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? 

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.
$$\boxed{[DFE] = 2[ABC]} $$
$\blacksquare$
