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# 三角形重心與中線 (2D 英語證明) : 重心位於每條中線的2/3。

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- In the video on triangle medians and centroids
- I did essentially the proof that the centroid is 2/3 along the way of the 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
- Ah, but someone mentioned that they'd be interested in seeing the two dimensional version of the proof,
- so, so why not do that. So let's just draw an arbitrary triangle,
- and I'll make one, 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, let's put kind of the height of the triangle, let's put it on the y-axis.
- So this triangle right here, this is y-axis, that is the y-axis, and then
- this right here. This right here,is our x-axis. Now, we said that this is an arbitrary triangle.
- So let's make, let's just make this distance equal to A, so this point right here, this vertex would be the point (A,0), so that distance is A, let's make
- this distance, 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 (-B,0)
- and then let's make this distance, this is the point Y=Z, so this is the point (0,Z) So this is an arbitrary triangle, any triangle can be represented this way
- Now let's think about it's medians, let's think about medians and the centroid And I'm only gonna do for two of the medians, because we know
- that the third median will also intersect at the same centroid
- So let's just find, so we have a median on this, we have a midpoint,
- I should say, on this side of the triangle
- And it's coordinates are just gonna be
- midpoint of those two points or,
- zero plus A over 2, which would just be A over 2
- And then C plus 0 over 2, which would just be C over 2
- And let's do the same thing over here
- The midpoint of this side, right over here, is going to be,
- negative B plus 0 over 2
- So it's negative B over 2
- And then, 0 plus C over 2
- So it's just going to be, it's just going to be C over 2
- Now that we know all the coordinates for the vertices
- and the midpoints,
- at least for this two sides, we can find the equations
- for the lines that the medians are part of
- So we could find the equation for this line, we could find
- the equation in this line in purple, then we could find the equation
- for the other lines that connect this 2 dots, find the intersection,
- and we'll essentially have the coordinates of our centroid,
- of the intersection of the medians
- And we could 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, what would be the equation
- for this line right over here?
- Well our slope, our slope is going to be changed in Y,
- so C over 2 minus 0,
- so that's just C over 2, over changing in X
- So, A over 2 minus, minus B, minus B
- So it's a minus -B I should say
- So, it's plus B, could write a plus B over here,
- but just to make the math simple,
- I'll write plus 2B, plus 2B over 2, right?
- I just subtracted a -B, which is the same thing as adding a B,
- and I just, and adding B is the same thing as adding a 2B over 2
- And I did that so these have the same,
- these have the same denominator
- I could add them, or I could just multiply the numerator
- and the denominator both by 2,
- 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 -B comma 0 is on the line,
- so we know that the equation of this line in purple,
- let me state in the purple is going to be, Y minus 0
- is equal to, is equal to X minus negative B,
- or I could say X plus B times our slope,
- times C over A plus 2B, A plus 2B
- And obviously, this would just simplify that Y
- is equal to all of these business over here
- So this is essentially the equation of this median right here,
- if the line just kept going, or this line
- The median is a line segment that is part of this line
- Now let's do the same thing for, let us do the same thing for
- this line right over here
- This median right over here
- So once again, its slope is changed over,
- change in Y over change in X
- So the change in line C over 2 minus 0
- So the slope here is equal to C over 2 minus 0,
- which is just C over 2,
- over -B over 2, -B over 2 minus A
- Or I could say, minus 2A over 2,
- that's the same thing as minus A
- And do the same thing, multiply the numerator
- and the denominator by 2
- We get C over, C over -B minus, -B minus 2A
- So the equation of this line right here, the other median,
- the equation of this other median right over here is,
- we have this point over here so, we could use the point slope again
- We get Y minus 0, Y minus 0, is equal to X minus A,
- is equal to X minus A times the slope
- Time C over -B minus, -B minus 2A
- So we have the 2 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 0 is the same thing as Y, Y minus 0 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, is equal to
- X minus A times C over -B minus 2A
- Well both sides are divisible by C,
- we could assume that their C is non-zero,
- so this triangle actually exist
- It actually exist in 2 dimension
- So we can divide both sides by C and we get that
- Now 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
- And so we can just distribute it, so it's going to be
- X times all of this business,
- so it's going to be X times -B minus 2A
- plus B times all of these
- So that gonna be minus B squared, minus 2AB
- So i just multiplied that times that,
- is going to be equals to that times that
- So it's gonna be X times all of this business
- X times A plus 2B, minus A times that
- So minus, distribute the A, minus, A squared minus to AB,
- see if we can simplify this
- So it's, we have a negative 2AB on both sides,
- so let's just subtract that out
- And then if we, let's, so we cancel things out,
- let's subtract this from both sides of the equation,
- so we'll have a minus, minus X times A minus 2B do,
- and I'll subtract that from here,
- but I'll write it a little different,
- it'll simplify things
- So this will become, this will become a 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 the the constants on the other
- And so this becomes, 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,
- these guys cancel out, is equal to
- B squared minus A squared
- Is equal to B squared minus A squared
- Let's see if we can fact,
- 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 our negative 3
- or we can actually factor out a 3,
- we'll get 3 times B minus A, 3 times, actually,
- let's factor a negative 3
- So negative 3, negative 3 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 3 times X
- is equal to B minus A, we'll just divide both side by negative 3
- We get X is equal to B minus A over negative 3,
- which is the same thing asA minus B, A minus B over 3
- So we have the X coordinate,
- we have the X coordinate so far for our centroid
- It is A minus B over 3
- It is A minus B over 3
- And you kinda see hints of the solution over here
- A minus B is its entire distance
- It's one third of this entire distance, although,
- I don't want to jump the gun too much because
- it complicates it just thinking about, just the X coordinated
- Let's figure out the Y coordinate
- To do that, we could just substitute back
- into one these equations up here
- This was the, this was the equation of this median,
- the gray median that we had done
- So let's substitute back in, we have Y is equal, Y, I did over here
- Y is equal to X minus A times C
- X is this thing over here
- So it's A minus B over 3 minus A
- So I'm gonna subtract this, instead of just writing
- a minus A over here, I'm gonna write negative 3 over 3
- So minus 3A over 3
- So that's just minus A right over here
- I just multiplied it and divided it by 3
- It's gonna be all of that times C over, over negative B minus 2A
- So this is going to be equal to C
- In the numerator up here, we have an A minus 3A,
- so this equal to negative B, negative B
- We have a negative B there
- And then we have aA minus 3A is minus 2A
- So this becomes negative B minus 2A times,
- we could factor out the one third,
- times C over 3, 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 3
- So our Y coordinate here is C over 3
- Now could just use the distance formula now
- We know all the coordinates and,
- and you 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, we know that the height, we know that the Y coordinate,
- we know that the Y coordinate of the centroid is C over 3
- So this right here, this point over here is C over 3
- So we could actually use, maybe an easier argument, cause remember,
- our whole point was to show, our whole point was to show
- that this point is two thirds along the median away from the vertex
- So 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, this right here is
- two thirds along the way
- Or that this length is twice this length
- Or that this length, this length is two thirds
- of the entire length of the median
- So how can we do that?
- Well that simplest way I can do that, I mean you actually,
- you could actually just use the coordinates
- and use the the distance formula
- But a simpler way, just the way that we've arranged this,
- is just to use an argument
- of similar triangles
- Cause this right here, this right here is,
- we know that this entire height right here in blue is C
- We know this entire height over here in blue is C
- We know that, we know that this height
- right over here is C over 3 or,
- you could do that as one third times C
- And so, we know that this height,
- we know that this height over here,
- we know that, that distance is two thirds
- Is two thirds, is two thirds C
- So we can use a similar triangle argument
- Let's say, let's say that this entire thing right here
- is the length of the median
- And let's just call that, I dunno, let's call that L
- And let's say we wanna figure out how far along the median away
- from the vertex that is
- And let's just call that, I dunno,
- let me use another, another,
- let me just call that distance right over,
- let me do, use a different color
- Let's call, let's call this distance right over here,
- you see, I've already used A, B's, and C's,
- let's call that D
- So I, if we can show that the ratio
- of D to L is two thirds
- So this is two thirds of the distance then, we're done!
- And use that by similar triangles
- we have two similar triangles
- We have this, lemme use a new color
- We have this triangle, which is kind of embedded inside
- of this larger triangle right over here
- The both share the same vertex, this angle over here
- They both have right angles over here
- It's as they have two of the same angles,
- the third angle has to be the same
- So they're definitely similar triangles
- so we could just set up a ratio here
- Two thirds C, which is this length right over here
- That's the, that's just kind of the,
- the vertical side of the smaller triangles
- We could write two thirds C over this entire length,
- over the length of the larger that's D
- This corresponding side of the larger similar triangle
- over C is equal to, is equal to
- The hypotenuse of the smaller similar triangle D
- over the hypotenuse of the longer,
- of the longer similar triangle, or the bigger similar triangle
- D over L
- Well this is clearly, just divide the numerator and denominator by C
- this clearly becomes two thirds
- So the ratio from, of this, of this, of this length
- to this larger length, is two thirds
- Or this is two thirds a long way,
- a long the median away from the vertex
- So there you go
- That's the two dimensional proof that the centroid
- is two thirds along the way
- of any median from the vertex, or one third along the way,
- along the median
- Or one third of the distance of the median
- from the opposite midpoint
- Anyway, hopefully you enjoyed that

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