Proving Parallelogram Diagonals Bisect Each Other

[danago]

Hey. I've got a vectors test coming up soon, so I've been going through everything. At the moment I am working on vector proofs. here's the question:

OABC is a parallelogram where $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. M is the midpoint of the diagonal OB. Prove that the diagonals of this parallelogram bisect each other.

I drew a diagram
http://img102.imageshack.us/img102/7223/vectorproof0rg.gif

I already know that M is the midpoint of OB, so i just need to show that its also the midpoint of AC. This means that $\overrightarrow{AM}=\overrightarrow{MC}$, which is what I am proving.

$$\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}$$
$$\overrightarrow{OB}=\mathbf{c}+\mathbf{a}$$


$$\overrightarrow{AM}=\overrightarrow{AO}+\overrightarrow{OM}$$

$$\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}\overrightarrow{OB}$$

$$\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}(\mathbf{c+a})$$

$$\overrightarrow{AM}=\frac{1}{2}(\mathbf{c-a})$$


So now i know $\overrightarrow{AM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

$$\overrightarrow{MC}=\overrightarrow{MO}+\overrightarrow{OC}$$

$$\overrightarrow{MC}=-\frac{1}{2}\overrightarrow{OB}+\mathbf{c}$$

$$\overrightarrow{MC}=-\frac{1}{2}(\mathbf{c+a})+\mathbf{c}$$

$$\overrightarrow{MC}=\frac{1}{2}(\mathbf{c-a})$$

So i now also have $\overrightarrow{MC}$. Since they equal the same, as required, i can conclude that:
$$\overrightarrow{AM}=\overrightarrow{MC}$$

Is that all i would need to do to answer the question? is the information i provided sufficient to prove that the two diagonals bisect each other?

Thanks,
Dan.

 

[Curious3141]

Hey. I've got a vectors test coming up soon, so I've been going through everything. At the moment I am working on vector proofs. here's the question:

OABC is a parallelogram where $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. M is the midpoint of the diagonal OB. Prove that the diagonals of this parallelogram bisect each other.

I drew a diagram
http://img102.imageshack.us/img102/7223/vectorproof0rg.gif

I already know that M is the midpoint of OB, so i just need to show that its also the midpoint of AC. This means that $\overrightarrow{AM}=\overrightarrow{MC}$, which is what I am proving.

$$\overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}$$
$$\overrightarrow{OB}=\mathbf{c}+\mathbf{a}$$


$$\overrightarrow{AM}=\overrightarrow{AO}+\overrightarrow{OM}$$

$$\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}\overrightarrow{OB}$$

$$\overrightarrow{AM}=\mathbf{-a}+\frac{1}{2}(\mathbf{c+a})$$

$$\overrightarrow{AM}=\frac{1}{2}(\mathbf{c-a})$$

At this point instead of doing the rest of what you did, finish up like so :

$$\overrightarrow{AM}=\frac{1}{2}(\overrightarrow{OC} - \overrightarrow{OA}) = \frac{1}{2}(\overrightarrow{AC})$$

You can actually end it here. Think about it. You're given a point M that is the midpt of OB. You've proved that the vector from point A to M is parallel to the vector from A to C (since this the property of scalar multiplication of a non-null vector) meaning that A, M and C are collinear. You've also established that the magnitude of $$\overrightarrow{AM}$$ is half that of $$\overrightarrow{AC}$$ meaning that M is also the bisector of AC which is the other diagonal of the parallelogram. In other words, the bisector of one diagonal is coincident with the bisector of the other, and the result is proved.

I would caution against drawing the other diagonal AC in your initial figure like its passing through M, because by right you cannot assume that (it's one of the things you have to prove). Better to just omit the line segment AC from the diagram altogether to avoid confusion.

 

[danago]

ok thanks for that. And yea, now that i think about it, it does make sense to omit line AC.

 

[danago]

What about this question:

http://img223.imageshack.us/img223/486/vectorproof23up.gif

I need to prove that if line AC and DE are parallel, point D cuts line AB in the same ratios as E cuts CB.

I did it by first saying that:
$$\overrightarrow{CE}=k\overrightarrow{CB}$$
$$\overrightarrow{AD}=h\overrightarrow{AB}$$

And then ill prove that h=k.

If they are in the same ratio, then the following equation should prove true (please correct me if I am wrong):

$$\frac{\overrightarrow{AD}}{\overrightarrow{AB}}=\frac{\overrightarrow{CE}}{\overrightarrow{CB}}$$

I then made all the vectors in terms of vectors a and b:

$$\frac{h\mathbf{a}}{\mathbf{a}}=\frac{k(\mathbf{-b+a})}{\mathbf{-b+a}}$$

Next, i simply simplified the equation, which left me with:

$$h=k$$

Is that how i would prove it?

 

[pavadrin]

yeah along those lines will be fine

 

[pizzasky]

With regard to the second question, the working starts off with the assumption that AB and CB are split in the same ratio. This means that you assume that h=k. Having already assumed this, why do you go on to prove it? Instead, you should use the assumption that h=k to try to prove that AC and DE are parallel.

Also, a vector cannot be divided by another vector, so expressions like $$\frac{\overrightarrow{AD}}{\overrightarrow{AB}}= \frac{\overrightarrow{CE}}{\overrightarrow{CB}}$$ are not really correct. Instead, try expressing it like this... $$\frac{\mid\overrightarrow{AD}\mid}{\mid\overrightarrow{AB}\mid}= \frac{\mid\overrightarrow{CE}\mid}{\mid\overrightarrow{CB}\mid}$$,
where $$\mid\mathbf{a}\mid$$ is the magnitude of $$\mathbf{a}$$.

Actually, the question wants us to assume that AC and DE are parallel and to use this assumption to prove that AB and CB are split in the same ratio. However, if we do it this way, we may end up making too many assumptions (as we also have to assume the ratio is valid for the lengths of the parallel lines). So, I think it's acceptable if we do it the other way round.

 

[ndbunker]

Hi! I'm not sure if this is the correct forum or not, but anyway... I can't get the following question out, so I require your assistance.

Show that |a + b| =< |a| + |b| and interpret this result geometrically.

=< means less than or equal to.

Thanks in advance.

 

[danago]

Wow...4 years ago i made this thread

ndbunker, in general it would be better to create a new thread rather than bring a very old thread back alive. For your question, that inequality is called the "triangle inequality". Try doing a google search and you will find heaps.