Problem: Let $ABC$ be an equilateral triangle and $P$ a point on its circumcircle.
Set $D = \ol{PA} \cap \ol{BC}$, $E = \ol{PB} \cap \ol{CA}$, $F = \ol{PC} \cap \ol{AB}$.
Prove that the area of triangle $DEF$ is twice the area of triangle $ABC$.

Solution: \paragraph{First solution (barycentric).}
We invoke barycentric coordinates on $ABC$.
Let $P = (u:v:w)$, with $uv+vw+wu = 0$ (circumcircle equation with $a=b=c$).
Then $D = (0:v:w)$, $E = (u:0:w)$, $F = (u:v:0)$.
Hence
\begin{align*}
  \frac{[DEF]}{[ABC]}
  &= \frac{1}{(u+v)(v+w)(w+u)}
  \det
  \begin{bmatrix}
    0 & v & w \\
    u & 0 & w \\
    u & v & 0
  \end{bmatrix}
  \\
  &= \frac{2uvw}{(u+v)(v+w)(w+u)} \\
  &= \frac{2uvw}{(u+v+w)(uv+vw+wu)-uvw} \\
  &= \frac{2uvw}{-uvw} = -2
\end{align*}
as desired (areas signed).

\paragraph{Second solution (``nice'' lengths).}
WLOG $ABPC$ is convex.
Let $x = AB = BC = CA$.
By Ptolemy's theorem and strong Ptolemy,
\begin{align*}
  PA &= PB + PC \\
  PA^2 &= PB \cdot PC + AB \cdot AC = PB \cdot PC + x^2 \\
  \implies x^2 &+ PB^2 + PB \cdot PC + PC^2.
\end{align*}
Also, $PD \cdot PA = PB \cdot PC$ and similarly
since $\ol{PA}$ bisects $\angle BPC$ (causing $\triangle BPD \sim \triangle APC$).

Now $P$ is the Fermat point of $\triangle DEF$,
since $\angle DPF = \angle FPE = \angle EPD = 120\dg$. Thus
\begin{align*}
  [DEF] &= \frac{\sqrt3}{4} \sum_{\text{cyc}} PE \cdot PF \\
  &= \frac{\sqrt3}{4} \sum_{\text{cyc}}
  \left( \frac{PA \cdot PC}{PB} \right)
  \left( \frac{PA \cdot PB}{PC} \right) \\
  &= \frac{\sqrt3}{4} \sum_{\text{cyc}} PA^2 \\
  &= \frac{\sqrt3}{4} \left( (PB+PC)^2 + PB^2 + PC^2 \right) \\
  &= \frac{\sqrt3}{4} \cdot 2x^2 = 2[ABC].
\end{align*}