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

Proving triangle medians intersect at a point

CCSS.Math:

Video transcript

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 it's pretty neat and this is true for all triangles and so to set up this proof I've put an arbitrary triangle here but I've 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 0 0 this one over here we're just saying that the x-coordinate is a and so it's a comma 0 and then this one up here has some X in 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 because 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 medians 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 at 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 R 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 it's going to be the average of B and a so you could just write that as a plus B over 2 and then the y-coordinate is going to be the average of C and 0 that would be C plus 0 over 2 or just C over 2 and we could do that for each of these points so at this point right over here it's x-coordinate is going to be the average of zero and a so that's just a over two and it's why corn 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 0 and then last but not least what's the coordinate of this point pause the video try to figure that out all right well it's going the x coordinate so it's going to be the average of B and 0 which is just going to be B over 2 and then the y-coordinate is going to be the average of C and 0 which is just going to be C over 2 so the way that I'm going to 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 two-thirds 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 2 plus 1/3 times 0 and then the y-coordinates going to be 2/3 times C over 2 plus 1/3 times the y-coordinate here which is just going to be 0 now once again why do we have this 2/3 and this 1/3 waiting 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 2 would cancel with that too so and this is 0 so we would get a plus B over over 3 for the x-coordinate and for the y-coordinate this is that - cancels with that - and we get see over three so we just found a point that foreshore 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 wait these coordinates so the x coordinate would be two thirds 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 one three times zero and what does that get us well let's see this 2 cancels out with that 2 and we are left with B over 3 plus a over 3 so that's the same thing as a plus B over 3 and over here that's a zero that cancels out that is C over 3 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 point 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 will doubly weight these points right over here so the x coordinate would be 2/3 times a over 2 plus 1/3 times B the y coordinate would be 2/3 times 0 2/3 times 0 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 3 plus B over 3 well that's the same thing as a plus B over 3 and then over here - 2 0 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 as 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