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Current time:0:00Total duration:7:22

Proving triangle medians intersect at a point

CCSS.Math:

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.