Problem: Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane
of triangle $ABC$) such that the intersection of the interior of
triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has
area more than $\frac23$ the area of triangle $ABC$.

Solution: All that's needed is:
\begin{claim*}
  If $ABC$ is a triangle where $\half < \frac{AB}{AC} < 1$, then
  the $\angle A$ bisector works.
\end{claim*}
\begin{proof}
  Let the $\angle A$-bisector meet $BC$ at $D$.
  The overlapped area is $2[ABD]$ and
  \[ \frac{[ABD]}{[ABC]} = \frac{BD}{BC} = \frac{AB}{AB+AC} \]
  by angle bisector theorem.
\end{proof}

In general, suppose $x < y < z$ are sides of a triangle.
Then $\half < \frac yz < 1$ by triangle inequality as needed.