What's the general idea behind how to prove concurrency using vectors?
More specifically, how do you prove that the medians of a triangle are concurrent?
And from there, how do you prove that the centroid divides each median in the ratio 2:1? (Vectors or an easier way using simple geometry)
Recall in a triangle a median is a line connecting a vertex to the midpoint of the edge opposite the vertex, so that a triangle has 3 median lines. We should make sure we know what it means for the medians of a triangle to be concurrent. It means that the 3 medians intersect at a single point, for example we might imagine that each pair of medians intersects at a different point, so that there are 3 possible intersection points (3 ways to form a pair of medians), we have to prove that this cannot happen.
Let's set things up. First let ABC be a triangle and let A', B', C' be the midpoint of the edges opposite A, B and C respectively, so that the medians are AA', BB', CC'. Furthermore let's call the point at which the medians AA' and BB' intersect P. I've sketched a diagram of our setup below.
What would we need to show? Well, we need to show that the lines AA' and CC' intersect at P, and the lines BB' and CC' intersect at P. Both of these are true if and only if the median CC' passes through P, and that's what we'll aim to show.
Next we should convert the statement
"the median CC' passes through P" into a statement about vectors if we hope to prove it using vectors. Another way to phrase the statement is that the points C, P and C' are collinear, and this is something we should be familiar with and be able to recognise as being the same as the (vector) statement that CC' = k CP for some constant k (i.e. CC' and CP are parallel).
We've now set up the framework and we know what we need to prove. The next bit, which is actually proving the statement relies only on technique (if you have experience there isn't much to think about).
Our aim is to write CC' and CP in terms of only AB' and AC' (i.e. the minimum number of variables which completely define the problem, we could've chosen other quantities like AB and CB, as long as there are two).
We have CC' = -2 AB' + AC'
However it seems more difficult to get CP, at this stage we should bear in mind the information we have and
must use if we are to solve the problem, which is that AA' = a AP and BB' = b AP for some constants a, b (which is to say the points A, P, A' and B, P, B' are each collinear).
Using this we find that: CP = CA + AP = -2 AB' + a AA' = -2 AB' + a (AC' + AB') = (a - 2) AB' + a AC'
Even having done this we note that we haven't got enough information to deduce that CC' = k CP for some constant k since we have this problematic constant of a. The problem is that we haven't used BB' = b AP in any of our equations and if we think about it a little we find that we could've found CP in another way by going through the path C-A-B-P.
If we do that we get CP = -2 AB' + 2 AC' + BP = -2 AB' + 2 AC' + b (-2 AC' + AB') = (b - 2) AB' + (2 - 2b) AC'
Now we have two different expressions for CP which are CP = (b - 2) AB' + (2 - 2b) AC' = (a - 2) AB' + a AC'. Since AB' and AC' are not parallel we can equate the coefficients of AB' and AC' (this is to say that AB' and AC' are linearly independent, intuitively they form a coordinate system for the plane). Hence we have the simultaneous equations b - 2 = a - 2 and 2 - 2b = a, which we solve to get a = b = 2/3.
Going back to our expression for CP but putting in a = 2/3 or b = 2/3 we have CP = -4/3 AB' + 2/3 AC', comparing this to our expression for CC' we have CC' = 3/2 CP, so that our constant k from before is 3/2. And we've proved what we wanted, and along the way we've essentially obtained the fact that the centroid, which we can now confidently say is the well-defined point P, divides each median in the ratio 2 : 1 (follows from our constant k = 3/2, do you see how?).
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