Proving two lines trisects a line A question from my vector calculus assignment. Geometry, anything visual, is by far my weakest area. I've been literally staring at this question for hours in frustrations and I give up (and I do mean hours). I don't even now where to start... not feeling good over here.

Question:
In the diagram below $ABCD$ is a parallelogram with $P$ and $Q$ the midpoints of the the sides $BC$ and $CD$, respectively. Prove $AP$ and $AQ$ trisect $BD$ at the points $E$ and $F$ using vector methods.

Image: 

Hints: Let $a = OA$, $b = OB$, etc. You must show $ e = \frac{2}{3}b + \frac{1}{3}d$, etc.


I figured as much without the hints. Also I made D the origin and simplified to $f = td$ for some $t$. And $f = a + s(q - a)$ for some $s$, and $q = \frac{c}{2}$ and so on... but I'm just going in circles. I have no idea what I'm doing. There are too many variables... I am truly frustrated and feeling dumb right now.
Any help is welcome. I'm going to go watch Dexter and forget how dumb I'm feeling.

The statement is invariant under affine transformations of the plane, so we can assume that $A=(0,1)$, $B=(1,1)$, $C=(1,0)$ and $D=(0,0)$. Then of course $Q=(1/2,0)$ and $P=(1,1/2)$, and we can compute. The line through $A$ and $P$ is $$x+2y=2;$$ the line through $A$ and $Q$ is $$2x+y=1;$$ finally the line through $B$ and $D$ is of course $$x-y=0.$$ Computing the intersection of $AP$ and $BD$, we get the point $E=(2/3,2/3)$, and the intersection of $AQ$ and $BD$ is $F=(1/3,1/3)$. It is clear, then, that $E$ and $F$ trisect the segment $BD$.
 
The thing to takehome from this is that sometimes symmetry allows to reduce a theorem to a computation.
