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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: enter image description here

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.

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    Depending on which "Dexter" show you intend to watch that might not make you feel any better.2011-10-04
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    The one that kills people. I will make me feel better lol2011-10-04
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    I'm not sure how you can adapt this approach to your "vector" requirement, but: letting $D:(0,0)$, $C:(c,0)$, $A:(a,b)$, and $B:(a+c,b)$ (why?), use the two-point form of the equation of a line to get the equations of the lines $\overline{AQ}$ and $\overline{AP}$, where e.g. $Q:(c/2,0)$ (why?). Find the intersection points of those lines with $\overline{DB}$. Check that those two intersection points are in fact trisection points for $\overline{BD}$.2011-10-04
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    (I couldn't resist; here's a *Mathematica* "proof": `(({x, y} /. First@Solve[{y == InterpolatingPolynomial[{{0, 0}, {a + c, b}}, x], y == InterpolatingPolynomial[{{a, b}, #}, x]}, {x, y}]) & /@ {{c/2, 0}, {a + 2 c, b}/2}) === Map[{a + c, b} # &, {1, 2}/3]`)2011-10-04
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    @J.M I already tried this and I keep getting anything but the correct answer. I get point of intersection (-4c, 0), which makes no sense. This is the most frustrating thing ever. I'm just going to leave this question blank. I've never been so frustrated by something. — And there's no way I could come up with any of the answers below. I barely understand them.2011-10-04
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    Let's check your algebra: $\overline{BD}$ is $y=\frac{bx}{a+c}$, $\overline{AQ}$ is $y=\frac{b(2x-c)}{2a-c}$, and $\overline{AP}$ is $y=\frac{b(x-2c)}{a-2c}$. If these weren't the lines you got, something went wrong in your algebra...2011-10-04

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