<s>[INST] Finding Angle using Geometry In an equilateral triangle $ ABC $ the point $ D $ and $ E $ are on sides $ AC $ and $AB$  respectively, such that $ BD $ and $ CE $ intersect at $P$ , and the area of the quadrilateral $ ADPE $ is equal to area of  $ \Delta BPC $ find $ \angle BPE $. 
This question when I first tried looked easy and I was also able to guess the answer but when I tried to proof, was not able to work it out. I want some help. Thank you. No image was provided in question. I am attaching my drawing.

 [/INST] 
Let $a$ be the side length of the equilateral $\triangle ABC$,
$\angle BPE=\theta$, $\angle PBC=\phi$,
and let $[\cdot]$ denote the area. 
Note that
\begin{align} 
[AEPD]&=[ABC]-[BCD]-[BCE]+[BCP]
,
\end{align} 
and since $[AEPD]=[BCP]$,
we must have
\begin{align}
[BCD]+[BCE]&=[ABC]
.
\end{align} 
Using the three-angles-and-a-side expression for the area, 
we get
\begin{align}
\frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\phi)}{\sin(120^\circ-\phi)}
+
\frac{\sqrt3}4\cdot a^2\cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)}
&=
\frac{\sqrt3}4\cdot a^2
,\\
\frac{\sin(\phi)}{\sin(120^\circ-\phi)}
+
\cdot \frac{\sin(\theta-\phi)}{\sin(120^\circ+\phi-\theta)}
-1
&=0
.
\end{align}
After expanding and refactoring we get
\begin{align}
\frac{(\sin^2\phi+\cos^2\phi)(\sqrt3\sin\theta-3\cos\theta)}{
(\sqrt3\cos\phi+\sin\phi)
(\sqrt3\cos\theta\cos\phi
+\sqrt3\sin\theta\sin\phi+\sin\theta\cos\phi
-\cos\theta\sin\phi)
}
&=0
,
\end{align}
and it follows that $\theta=60^\circ$.
</s>