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# Triangle medians and centroids (2D proof)

Showing that the centroid is 2/3 of the way along a median. Created by Sal Khan.

## Want to join the conversation?

- I wonder if someone can prove a theorem that the median of any triangle (not just isosceles)divides the triangle into two triangles with equal areas. Thanks(18 votes)
- Since the median, by definition, divides a base in half, then you will have two triangle with the same base, as well as the same height. Therefore, the triangles will have the same area.(59 votes)

- I wish could have more practice or at least some mini tests here to see how much we have learned from these videos.(12 votes)
- Is it possible for you to give us the equation for the median line in an equation without using 0 as one of the coordinate points?(5 votes)
- I don't think x=(a-b)/3. Cause b is positive.

Say a=6, b=3, then the length of line AB should equal to a+b, which means 9.

So the centroid should be (a-(-b)/3 and c/3.

Does anyone agree?(5 votes)- It's already correct that the x-coordinate of the centroid is (a-b)/3.

The false thing is to say that it is 1/3 of the entire base length ;)(1 vote)

- Will this be used in further math progression? Say, Algebra II for example?(4 votes)
- Well, maybe not. But this comes in helpful with much more difficult problems.(1 vote)

- Is there a way to prove this without graphing it on a coordinate grid?(4 votes)
- I still don't understand the theorem. Can anyone help me understand? Thank you, in advance. :)(2 votes)
- If you connect a line from the midpoint of one side to the vertex opposite to that side (which is a median), then the centroid is where all 3 medians intersect.

The theorem basically says that:

The length of the centroid to the midpoint of the opposite side is 2 times the length of the centroid to the vertex.

Hope this helps!(3 votes)

- How does this relate to the centroid formula? (x1+x2+x3)/3 , (y1+y2+y3)/3

I recently found some information on this formula but I'm not exactly sure why it works.(2 votes) - At9:37x-coordinate is (a-b)/3. I worked a problem with a triangle with vertices (7,0), (-2,0), and (0,9) (a=7, b=-2, and c=9). Then x-coordinate centroid (7-(-2))/3 = 3. This, however, is incorrect (a simple drawing shows that the x-coordinate is less than 2). Can you elaborate?(2 votes)
- At12:00Sal draws a green segment from the vertex at the top of the triangle through the centroid to the side of the triangle at the bottom (on the x axis). He calls it a median but am I wrong in thinking that he does not prove that it is a median? He proves that d is 2/3 of l, but doesn't quite prove that l is the median. I don't think he quite proves that that green segment meets the triangle side at the bottom at its midpoint. He comes close; he just needs to show, using those similar triangles at13:00that (a-b)/3 is 2/3 of the distance from the origin to the midline which is (a-b)/2.(1 vote)

## Video transcript

- [Voiceover] In the video on
triangle medians and centroids I did, essentially, the
proof that the centroid is 2/3 along the way of a median. I did it using a two-dimensional triangle in three dimensions, and I mentioned that I thought, at least it made the math a little simpler, but someone mentioned
that they'd be interested in seeing the two-dimensional
version of the proof, so why not do that? So let's just draw an arbitrary triangle, and I'll make one side of the triangle right along the x-axis just to make, I think it'll make the
math a little bit easier, maybe there's easier ways of doing it. So that's one side of the triangle, that's the other side of the triangle, and let's put the height of the triangle, let's put it on the y-axis, so this triangle right here, this is the y-axis that is the y-axis, and then this right here, is our x-axis. Now, we said that this
is an arbitrary triangle, so let's just make this distance equal to a, so this point right here, this vertex would be the point a comma zero, so that distance is a. Let's make this distance right here be equal to, I don't know,
let's make it equal to b, so this is the point
negative b comma zero, and then let's make this distance, this is the point y is equal to c, so this is a point zero comma c, so it's an arbitrary triangle, any triangle can be represented this way. Now, let's think about its medians. Let's think about its
medians and the centroid. I'm only going to do it
for two of the medians, because we know that the third median will also intersect at the same centroid, so we have a median on this, we have a midpoint, I should say, on this side of the triangle, and its coordinates are just going to be the midpoint of those two points, or zero plus a over two, which
would just be a over two, and then c plus zero over two, which would just be c over two. Let's do the same thing over here. The midpoint of this side right over here is going to be negative
b plus zero over two, so it's negative b over two, and then zero plus c over two, so it's just going to be c over two. Now that we know all the coordinates for the vertices and the midpoints, at least of these two sides, we can find the equations for the lines that the medians are part of, so we can find the equation for this line, we could find the equation
for this line in purple, then we can find the
equation for the other line that connects to these two dots, find their intersection, and we'll essentially have the
coordinates of our centroid, of the intersection of the medians, and we can do it for this one, too, but that would be redundant, because it's going to
intersect in the same point. So what would be the equation
for this line right over here? Well, our slope is going
to be change and y, so c over two minus zero, so that's just c over two, over change and x, so a over two minus minus b, so minus negative b, I should say, so it's plus b, I could
write a plus b over here, but just to makes the
maths involved right, plus 2b over two. I just subtracted a negative b, which is the same thing as adding a b, and adding a b is the same
thing as adding a 2b over two, and I did that so these
have the same denominator. I could add them, or I could just multiply the numerator and the
denominator both by two and I'll get the slope is equal to, c over a plus 2b, so that's the slope of
this line right over here, and then we know some points here. We could just use the point slope formula for the equation of a line. We know the point negative
b comma zero is on the line, so we know that the equation
of this line in purple, let me stay in the purple, is going to be y minus zero, is equal to x minus negative b, or I could say x plus b, times our slope, times c over a plus 2b. And obviously, this would just simplify to y is equal to all of
this business over here, so this is essentially the equation of this median right here
if the line just kept going, or the median is a line segment
that is part of this line. Let us do the same thing for
this line right over here, this median right over here. So once again, its slope is change and y over change and x, so the change and y,
c over two minus zero, so the slope here is equal
to c over two minus zero, which is just c over two, over negative b over two, minus a, or I could say minus 2a over two, that's the same thing as minus a, and do the same thing, multiply the numerator and
the denominator by two. We get c over negative b minus 2a, so the equation of this line
right here, the other median, the equation of this other
median right over here, well we have this point over here, so we could use the point slope again. We get y minus zero is equal to x minus a times the slope, times c over negative b minus 2a, so we have the two
equations for these lines. We can now use that information
to find their intersection, which is going to be the centroid. Well, to do that, we have both of these in terms of y, right? Y minus zero is the same thing as y, so we could just set this as being equal to that, so let's do that. We get x plus b times c, over a plus 2b is equal to x minus a times c over negative b minus 2a. Well, both sides are divisible by c, we can assume that c is non-zero, so this triangle actually exists, it actually exists in two dimensions, so we can divide both
sides by c and we get that. Let's see, now we can cross-multiply. We can multiply x plus b times
this quantity right here, and that's going to be equal to a plus 2b times this quantity over here. We can just distribute it, so it's going to be x
times all of this business, it's going to be x times
negative b minus 2a, plus b times all of this, so that's going to be
minus b squared minus 2ab, so I just multiply that times that, is going to equal to this times that, so it's going to be x
times all of this business. X times a plus 2b, minus a times that, so distribute the a, minus a squared minus 2ab. Let's see if we can simplify this. We have a negative 2ab on both sides, so let's just subtract that out, so we cancelled things out, let's subtract this from
both sides of the equation, so we'll have a minus x times a plus 2b, and I'll subtract that from here, but I'll write it a little
different, it'll simplify things, so this will become x times, I'll distribute the
negative inside the a plus 2b, so negative a minus 2b, and let's add a b squared to both sides, so plus b squared, plus b squared. I want to collect all
the x terms on one side and all of the constants on the other. This becomes on the left-hand side, the coefficient on x, I have negative b minus 2b is negative 3b negative 2a minus a is negative 3a, times x is equal to,
these guys cancel out, and these guys cancel out, is equal to b squared minus a squared. Let's see if we can factor, this already looks a little
suspicious in a good way, something that we should be able to solve. This can be factored. We can factor out a negative three. We can actually factor out a three. We'll get three times b minus a, actually, let's factor
out a negative three, so negative three times b plus a times x is equal to, now this is
the same thing as b plus a times b minus a. So we can divide both sides by b plus a, we have negative three times
x is equal to b minus a, let's just divide both
sides by negative three. We get x is equal to b minus a over negative three, which is the same thing as a minus b over three, so we have the x-coordinate
so far for our centroid. It is a minus b over three. You kind of see hints of
the solution over here. A minus b is this entire distance. It's one third of this entire distance, although I don't want
to jump the gun too much because it complicates
just the x-coordinate. But let's figure out the y-coordinate. To do that, we can just substitute back into one of these equations up here. This was the equation of this median, the grey median that we had done, so lets substitute back in. We have, I'll do it over here, y is equal to x minus a times c. X is this thing over here, so it's a minus b over three minus a, so I'm going to subtract, instead of just writing
a minus a over here, I'm going to write a
negative three over three, so minus three a over three. That's this minus a right over here. I just multiplied and divided it by three. It's going to be all of that times c over negative b minus 2a, so this is going to be equal to c and the numerator up here, we have an a minus 3a, so this is equal to negative b, we have a negative b there and then we have a minus 3a is minus 2a, so this becomes negative b minus 2a, we can factor out the 1/3, times c over three, so
that's what the numerator is. All of that over negative b minus 2a, well, that's going to cancel with that, and so our y-coordinate is going to be equal to c over three, so our y-coordinate over
here is c over three. Now, we could just use
the distance formula now. We know all the coordinates, and feel free to do it if you like, but there's a slight simplification, or at least in my mind
a slight simplification. We know that the height, we know that the y-coordinate of the centroid is c over three, so this right here, this point
over here, is c over three. We could actually use,
maybe an easier argument, because remember, our
whole point was to show that this point is 2/3 along the median away from the vertex. Let me draw this vertex now here. Our whole point of this video as we said this is the
midpoint of this side over here is to show that this right here is 2/3 along the way, or that this length is twice this length, or that this length is 2/3 of the entire length of the median. So how can we do that? Well, the simplest way I can do that, I mean you could actually
just use the coordinates and use the distance formula, but a simpler way, just the
way that we've arranged this, is just to use an argument
of similar triangles. We know that this entire
height right here in blue is c. We know that this height right
over here is c over three. Or you could do that as 1/3 times c, so we know that this height over here we know that that distance is 2/3 c, so we can use a similar triangle argument. Let's say that this
entire thing right here is the length of the median, and let's just call that, I don't know, let's call that l, and let's say we want to figure out how far along the median
away from the vertex that is and let's just call that, let me use another, let me just call that distance, let me do this in a different color. Let's call this distance, right over here, let's see, I've already used a, b, and c, let's call that d. So if we can show that the ratio of d to l is 2/3, that this
is 2/3 of the distance, then we're done. We can do that by similar triangles. We have two similar triangles. Let me use a new color. We have this triangle,
which is kind of embedded inside of this larger
triangle right over here. They both share the same
vertex, this angle over here, they both have right angles over here, and since they have
two of the same angles, their third angle has to be the same, so they're definitely similar triangles. So we can just set up a ratio here. 2/3 c, which is this
length right over here, that's just kind of the vertical side of these smaller triangles, we could write 2/3 c
over this entire length, over the length of the larger, this corresponding side of
the larger similar triangle, over c is equal to the hypotenuse of the
smaller similar triangle, d over the hypotenuse of
the longer similar triangle, or the bigger similar triangle. D over l. This is clearly just dividing the numerator and denominator by c, this clearly becomes 2/3, so the ratio of this length to this larger length is 2/3, or this is 2/3 along the median away from the vertex. So there you go, that's the two-dimensional proof that the centroid is 2/3 along the way of any median from the vertex, or 1/3 along the median, or 1/3 the distance of the median from the opposite midpoint. Anyway, hopefully you enjoyed that.