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.
 A: Note that EBP and EDA are similar triangles. Since 2BP=AD, it follows that 2EB=ED, and thus 3EB=BD. Which is to say, AP trisects BD.
A: There are as usual many approaches.  We deal with $F$ only.  Basically the same method will work for $E$.  Actually, we don't need to do anything for $E$, just a little reflection, and twisting our necks around. Or else we can simply rename our points: interchange $B$ and $D$.  Recycling is a good thing.
Let $u$ be the vector $DC$, and $v$ the vector $DA$.  Then $DB$ is the vector $u+v$.
We want to show that $DF=(1/3)(u+v)$.
To do this, it is enough to show that with this choice of $F$, the vector $AF$ is parallel to the vector $FQ$.  Compute.  For our choice of $F$, we have 
$$AF=(1/3)(u+v)-v=(1/3)u -(2/3)v.$$
Also,
$$FQ=(1/2)u-(1/3)(u+v)=(1/6)u -(1/3)v.$$
The parallelism is obvious, the vector $AF$ is twice the vector $FQ$.  As a little bonus we get therefore an additional geometric result, that $AF=2FQ$.
Comment: Since we were given the answer, we were able to save a few steps.    Clearly $DF=\lambda(u+v)$ for some $\lambda$. We verified that $\lambda=1/3$.  But what if we had not been given the answer? And what if $Q$ was not the midpoint of $DC$, but a division point of $DC$ in some ratio other than $1:1$?  We sketch how to do the same problem, without being supplied the $1/3$.  Minor modification will take care of other division ratios.
The idea is much the same as before.  We have $DF=\lambda(u+v)$ for some now unknown $\lambda$. And $AF=\kappa FQ$ for some unknown constant $\kappa$.
Thus we have
$$AF=\lambda(u+v)-v=\lambda u +(\lambda-1)v,$$
$$FQ=(1/2)u-\lambda(u+v)=(1/2-\lambda)u -\lambda v.$$
The condition $AF=\kappa FQ$ comes down to the two equations
$$\lambda =\kappa(1/2-\lambda) \text { and } \lambda-1=-\kappa\lambda.$$ 
To solve, substitute $\lambda-1$ for $-\kappa\lambda$ in the first equation. Quickly we get $\kappa=2$, and then $\lambda=1/3$.      
A: 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.
A: Since we know what exactly we want to prove, I will first give a "cheating solution" that makes a neat homework answer but provides little insight. After all, such tricks are also useful sometimes. :)
I will prove that $E$ divides $BD$ in the ratio $2:1$. It seems convenient to fix the origin at $C$. Then the vertices $B$, $D$ and $A$ are respectively at $\mathbf b$ and $\mathbf d$ and $\mathbf b + \mathbf d$. Now, let $X$ be the point $\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d$. (This is the part that is most unsatisfactory.) 
We will show that $A$, $X$ and $C$ are collinear. To verify this, just compute:
$$
\overline{XA} = (\mathbf b +\mathbf d) - (\frac{2}{3} \mathbf b + \frac{1}{3} \mathbf d)  = \frac{1}{3} \mathbf b + \frac{2}{3} \mathbf d = \frac{1}{3} (\mathbf b + 2\mathbf d),
$$ 
and
$$
\overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b  = \frac{1}{2} (\mathbf b + 2\mathbf d).
$$ 
Now is it evident that $\overline{PA}$ is parallel to the vector $\overline{XA}$? What does this mean? How does this fact help you? 

To make this into a more systematic proof, we will proceed roughly as J.M. suggests, but using the vector notation. Let $E$ be the point of intersection of $AP$ with $BD$. We want to find $E$. 
Since $E$ lies inside the segment $BD$, $E$ can be written as $\alpha \mathbf b + (1-\alpha) \mathbf d$ for some $\alpha \in [0,1]$ (the exact range in which $\alpha$ lies is not that important). Then, $E$ will lie in $AP$ iff $\overline{EA}$ is parallel to $\overline{PA}$. But we can compute these vectors:
$$
\overline {EA} = (\mathbf b +\mathbf d) - (\alpha \mathbf b + (1-\alpha) \mathbf d) = (1-\alpha) \mathbf b + \alpha \mathbf d, 
$$
and
$$
\overline{PA} = (\mathbf b +\mathbf d) - \frac{1}{2} \mathbf b  = \frac{1}{2} \mathbf b + \mathbf d.
$$ 
Now under what condition will these two vectors be parallel to each other? 
