<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] 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
$$\frac{1}{2}(1)^2\sin60^\circ=\frac{\sqrt{3}}{4}$$
</s>