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## High school geometry

### Course: High school geometry > Unit 6

Lesson 2: Dividing line segments# Proving triangle medians intersect at a point

CCSS.Math:

Prove that for any triangle, the medians intersect at a point 2/3 of the way from each vertex to the midpoint of the opposite side.

## Want to join the conversation?

- This proof begins with the assumption that the ratio between vertex to triangle midpoint and vertex to opposite angle midpoint is 2:3.

How can you prove this without knowing the 2:3 ratio from the offset?(23 votes)- Hey everyone, @jamesclarks posts this helpful video to answer the question in the comments. Here's a repost for you all to see:

@jamesclarks

"I've looked at some other videos and see the 2/3 ratio justified.

I also found some explanation for weighted averages here:

https://www.khanacademy.org/computing/pixar/environment-modeling-2/mathematics-of-parabolas2-ver2/v/weighted-average-two-points"(1 vote)

- isn't that i use 2/3 time the a+b/2 enough to find out the x coordinate of centrod, what add teh 1/3 time the x coordinate of the origin ?

confused .(8 votes)- I believe you are asking why Sal is adding 1/3 times the x-coordinate of the origin when he is already adding 2/3 times a+b/2.

Let me start with an example. When you have to find the midpoint of two points (a,b) and (x,y) to find the midpoint you must do (a+x)/2 and (b+y)/2. That is exactly the same as doing (1/2a + 1/2x) and (1/2b+1/2y) if you distribute the terms. This is why you have to do 2/3 times a+b/2 and 1/3 times the x-coordinate of the origin to find the x-coordinate of the centroid; its just like finding the midpoint but with 1/3 and 2/3. If you are familiar with expected value/weighted average, it is exactly like calculating that.

If it still does not make sense, then I'll explain it this way. I think we can agree that the centroid is twice as close to a+b/2 as it is to 0 (the x-coordinate of the origin) because the centroid is 2/3 the way from origin, so it is closer to the point with the x-coordinate of a+b/2 (4:00-4:10). However, since the ratio is 2:1, a+b/2 is 2 OUT OF THREE PARTS, because if the distance from one point of a median to the other is 3, then a+b/2 is 2 (2/3 of 3) and the x-coordinate of the origin is 1. Essentially what Sal is doing is that he is setting up a comparison. By multiplying a+b/2 x 2/3 and the x-coordinate of the origin by 1/3, he is setting up a comparison/ratio of the distance that the centroid is 2/3 of the distance to a+b/2 as it is from the origin.

In this case, it does not matter if Sal multiplied by 1/3 or not, as the origin's x-value is 0. I think he did that to be thorough. If the origin were some other point that has non-zero x and y-values, multiplying by 1/3 would matter.

Hope this helps.(12 votes)

- Nothing makes sense about the multiplying part ._.(7 votes)
- I assume you're talking about the
`(2/3*((a+b)/2) + 1/3*(0), 2/3*(c/2) + 1/3*(0))`

, which is at4:03of the video. Also the ones at5:20and6:28.

Before we dive in as to why they were multiplied that way, let's first talk about the difference between a "normal average" and a "weighted average."

---------------------------------------------------

What is the normal average? That's your typical way of doing the average, which is also used for finding the midpoint of a line segment.

For example, you have two numbers

and**7**

, and you want to find their average. What do you do? Of course you do the sum of those two and divide them by two, in other words,**5**`(7+5)/2 = 6`

.

Here the normal average is`6`

.

However, what if, for some reason, the number`7`

weighs*three times*than`5`

? In other words, what if the ratio between them is 3:4 where`7`

weighs more? This is where weighted average comes in.

---------------------------------------------------

What is the weighted average? It's still your average but taking into account that now your number`7`

is more important than your number`5`

. (Remember that we're weighing`7`

*three times more*than`5`

)

Now what's your*"weighted average"*going to be? It's going to be`(7+7+7+5)/4`

or`(3*7 + 5*1)/4`

which equal to`6.5`

.

Notice that we divided by four and not two because remember we said that`7`

is**three**times more than`5`

, and there's only**one**5, hence,**three**plus**one**is four, which is why we divided them by 4.

Also, notice that our weighted average is now`6.5`

, which is`0.5`

more than our normal average. Why? That's because we made it so that the number`7`

is*more important*than our number`5`

hence the weighted average is**closer**to`7`

.

---------------------------------------------------

How does this apply to the video?

Let's go back at4:03on the video and let's try to breakdown`(2/3*((a+b)/2) + 1/3(0), 2/3*(c/2) + 1/3*(0))`

.

And remember that Sal said that*the intersection point of the medians have the ratio of 2:3*.

What does that mean? As you can see from the video, the intersection point*is near*to the point`((a+b)/2, c/2)`

. That would mean that that point weighs 2 times than point`(0, 0)`

.

Let's focus on the x-coordinate first. Since we know that`(a+b)/2`

weighs*two more*than`0`

then the weighted average of the x-coordinate would be:`((a+b)/2 + (a+b)/2 + 0) / 3`

Combine like, or same, terms (imagine that`(a+b)/2`

is a term then our expression would now be):`(2*((a+b)/2) + 0) / 3`

Let's distribute the`3`

:`2/3*((a+b)/2) + 0/3`

Notice that we can`0/3`

is the same thing as`1/3*(0)`

so now our expression would be:`2/3*((a+b)/2) + 1/3*(0)`

You might be thinking why we divided by`3`

and that's because`(a+b)/2`

weighs*two times*than`0`

. In other words, we have**two**of (a+b)/2 and**one**of zero. This also explains why we can say`0/3 = 1/3*(0)`

because we have**one**zero.

And there we have it, we derived the expression that Sal made himself. How about the y-coordinate? Same drill.`(c/2 + c/2 + 0) / 3`

Notice we have**two**of`c/2`

and**one**of`0`

so we can rewrite the expression to:`(2*(c/2) + 1*(0)) / 3`

Distribute the`3`

and we have:`2/3*(c/2) + 1/3*(0)`

And there we have it, we've derived the expressions that Sal did.

Thank you for reading this far, I hope I somehow enlightened you.(9 votes)

- How do you know that the centroid 2/3 of the way from each vertex?(6 votes)
- Here's a proof that explains why: http://jwilson.coe.uga.edu/EMAT6680Fa09/Pringle/Pringle%20Assignment%204/PringleAssignment4.html

Basically, you can construct a parallelogram which allows you to prove that two triangles are similar, and that the scale factor between them is 1/2. The centroid is located in the middle of two of the lines that have this scale factor, which means that it is 2/3 of the total line.(5 votes)

- How does Sal know which coordinates of a median to multiply by 2/3 and 1/3? If there a two coordinates for each median, then how do we know which coordinate's x and y values need to be multiplied by 2/3 and which ones needs to be multiplied by 1/3? An answer would be greatly appreciated.(8 votes)
- It is 2/3 from the vertex and 1/3 from the side.(1 vote)

- I have the same question about why it comes to times 2/3 first then times 1/3 to another.

But the interesting thing is I found some clues from seesaw balance equation. I still can't figure out why it goes like that, but I know how to use this.(2 votes)- Hi Cheng! Sorry I am answering a bit late. I believe you “which point do we multiply 2/3 and 1/3 with and why?” Actually there is a proof for that. For this proof, it was a given. For the video of that proof on KA, here it is. I could explain it myself, but honestly, I can’t do better than Sal.

Video:

https://www.khanacademy.org/math/geometry-home/triangle-properties/medians-centroids/v/proving-that-the-centroid-is-2-3rds-along-the-median

This video helps too:

https://www.youtube.com/watch?v=L7mfmg0VAU0

In these proofs, especially Sal’s, you can see that the centroid is specifically 2/3 from the vertex and 1/3 from the side. I believe there was a similar question from another user regarding this question.(1 vote)

- why does the mid points have 2/3 and co-mid points have 1/3?(1 vote)
- I'd love to answer your question. It's mesmerizing because it seems like magic that they looked like they appeared out of nowhere.

To fully understand how they got there, let's first talk about the difference between a "normal average" and a "weighted average."

---------------------------------------------------

What is the normal average? That's your typical way of doing the average, which is also used for finding the midpoint of a line segment.

For example, you have two numbers

and**7**

, and you want to find their average. What do you do? Of course you do the sum of those two and divide them by two, in other words,**5**`(7+5)/2 = 6`

.

Here the normal average is`6`

.

However, what if, for some reason, the number`7`

weighs*three times*than`5`

? In other words, what if the ratio between them is 3:4 where`7`

weighs more? This is where weighted average comes in.

---------------------------------------------------

What is the weighted average? It's still your average but taking into account that now your number`7`

is more important than your number`5`

. (Remember that we're weighing`7`

*three times more*than`5`

)

Now what's your*"weighted average"*going to be? It's going to be`(7+7+7+5)/4`

or`(3*7 + 5*1)/4`

which equal to`6.5`

.

Notice that we divided by four and not two because remember we said that`7`

is**three**times more than`5`

, and there's only**one**5, hence,**three**plus**one**is four, which is why we divided them by 4.

Also, notice that our weighted average is now`6.5`

, which is`0.5`

more than our normal average. Why? That's because we made it so that the number`7`

is*more important*than our number`5`

hence the weighted average is**closer**to`7`

.

---------------------------------------------------

How does this apply to the video?

Let's go back at4:03on the video and let's try to breakdown`(2/3*((a+b)/2) + 1/3(0), 2/3*(c/2) + 1/3*(0))`

.

And remember that Sal said that*the intersection point of the medians have the ratio of 2:3*.

What does that mean? As you can see from the video, the intersection point*is near*to the point`((a+b)/2, c/2)`

. That would mean that that point weighs 2 times than point`(0, 0)`

.

Let's focus on the x-coordinate first. Since we know that`(a+b)/2`

weighs*two more*than`0`

then the weighted average of the x-coordinate would be:`((a+b)/2 + (a+b)/2 + 0) / 3`

Combine like, or same, terms (imagine that`(a+b)/2`

is a term then our expression would now be):`(2*((a+b)/2) + 0) / 3`

Let's distribute the`3`

:`2/3*((a+b)/2) + 0/3`

Notice that we can`0/3`

is the same thing as`1/3*(0)`

so now our expression would be:`2/3*((a+b)/2) + 1/3*(0)`

You might be thinking why we divided by`3`

and that's because`(a+b)/2`

weighs*two times*than`0`

. In other words, we have**two**of (a+b)/2 and**one**of zero. This also explains why we can say`0/3 = 1/3*(0)`

because we have**one**zero.

And there we have it, we derived the expression that Sal made himself. How about the y-coordinate? Same drill.`(c/2 + c/2 + 0) / 3`

Notice we have**two**of`c/2`

and**one**of`0`

so we can rewrite the expression to:`(2*(c/2) + 1*(0)) / 3`

Distribute the`3`

and we have:`2/3*(c/2) + 1/3*(0)`

And there we have it, we've derived the expressions that Sal did.

I know that this took a long time to read but thank you for reading thus far.(1 vote)

- I know that this is the best proof for now, but what about a geometric proof, I mean, without analytic geometry?(1 vote)
- Couldn't you also figure out the point-slope equation of each line and then prove that they all meet at the same point, maybe using systems of equations?(1 vote)

## Video transcript

- [Instructor] So the goal
of this video is to prove that the three medians of
a triangle always intersect at one point, which is pretty interesting, because you would expect
two different lines with different slopes to
intersect in one point, but three lines intersecting
in one point is pretty neat. And this is true for all triangles. And so to setup this proof, I put an arbitrary triangle here, but I put one vertex at the origin. That'll simplify the math. And then I put another
vertex on the x-axis. And I've given them coordinates. So this one right over
here is at zero, zero. This one over here, we're just
saying that the x-coordinate is a, and so it's a comma zero. And then this one up here has
some x and some y-coordinate. We're just calling them b and c. This is some arbitrary triangle. And if you had some other triangle that had the same dimensions as this one, and this one can have any dimensions 'cause we haven't defined a, b, and c. You could go from this triangle
to any of those other ones using rigid transformations. So if we can prove that the
median of this triangle, this general triangle, always
intersect at one point, this will be true for all triangles. So let's do a little bit more. Let's draw the medians. So what we're going to do is draw lines from each of the vertex to the midpoint of the opposite side. So if we do that, we've
drawn all the medians. And it for sure looks like
they intersect in one point. But to prove that, let's think about what the coordinates are of the midpoints of each of these sides. So what is the coordinate right over here? Pause this video and think about that. Well this is going to be the
midpoint of this top point and this bottom right point. So this length is equal to that length. And for the midpoint, you can
really just think about it as you're taking the average
of each of the coordinates. So the x-coordinate here's
going to be the average of b and a. So you could just write
that as a plus b over two, and then the y-coordinate
is going to be the average of c and zero. That would be c plus zero
over two, or just c over two. And we could do that for
each of these points. So this point right over here, its x-coordinate is going to
be the average of zero and a. So that's just a over two. And its y-coordinate is
going to be the average of zero and zero. You can see that it sits on the x-axis, so its y-coordinate is zero. And then last but not least, what's the coordinate of this point? Pause the video and
try to figure that out. All right, well it's going be x-coordinate who's going to be the
average of b and zero, which is just going to be b over two. And then the y-coordinate
is going to be the average of c and zero, which is
just going to be c over two. So way that I'm gonna prove
that all three of these medians intersect at a unique point,
is by showing you a coordinate that sits on all three lines. If it sits on all three lines, that must be the point of intersection. And that interesting
point is 2/3 along the way of any one of the medians. So one way to think
about it is the distance between the vertex and that
point is 2/3 of the length of the median. So if we just look at this blue median, the coordinate of this point
that is twice as far away from the vertex as it is
from the opposite side, it will be based on a weighted average of the x and y-coordinates. When we did a midpoint and
things were equally far away, you equally weighted the coordinates. So you just took their average. But when you're, if you
are closer to this side, you will take a weighted
average accordingly. So it's going to be 2/3
times a plus b over two, plus 1/3 times zero, and
then the y-coordinate is going to be 2/3 times c over two, plus 1/3 times the y-coordinate here, which is just going to be zero. Now once again, why do we have this 2/3 and this 1/3 weighting? Because we are twice
as close to this point as we are to that point. Now if we wanted to simplify
it, what would we get? So this two would cancel with that two. So and this is zero, so we
would get a plus b over three for the x-coordinate. And for the y-coordinate, this is zero. That two cancels with that
two, and we get c over three. So we just found a point
that for sure sits on this blue median. Now let's do a similar exercise with this pink colored median. So that pink colored median,
what is the coordinate of the point that sits on that median, that is twice as far from the vertex as it is from the opposite side? Well it would be the exact same exercise. We would doubly weight these coordinates. So the x-coordinate would
be 2/3 times b over two plus 1/3 times a, 1/3 times a. And then the y-coordinate
would be 2/3 times c over two. 2/3 times c over two plus 1/3 times zero. 1/3 times zero. And what does that get us? Well let's see. This two cancels out with that two and we are left with b over
three plus a over three. So that's the same thing
as a plus b over three and over here, that's
zero, that cancels out, that is c over three. So notice, this exact coordinate sits on both the blue median and this pink median. So that must be the place
that they intersect. Let's see if that's also
true for this orange median. So in the orange median,
same exact exercise. And I encourage you to pause this video and try to calculate the value
of this point on your own. What's the coordinate of that
point on the orange line, where this distance is twice
as large as this distance? Well, same idea. We would doubly weight these
points right over here. So the x-coordinate would
be 2/3 times a over two, plus 1/3 times b. The y-coordinate would be 2/3 times zero. 2/3 times zero plus 1/3 times c. 1/3 times c, and what does that get us? Let's see, that cancels with that. So we have a over three plus b over three. Well that's the same thing
as a plus b over three. And then over here, that's just zero. And we're left with c over three. So notice, we have just shown
that this exact coordinate sits on all three medians. And so therefore, all three
medians must intersect at that point because that
point exists on all the lines. We've just shown that. And that's true for
this arbitrary triangle. You could make this triangle
have arbitrary dimensions by changing your value for a, b, or c. And if you see a triangle
that has the same dimensions but just, it's shifted or it's
in a different orientation, you can do a rigid transformation,
which doesn't change any of the dimensions. And you can show that that would be true for that triangle as well.