# The three medians meet where in relationship to triangle

### All about medians Using the standard notations, in ΔABC, there are three medians: AMa, BMb, CMc. Three medians of a triangle meet at a point - centroid of the triangle. Below are. In a triangle, a median is a line joining a vertex with the mid-point of the opposite side. Every triangle have 3 medians. The three medians meet at one point. a b a + b We will assume that our common laws for algebra hold for vector algebra, and try to prove our theorem. After testing our theorem on real world triangles.

We notice that the second median divides the green and yellow triangles in the ratio 1: We can write a conjecture here, If a median one median of a triangle is drawn, the second median to be drawn will divide the areas of the two triangles formed by the first median in the ratio 1: We will postpone the proof for now as we proceed to the case where there are three medians.

Three medians Figure 3 From the areas calculated it is easy to see that after drawing all the three medians the original triangle is divided into six triangles that are all of the same area. Another thing we have noticed is that the three medians are meeting at the same point All they are concurrent. Conjecture Three medians of a triangle divide the triangle into six triangles that are all equal in area.

Earlier on when we considered the case of two medians we saw that the second median divided the two triangles formed by the first median in the ratio 2: The rest of the proof is trivial. The medians are concurrent. The medians of a triangle intersect each other in the ratio 2: Here I will simply state the theorems formal proofs are omitted, but are part of secondary school mathematics 1.

Mid-Segment theorem A line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. If a pair of opposite sides are equal and parallel, the quadrilateral is a parallelogram. Diagonals of a parallelogram bisect each other. We start the proof as follows. Let the three medians meet in G.

### Triangle medians & centroids (video) | Khan Academy

Also we know that diagonals of a parallelogram bisect each other. From above we know that the median BO intersects with the median AN in G, therefore G must be the common point where all the three medians are meeting. So the three medians are concurrent. Well, I think you get the idea. These are all medians of this triangle.

## Triangle medians & centroids

And what's neat about medians is that all three medians always intersect in one point. And that by itself is a pretty neat property. And that one point that they intersect in is called the centroid. And if this was actually a physical triangle, let's say you made it out of iron, and if you were to toss it-- well, even before you toss it, the centroid would actually be the center of mass.

So let's say this is an iron triangle. Let's say that this right here is an iron triangle that has its centroid right over here, then this iron triangle's center of mass would be where the centroid is, assuming it has a uniform density. And if you were to throw that iron triangle, it would rotate around this point. Assuming that it had some rotational motion, it would rotate around that centroid, or around the center of mass. But anyway, the point of this video is not to focus on physics and throwing iron triangles.

The point here is I want to show you a neat property of medians. And the property is that if you pick any median, the distance from the centroid to the midpoint of the opposite side-- so this distance-- is going to be half of this distance.

So if this distance right here is a, then this distance right here is 2a. And let's just prove it for ourselves just so you don't have to take things on faith.

And to do that, I'll draw an arbitrary triangle. I'll do a two-dimensional triangle, and I'll do it in three dimensions because at least in my mind, it makes the math a little bit easier. In general, whenever you have an n-dimensional figure and you embed it in n plus 1 dimensions, it makes the math a little bit easier.

The actual tetrahedron problem that we did, you could actually embed it in four dimensions and it would make the math easier. It's just much harder to visualize, so I didn't do it that way. But let's just have an arbitrary triangle. And let's say it has a vertex and there, a vertex there, and a vertex there. So I'm not making any assumptions about the triangle. I'm not saying it's isosceles, or equilateral or anything.

It's just an arbitrary triangle. And so let's say this coordinate right over here is-- I'll call this the x-axis. So, this is the x-axis, the y-axis, and the z-axis. I know some of y'all are used to swapping these two axes, but it doesn't make a difference. So let's call this coordinate right here a, 0, 0. So it's a along the x-axis. Let's call this coordinate 0, b, 0.

### Median (geometry) - Wikipedia

And let's call this coordinate up here, 0, 0, c. And if you connect the points, you're going to have a triangle just like that. Now, the centroid of a triangle, especially in three dimensions. The centroid of a triangle is just going to be the average of the coordinates of the vertices.

## Median (geometry)

Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. So we have three coordinates. They add up to a, and we have to divide by 3. So it's a over 3. The y-coordinate is going to b plus 0 plus 0. They add up to b, but we have three of them, so the average is b over 3. And then same thing-- we do it for the z-coordinate. The average is going to be c, is c over 3. And I'm not proving it to you right here.

You could verify it for yourself. But it's going to be the average, that if you were to figure out what this line is, this line is, and this line is, this centroid, or this center of mass of this triangle, if it had some mass, is just the average of these coordinates. Now, what we want to do is use this information.