# Prove Parallelogram Area Is Twice Triangle Area

I thought this would be easy but I can't seem to find the answer.

Edit: I did my best to draw the diagram:

$\overline{EC}=\frac{1}{3} \overline{AC}, \overline{AF}=\frac{1}{3} \overline{AB}, \overline{BD}=\frac{1}{3} \overline{BC}$

I drew a line parallel to $\overline{BE}$ through $R$. I called the intersection of that line and $\overline{AB}$ point $M$. I drew a line parallel to $\overline{AD}$ through $B$, and called the intersection of that line and my previous line $G$. Then I want to prove $BGRP = 2\triangle PRQ$. This is for the ultimate goal of solving the $1/7$ area triangle problem. In addition I need to prove $\triangle BGM = \triangle ARM$, and to prove that I think I need to prove $M$ is the midpoint. I think I'm close with similar triangles $\triangle ARM$ and $\triangle APB$ but I can't get the relationship. Thanks if you managed to read all this.

Edit: A full proof is given here, I just need someone to clarify the dialations and transformations.

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$DPR$ are collinear, right? So, what's $BDRP$? – Gerry Myerson Mar 2 '14 at 2:46
@GerryMyerson Oops, I didn't realize $D$ was already used. I've changed it to $G$ – qwr Mar 2 '14 at 2:47
OK, so several of the important objects do not appear in the diagram. Pity. – Gerry Myerson Mar 2 '14 at 2:50
@GerryMyerson I could try out my MS paint skills :P – qwr Mar 2 '14 at 2:58
– Sawarnik Mar 2 '14 at 5:26

In the barycentric coordinates associated with $\triangle ABC$, we have $$P = [1,4,2] \qquad\text{and}\qquad Q = [2,1,4]$$ Of course I solved a linear system to get this, but it's easy to check after the fact: $[1,4,2]$ is on the line joining $A[1,0,0]$ and $D[0,2,1]$ (since it's $[1,0,0] + 2[0,2,1]$) and on the line joining $B[0,1,0]$ and $E[1,0,2]$ (since it's $4[0,1,0]+[1,0,2]$), so it must be $P$. Then we get the coordinates of $Q$ by symmetry.

Thus $P$ is the midpoint of $BQ$. (Take $B = [0,7,0]$.)

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I thought barycentric coordinates had fraction values for points inside the triangle? – qwr Apr 9 '14 at 18:05
If you like, you can scale everything I wrote so that each point's coordinates add up to 1: $P=[\frac17,\frac47,\frac27]$, and so on. – user21467 Apr 10 '14 at 11:44

Proving $\triangle BGM ~ \triangle ARM$ is easy.
$\angle GMB = \angle AMR$ because they are vertically opposite angles.
$\angle GBM = \angle MAR$ because line $BG$ is parallel to line $AR$, and the angles are corresponding angles made by intercept $GR$.
For proving $BGRP=2PRQ$, you need to show $BP=PQ$ which is not clear to me using this information.

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The triangles are similar, but to prove they're equal you need to prove $M$ is the midpoint of $AB$? – qwr Mar 2 '14 at 5:19
I thought you were referring to the angles, it seems you were talking about the triangles. My bad. – taninamdar Mar 2 '14 at 5:45

All we need is to prove $\overline{BP} = \overline{PQ}$

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Can you add detail? As in, how you would prove that, because that's a large part of solving the problem. – qwr Mar 19 '14 at 4:23
I would you like to.. But not yet. – sidneimv Mar 19 '14 at 16:20