<s>[INST] Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$ Let $\triangle ABC$  be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ , let lines $PB$ and $CA$ intersect at $E$, and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF $ is twice the area of $\triangle ABC$.

I had used my method and it required that $\triangle DEF $ have to be isosceles triangle which means $ P $ should be the mid point of $\widehat {BC}$ or $\widehat{AC}$ or $\widehat {BC}$. But the problem does not said so. So, how to solve it correctly? 
 [/INST] It is very interesting to solve this problem by only using our greek friends Pythagoras and Menelaus, together with some simple algebra.

Initial Observations
Without loss of generality assume that $P$ lies on the half-plane determined by $BC$ and not containing $A$. Note that there is no need to use the circle, once we fix $\angle BPC = 120°$. In the figure above I added point $G$ given by the intersection of line $FD$ with $AC$. Suppose also $\overline{AB}=1$.
Once $\overline{BF} = x$ is chosen, the entire Figure is defined. We aim therefore at showing that, independently of $x$,
$$[DEF] = 2[ABC].$$
Characterization of $\triangle BFC$
Below the triangle in question has been isolated, to better help you in the demontrations. 

Recall that $\overline{BC} = 1$ and $\overline{BF} = x$. 
Use Pythagorean Theorem on $\triangle CKF$ in oder to determine
$$\overline{CF} = \sqrt{x^2+x+1}.$$
From the similarity 
$$\triangle BPC \sim \triangle BCF,$$ 
show that
$$\overline{CF}\cdot\overline{CP} = 1,$$
so that, in the end, you have
$$
\begin{cases}
\overline{CF} = \sqrt{x^2+x+1},\\
\overline{CP} = \frac{1}{\sqrt{x^2+x+1}},\\
\overline{FP} = \frac{x(x+1)}{\sqrt{x^2+x+1}}.
\end{cases}
$$
Four Applications of Menelaus's Theorem (MT)
MT on $\triangle ACF$ with the line $BE$, gives
$$\frac{\overline{CE}}{\overline{AE}} = \frac{1}{x+1}.$$
Together with $\overline{AE}-\overline{CE} = 1$ this leads to
$$\overline{CE} = \frac{1}{x}$$
and
$$\overline{AE} = \frac{x+1}{x}.$$
Similarly, MT on $\triangle BFC$ and line $AP$, with the known fact that $\overline{BD} + \overline{CD} = 1$ yields
$$\overline{CD} = \frac{1}{x+1}$$
and
$$\overline{DB} = \frac{x}{x+1}.$$
MT on $\triangle ABC$ and line $FG$, with $\overline{AG} + \overline{CG} = 1$, gives you
$$ \overline{AG} = \frac{x+1}{x+2}$$
and
$$ \overline{CG} = \frac{1}{x+2}.$$
Finally, MT on $\triangle CGF$ with line $AP$ yields
$$ \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$ 
Areas Computation
Now we only need to compute areas of triangles with fixed altitude and a given ratio between bases.
Firstly we have
$$[ACF] = [ABC](1+x).$$
Then
$$[AFE] = [ACF] \frac{\overline{AE}}{\overline{AC}},$$
that is
$$[AFE] = [ABC]\frac{(x+1)^2}{x}.$$
We also have
$$[GFE] = [AFE]\frac{\overline{GE}}{\overline{AE}},$$
yielding
$$ [GFE] =2[ABC]\frac{(x+1)^2}{x(x+2)}.$$
Finally observe that
$$\frac{[DFE]}{[GDE]} = \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$
Thus we have the system of equations
$$
\begin{cases}
\frac{[DFE]}{[GDE]} = \frac{1}{x(x+2)}\\
[DFE] + [GDE] =2[ABC]\frac{(x+1)^2}{x(x+2)}. 
\end{cases}
$$
leading, once solved, to the desired result, i.e.
$$\boxed{[DFE] = 2[ABC]} $$
$\blacksquare$
</s>