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.
The hint.

Prove that $$\left(S_{\Delta BPC}\right)^2=S_{\Delta APB}S_{\Delta APC}.$$
I got $$\measuredangle BPE=60^{\circ}.$$

Let $S_{\Delta BPC}=a$, $S_{\Delta PAC}=b$ and $S_{\Delta PAB}=c$.

Thus, $$\frac{S_{\Delta PEB}}{c}=\frac{BE}{AB}=\frac{BE}{BE+EA}=\frac{1}{1+\frac{EA}{BE}}=\frac{1}{1+\frac{b}{a}}=\frac{a}{a+b},$$
which gives $$S_{\Delta PEB}=\frac{ac}{a+b}.$$
Similarly, $$S_{\Delta PDC}=\frac{ab}{a+c}.$$
Thus, $$S_{AEPD}=b+c-\frac{ab}{a+c}-\frac{ac}{a+b}=\frac{bc}{a+c}+\frac{bc}{a+b}.$$
Id est, $$\frac{bc}{a+c}+\frac{bc}{a+b}=a$$ or
$$a^2=bc$$ or $$\frac{a}{b}=\frac{c}{a}$$ or
$$\frac{BE}{AE}=\frac{AD}{CD},$$ which gives $$DC=AE,$$
$$\Delta AEC\cong\Delta CDB,$$ which gives $$\measuredangle BPE=\measuredangle DBC+\measuredangle ECB=\measuredangle DBC+60^{\circ}-\measuredangle ACE=60^{\circ}.$$
IF there exists a set of points P for which the two areas are equal, then one such point P lies on the vertical altitude (axis of symmetry), thus we only need to find that point.  Dividing each of the two areas in half with the vertical altitude, the problem can be stated as Area(BPS) = Area (APE), where S is the foot of the altitude (the midpoint of BC).  The midpoint (centroid?) of the triangle (with altitudes) divides the area of the whole triangle into six similar triangles, including the two above.  So the midpoint of the triangle satisfies the condition, thus the angle is 60 degrees.