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. 
 A: 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.
A: Form the parallelogram $ABCG$ with centroid $F$, and write $BG=20a$. Comparing similar triangles $PAG$ and $PDB$ yields $BP=5a$ (and $PG=15a$). Likewise, comparing triangles $QAG$ and $QEB$ gives $BQ=8a$ (and $QG=12a$). Hence $PQ=BQ-BP=3a$ and so $BP/PQ=5/3$.
A: Trisect the side of an equilateral triangle will not trisect the corresponding opposite angle. If so, we can trisect any given angle.
A: The great thing about this is that it only says "triangle," so the answer is the same for all triangles. So why not take something that's very simple and suitable for the question - like an equilateral triangle with side length 3.
Consider the picture below:

For triangle ABP you know all angles and AB (length 3), so you can find BP by the law of sines. For triangle AQF you know all angles and AF (length 3/2), so you can find QF, also by the law of sines. The length of BF is $\frac{3}{2}\sqrt{3}$ since ABF is a 30-60-90 triangle, so then you can find PQ by subtraction. Now you have all that you need!
