# Thread: Proof using vectors

1. ## Proof using vectors

Prove the given theorem using vectors:

The midpoints of the sides of quadrilateral are the vertices of a parallelogram.

2. Originally Posted by kullgirl418
Prove the given theorem using vectors:

The midpoints of the sides of quadrilateral are the vertices of a parallelogram.
Maybe this will help: http://www.mathhelpforum.com/math-he...+parallelogram

3. Hello kullgirl418

Welcome to Math Help Forum!
Originally Posted by kullgirl418
Prove the given theorem using vectors:

The midpoints of the sides of quadrilateral are the vertices of a parallelogram.
I'm not quite sure how much you've been taught about representing plane figures like quadrilaterals using vectors. For instance, if the quadrilateral is $\displaystyle ABCD$, do you know that you can represent the position vectors of its vertices by vectors $\displaystyle \vec{OA}, \vec{OB},\vec{OC},\vec{OD}$, where $\displaystyle O$ is some arbitrary point (the origin)?

If you do, you've probably also seen that we can use lower-case letters to stand for these vectors: $\displaystyle \vec{a} = \vec{OA}$, and so on.

We can then use the vector law of addition to write down vectors representing each of the sides of the quadrilateral. For instance, we can say:
$\displaystyle \vec{OA}+\vec{AB}=\vec{OB}$

$\displaystyle \Rightarrow \vec{a}+\vec{AB}=\vec{b}$
So we get the vector representing the side $\displaystyle AB$ as:
$\displaystyle \vec{AB}=\vec{b}-\vec{a}$
This is an extremely important and useful result. Can you use it to find vectors representing the other three sides?

Another vital piece of information is the the position vector of the mid-point of a line. Suppose, for instance, that $\displaystyle M$ is the mid-point of the line segment $\displaystyle AB$. Then $\displaystyle \vec{AM}=\tfrac12\vec{AB}$, because $\displaystyle M$ is just half-way from $\displaystyle A$ to $\displaystyle B$.

So we can write down its position vector (from our origin $\displaystyle O$) as:
$\displaystyle \vec{OM}=\vec{OA}+\vec{AM}$

$\displaystyle \Rightarrow \vec{m}=\vec{a}+\tfrac12\vec{AB}$

$\displaystyle =\vec{a}+\tfrac12(\vec{b}-\vec{a})$, using the result above

$\displaystyle =\tfrac12(\vec{a}+\vec{b})$, when we simplify.
You'll see that this is just the 'average' of the position vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$. That makes sense, because $\displaystyle M$ is the 'average position' of $\displaystyle A$ and $\displaystyle B$. OK?

In the same way if we call N the mid-point of AD, its position vector is:
$\displaystyle \vec{n}=\tfrac12(\vec{a}+\vec{d})$
(I'll leave it to you to check this out.)

If we now use the first result above we can get the vector that represents the line joing $\displaystyle MN$ as:
$\displaystyle \vec{MN}=\vec{n}-\vec{m}$

$\displaystyle =\tfrac12(\vec{a}+\vec{d})-\tfrac12(\vec{a}+\vec{b})$

$\displaystyle =\tfrac12(\vec{d}-\vec{b})$
Now that's just one-half of the vector $\displaystyle \vec{BD}$ (again using that important result above).

Well now, if you do the same thing for the mid-points of BC and CD, you will get exactly the same result: $\displaystyle \tfrac12(\vec{d}-\vec{b})$. (I'll leave that to you.)

Can you see that this proves that these four mid-points form a parallelogram?