Finding Angle using Geometry

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 $$\measuredangle BPE=\measuredangle DBC+\measuredangle ECB=\measuredangle DBC+60^{\circ}-\measuredangle ACE=60^{\circ}.$$

From the condition, we add area of ∆BPC to both sides of the equation, this becomes

Area of ∆AEC = Area of ∆CDB

It is then easy to prove that ∆AEC is congruent to ∆CDB. So ∠BDC = ∠CEA.

Thus, quadrilateral ADPE is concyclic.

so ∠BPE = ∠BAC = 60˚