# Proof in Parallelogram using vectors

• November 13th 2006, 09:41 AM
meismo
Proof in Parallelogram using vectors
Hello everyone,
I need to proof that in a parallelogram, the lines connecting one corner to the middle of its opposite sides split the diagonal of the parallelogram into three equal parts.

I've uploaded an image to clarify see attachment.

(ABCD is the parallelogram, BD is the diagonal, and AE and AF are the lines splitting the diagonal. The rest are lines I thought might be useful, but haven't found out how.

I need to do this proof using vectors. Since it's a parallelogram I know that AB=DC,BC=AD, BD||EF||IJ.
So I need to proof that DH=HG=GB.
So far all my tries lead nowhere and I'm stuck. Any help would be appreciated.

meismo
• November 14th 2006, 09:37 AM
TriKri
Draw the line AC, it will cross DB at the middle because of the symetry of the parallelogram. Consider the two triangles BEA and ECA, their baces is the same (along the line BC) and so is their heights, hence also their area.

Now consider the two triangles BEG and ECG, by the same reason they have equal area as well. That means the two triangles GCA and GBA also has the same area, cause GCA + ECG = ECA, and GBA + BEG = BEA.

Now, also AC is split on the middle by BD, because of the symetry. If the midpoint (where they cross) is M, that means also triangles AMG and CMG has the same area, hence half the area of ACG. And since ACG and ABG has the same area, AMG has half the area of ABG. If we say their bases is along the line the line BM, GMA has the same height as BGA, hence the half base, hence GM is half the distance of BG and one third of BM.

If you do the same thing on the other half of the parallelogram, you'll se HM is one third of DM. And by adding two and two together you'll se DH equals HG equals GB.

/Kristofer
• November 14th 2006, 11:06 AM
earboth
Quote:

Originally Posted by meismo
Hello everyone,
I need to proof that in a parallelogram, the lines connecting one corner to the middle of its opposite sides split the diagonal of the parallelogram into three equal parts.
...

Hello, meismo,

if I understand your problem right you have to use vectors to prove the property:

a) As you have mentioned the vector $\overrightarrow{AB} = \overrightarrow{DC}$

b) You can prove by the property of the mid-parallel in a triangle that $\overrightarrow{AG} = \frac{2}{3}\overrightarrow{AE}$

You get 3 vectors:
$\overrightarrow{DH} = -\overrightarrow{AD}+\frac{2}{3}(\overrightarrow{AD }+\frac{1}{2}\overrightarrow{AB})$ = $-\frac{1}{3}\overrightarrow{AD}+\frac{1}{3}\overrig htarrow{AB}$

$\overrightarrow{BG} = -\overrightarrow{AB}+\frac{2}{3}(\overrightarrow{AB }+\frac{1}{2}\overrightarrow{AD})$ = $-\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrig htarrow{AD}$

$\overrightarrow{HG} = -\frac{2}{3}(\overrightarrow{AD}+\frac{1}{2}\overri ghtarrow{AB}) +\frac{2}{3}(\overrightarrow{AB}+\frac{1}{2}\overr ightarrow{AD})$ = $\frac{1}{3}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AD}$

So you can see that all three vectors have the same length and they only differ in their direction.

EB