1. ## Proof about medians of triangles using vectors

Prove the given theorem using vectors.

The medians of a triangle meet in a point whose distance from each vertex is two-thirds the length of the median from that vertex.

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

The medians of a triangle meet in a point whose distance from each vertex is two-thirds the length of the median from that vertex.
Continuing from my reply to your previous post, we can generalise the result about the position vector of the mid-point of a line segment, to get the position vector of a point dividing a line segment in a given ratio, and it's this:

The position vector of the point $R$ that divides the line segment $AB$ in the ratio $\lambda:\mu$ is given by:
$\vec{r} = \frac{\mu\vec{a}+\lambda\vec{b}}{\lambda+\mu}$
So suppose the triangle is $ABC, M$ is the mid-point of $BC$ and $G$ divides $AM$ in the ratio $2:1$. Then:
$\vec{m} = \tfrac12(\vec{b}+\vec{c})$, as before using the mid-point formula
and:
$\vec{g}= \frac{\vec{a}+2\vec{m}}{3}$, using the more general result above

$=\frac13\Big(\vec{a}+2\times\tfrac12(\vec{b}+\vec{ c})\Big)$

$=\tfrac13(\vec{a}+\vec{b}+\vec{c})$
Now you'll see that $G$ is the point that's two-thirds of the way down the median $AM$. I'll leave it to you to show that it's also two-thirds of the way down the other medians as well. (Although this is pretty obvious since the expression for $\vec{g}$ is symmetrical in $\vec{a}, \vec{b}$ and $\vec{c}$.)

3. Thanks this is very helpful! but i am still confused as to how you found vector m and what vector m would look like on the picture?

4. Hello kullgirl418
Originally Posted by kullgirl418
Thanks this is very helpful! but i am still confused as to how you found vector m and what vector m would look like on the picture?
$M$ is the mid-point of the side $BC$ of the triangle. So if the position vectors of $B$ and $C$ are $\vec{b}$ and $\vec{c}$, we use the 'mid-point formula' that I showed you in your first posting to get $\vec{m}=\tfrac12(\vec{b}+\vec{c})$.

It's not terribly helpful to try to imagine what the vector $\vec{m}$ looks like, but if you want to, you'll need to mark a point $O$ somewhere - it doesn't matter where - to represent the origin. Then join $O$ to $M$ - that's the vector $\vec{OM}$, or $\vec{m}$ for short.

5. I am still confused as to how you are labeling B and C as vectors when they are representing points not lines?

6. ## Position and displacement vectors

Hello kullgirl418
Originally Posted by kullgirl418
I am still confused as to how you are labeling B and C as vectors when they are representing points not lines?
It's clear that you're not really understanding vectors at all, so read this carefully.

I didn't label B and C as vectors - I referred to their position vectors:
... So if the position vectors of $B$ and $C$ are $\vec{b}$ and $\vec{c}$,...
...in other words, their positions in relation to some arbitrary origin - usually denoted by O.

If you're not clear what that phrase 'their positions in relation to some arbitrary origin' means, let me unpack it a little - but you'll have to do the work. So:

• Draw a triangle ABC - don't make it a special triangle; try not to make it right-angled or isosceles.

• Mark M as the mid-point of BC.

• Somewhere else on the paper mark a point O. It's easiest if it's outside the triangle and doesn't lie on any of its sides when they're produced.

• Join O to A, B, C and M.

Then the position vectors of the points A, B, C and M in relation to O are the displacement vectors $\vec{OA}, \vec{OB},\vec{OC}$ and $\vec{OM}$. (A displacement vector is a description of how you move from one point to another. So the displacement vector $\vec{OA}$ describes - using whatever language you'd like - how you would move from O to A.)

To make it simpler to write, we describe these position vectors - these 'movement descriptions' - using lower-case letters:
$\vec{OA}=\vec{a}$
$\vec{OB}=\vec{b}$
$\vec{OC}=\vec{c}$
$\vec{OM}=\vec{m}$
The triangle law of addition is then simply two movement descriptions' combined into one. So when we write:
$\vec{OA}+\vec{AB}=\vec{OB}$
we're just saying:
$\underbrace{\text{a movement from O to A }}_{\vec{OA}}$ $\underbrace{\text{ followed by }}_+$ $\underbrace{\text{ a movement from A to B }}_{\vec{AB}}$ $\underbrace{\text{ is the same as }}_=$ $\underbrace{\text{ a movement from O to B}}_{\vec{OB}}$
So when we use lower-case letters this becomes:
$\vec{a}+\vec{AB}=\vec{b}$
in other words
$\vec{AB}=\vec{b}-\vec{a}$
which I described in an earlier reply to you as a very important result.

Finally, let's show you again where the 'mid-point' formula comes from.

Writing this important result again, but using the points B and M this time:
$\vec{BM}=\vec{m}-\vec{b}$
But the movement from B to M is just one-half of the movement from B to C, since M is the mid-point of BC. So
$\vec{BM}=\tfrac12\vec{BC}$
Knowing that $\vec{BC}=\vec{c}-\vec{b}$ (there's that result again), we can say:
$\vec{BM}=\tfrac12\vec{BC}$

$\Rightarrow \vec{m}-\vec{b}=\tfrac12(\vec{c}-\vec{b})$

$\Rightarrow \vec{m}=\vec{b}+\tfrac12\vec{c}-\tfrac12\vec{b}$

$\Rightarrow \vec{m}=\tfrac12(\vec{b}+\vec{c})$
I hope that helps to make things clearer.

8. sorry but now i am confused on how you found vector g? because the answer doesn't make sense to me... how do i know that G is two-thirds of the way?

9. never mind i understand it! thanks for all of your help!!

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# prove that median of triangle divide in ratio 2:1 on vector method

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