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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

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?
• 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 .
• 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 (-). 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.
• Nothing makes sense about the multiplying part ._.
• I assume you're talking about the `(2/3*((a+b)/2) + 1/3*(0), 2/3*(c/2) + 1/3*(0))`, which is at of the video. Also the ones at and .

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 `7` and `5`, 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, `(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 at on 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.
• How do you know that the centroid 2/3 of the way from each vertex?
• 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.
• It is 2/3 from the vertex and 1/3 from the side.
• why does the mid points have 2/3 and co-mid points have 1/3?
• 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 `7` and `5`, 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, `(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 at on 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.
• this is hard