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# Dividing triangles with medians

Showing that the three medians of a triangle divide it into six smaller triangles of equal area. Brief discussion of the centroid as well. Created by Sal Khan.

## Want to join the conversation?

- Can someone please explain the difference between a median, an angle bisector, and a perpendicular bisector? Thank you.(26 votes)
- Median - A line segment that joins the vertice of a triangle to the midpoint of opposite side.

Angle bisector - A line segment that divides an angle of a triangle into two equal angles.

Perpendicular bisector - A line segment that makes an angle of 90 deg (right angle) with the side of a triangle.

The common point where the medians intersect is the centroid.

The common point where the angle bisectors intersect is the incenter.

The common point where the perpendicular bisectors intersects is the circumcenter.(63 votes)

- Is there somewhere on this site where I can practice this information?(7 votes)
- You should talk to your geometry teacher or the geometry teacher at your school. (assuming you're in school.) They could probably put something together for you.(4 votes)

- What exactly is the difference between a median and a midsegment?(3 votes)
- Here is a diagram of a triangle ABC with D, E, and F representing the midpoints of the three sides. AF is a median (a line segment connecting one vertex to the midpoint of the opposite side), and DE is a midsegment (a line segment connecting the midpoints of two sides).

http://www.khanacademy.org/cs/median-and-midsegment/1317725932(11 votes)

- Is there a proof that shows that the three medians of a triangle will all intersect at the centroid?(2 votes)
- Look at this video: http://www.khanacademy.org/math/geometry/triangle-properties/medians_centroids/v/triangle-medians-and-centroids--2d-proof

Sal derives the equations of 2 medians. You can use the same approach to deduce the equation of the third median. Then, you just need to make sure that the point of intersection of first two medians satisfies the equation of the third median. By doing so, you actually proving that all three medians intersect at one point.

Hope that helps.:)(4 votes)

- If you draw a another line in between that passes through a point in between E and C. Do the triangles formed still have the same area?(3 votes)
- From the middle of EC up to B?

or to the point 1/3 of the way above F on line FB?(2 votes)

- How do we find the median and the centroid ratio from shapes with more than 3 sides (triangles)? is it possible to do it with an even number of sides? - I couldn't see how.(2 votes)
- These are special points and lines in triangles, some don't exist the same way in higher order polygons. Even if we could construct them, they wouldn't behave the same way (as in, they wouldn't be concurrent at one point).(4 votes)

- how do we know that BE is the height of BEC and BEA if neither of those triangles is a right triangle? The medians of a triangle aren't necessarily perpendicular bisectors of the sides, are they?(2 votes)
- The altitude, or height of a triangle is defined to be perpendicular to the base. Therefore, the heights are equal.(2 votes)

- Just a quick question... is there a practice for this activity?(1 vote)
- If you go into the section "Special Properties And Parts Of Triangles" and look for the little circle with a star in it, you should find what you are looking for.(1 vote)

- does it mean anything if three lines going from each vertex to the opposite side of a triangle are concurrent if they are not medians, angle bisectors, and altitudes? kind of like the drawing at1:11(1 vote)
- Can I do this on the website?(1 vote)

## Video transcript

But what I think about now is the if we're given some triangle, and here we have triangle ABC What are the medians of the triangle and how do they relate to each other and do they have some interesting properties, and you might guess that they do So the median is, if we start at one of the vertices so let's start at this one right over here And then we bisect the opposite sides So this right here would be a median And so it started there, and it bisects this side right over here So the length from B to let's call this D is equal to the length from D to C Now, let's do that with every side I can draw a median over here, like that So, let's call this point E So the length from A to E is going to be equal to the length from E to C Although it's kind of a little lopsided, but that gets it pretty close And then we're gonna draw another median And I'm not gonna prove it in this video, but all the medians, and this is another neat thing, that, you know, when you have three lines that will always intersect at one point, that all the medians do intersect at one point They all are concurrent They all hit one common point in the center So, let's draw it that way I'm not gonna prove it in this video, so this length, let's call this point right over here F So this length right over here, this length right over there is equal to that length over there And the point at which these medians intersect is called the centroid, the centroid And when you start studying physics, if you're actually and this was a uniform triangle and you were to throw it, it would rotate around the centroid right over here But we'll study it geometrically for now So let's call this ten centroid well, we've already gone up to F So let's call this centroid G right over there Now, what I wanna that by itself is neat that you have a centroid, that if you were to throw it, if it was a uniform mass, it would rotate around the centroid But what's even neater about this, is that we can see that we've divided this triangle into six smaller triangles What's really cool, even though these aren't congruent triangles necessarily, but they do have the same area And that's what we're going to prove in this video That these six triangles all have the same area Now, to start off, I'm just going to look at two I'm gonna look at different pairs of triangles So let's look at let's look at these two triangles right over here Let's look at those two triangles over here And to show that those two have the same area, we're just going to invoke a very simple principle So, imagine rotating, just those two triangles over So just rotating, just those two triangles over It would look something like this It would look something like this Let me try my best to draw it Where this would be point G I'll even try to color code it the same that is point G That is that side right over there This is point C This is point B, that is point B And then this right over here would be the second part of that median right over there That over there would be point D Now, we know and I didn't draw it I should, at nicely over here We know that this length is equal to this length right over here And these two triangles, if we're starting to think about the area, they have the same base And we know area Area is equal to ? base * height So, they definitely have the same base What about their heights? Well, they also have the same heights Both of their heights, both of their height is exactly that tall They both have the same height So both of them have the same base They both have the same height On an ob on an obtuse triangle right here, the altitude sits outside of it So that might be a little counterintuitive So if you have an obtuse triangle like that I say obtuse cause this is more than 90 degrees Your altitude, your height will actually sit outside of the triangle, but that's okay Both of these triangles have the same base, and the same height, so they must have the same area So if this one right here has area x, this one right here will have area x as well And you can use the exact same logic to say, well, look This guy and this guy, they have the same base, and they both have the same height So if this one right over here is of area y, then this one over here is also going to be of area y They will have the same area And then finally, we could do the same thing for these two characters up here They both have the same base This was BF is equal to FA, and they both have the same height We drop an altitude like that And so if we call this area, if we call this area right over here Z, you could call that area Z, as well So, so far, we've shown that we can divide this into three pairs of triangles that have the same area But we wanna now show that they all have the same area And to do that, we can invoke this same principle, but we'll do it with different sets of triangles So now, let's look at triangle let's look at triangle BAE Look at triangle BAE, triangle BAE So the area of triangle BAE, area of BAE, is going to be equal to z + z + y z + z + y And let's look at the area of triangle BEC, BEC right over there That's going to be, this triangle right over here, is going to be x + x + y So the area of BEC is going to be x + x + y, but the same principle They both have the same base, and they both have the same height We could drop an altitude like this This one is obtuse, so that the altitude sits outside of it, but they have the exact same height So these two areas need to be equal to each other So you have Z, well, let me just add that up Now you have 2z + y is going to be equal to, is going to be equal to 2x + y 2x + y Subtract y from both sides You get 2z is equal to 2x Divide both sides by 2 You get z is equal to x And so we could say, we could write an x here, and an x there So we know that all of these will have the same area, but we still have to worry about these y's here And to do that, we just have to kind of rotate the way that we look at it And now look at triangle ADC Let me do that in a different color Triangle ADC, which I'm highlighting right over here Triangle ADC, whose area, so the area of ADC is going to be 2y + x, is equal to 2y + x And then look at triangle let's see, what color have I not used yet Let me use this green Triangle ADB, triangle ADB Triangle ADB is going to be equal to, well, you could say it's 2z + x, but we know that the z's are equal to x It's really just x + x + x It's really equal to ADB is equal to 3x And we have the same idea here ADB has this base, which is the same as ADC's base, and they both have the same height We can draw up an altitude like this, we can draw up an altitude like that They have the same height We're just invoking this principle over and over and over again So these things have to equal each other So we get 2y + x is equal to 3x, is equal to 3x Subtract x from both sides You get 2y is equal to 2x, or, or y is equal to x So that's a really neat result You go from each of the vertices of a triangle to the opposite side, and you bisect that opposite side, and you do that three times, you have three medians These lines are called medians, where they intersect as a centroid You know what's really cool is, it divides the triangle into six smaller triangles of equal area