see that $w-y = a-c$, $x-z = b-d$.
So the diagonals of $WXYZ$ have the same length as those
of $ABCD$ as well as the same directed angle between them.
This implies the areas are equal, too.