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? 
 A: 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.
$$\boxed{[DFE] = 2[ABC]} $$
$\blacksquare$
A: Proof (for the case when $P$ lies on arc $BC$):
Let $AB=BC=CA=1$ and $\angle PAC=\theta$.
$$\frac{PD}{\sin\angle PBD}=\frac{PB}{\sin \angle PDB} \implies \frac{PD}{\sin\theta}=\frac{PB}{\sin (120^\circ-\theta)} $$
$$\frac{PB}{\sin\angle PAB}=\frac{AB}{\sin \angle APB}\implies \frac{PB}{\sin(60^\circ-\theta)}=\frac{AB}{\sin 60^\circ}$$
So, $\displaystyle PD=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(120^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(60^\circ+\theta)}$.
$$\frac{PE}{\sin\angle PAC}=\frac{AE}{\sin \angle APE} \implies \frac{PE}{\sin\theta}=\frac{AE}{\sin 120^\circ} $$
$$ \frac{AE}{\sin\angle ABE}=\frac{AB}{\sin\angle AEB} \implies  \frac{AE}{\sin (60^\circ+\theta)}=\frac{AB}{\sin(60^\circ-\theta)}$$
So, $\displaystyle PE=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin120^\circ\sin(60^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin60^\circ\sin(60^\circ-\theta)}$.
$$\frac{PF}{\sin\angle PBF}=\frac{PB}{\sin \angle PFB} \implies \frac{PF}{\sin(120^\circ-\theta)}=\frac{PB}{\sin \theta}$$
So, $\displaystyle PF=\frac{\sin(120^\circ-\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}=\frac{\sin(60^\circ+\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}$.
The area of $\triangle DEF$ is
\begin{align*}
&\;\frac{1}{2}(PD\cdot PE+ PE\cdot PF+PF\cdot PD)\sin 120^\circ\\
=&\;\frac{\sqrt{3}}{4}\left[\frac{\sin^2\theta+\sin^2(60^\circ+\theta)++\sin^2(60^\circ-\theta)}{\sin^260^\circ}\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+(\sin60^\circ\cos\theta+\cos60^\circ\sin\theta)^2+(\sin60^\circ\cos\theta-\cos60^\circ\sin\theta)^2\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+2\sin^260^\circ\cos^2\theta+\cos^260^\circ\sin^2\theta\right]\\
=&\;\frac{1}{\sqrt{3}}\left[\frac{3}{2}\sin^2\theta+\frac{3}{2}\cos^2\theta\right]\\
=&\;\frac{\sqrt{3}}{2}
\end{align*}
The area of triangle $\triangle ABC$ is
$$\frac{1}{2}(1)^2\sin60^\circ=\frac{\sqrt{3}}{4}$$
