Regional Mathematics Olympiad(RMO-India) Geometry Problem How to do this problem? I drew the figure according to the given details but, I believe some extra lines should be drawn to solve this problem.
Applying Menelaus to triangle $ADC$ and transversal $BPF$, we get

$$ \frac{AP}{PD} \times \frac{DB}{BC} \times \frac{CF}{FA} = 1, $$

or that $\frac{ AP}{PD} = 3$.

APplying Menelaus to triangle $AEC$ and tranvsersal $BQF$, we get

$$ \frac{AQ}{QE} \times \frac{EB}{BC} \times \frac{CF}{FA} = 1, $$

or that $\frac{AQ}{QE} = \frac{3}{2} $.

Now, apply Menelaus to triangle $APQ$ and transversal $BDE$ to obtain $ \frac{PB}{BQ}$.  You should be able to take it from here.
Trisect the side of an equilateral triangle will not trisect the corresponding opposite angle. If so, we can trisect any given angle.