Show the sum of the medians of a triangle = 0.

Question

I am trying to show that the vector sum of the medians of a triangle is 0. I have seen the following question and answer: Sum of medians of a triangle

But I do not understand it. I'm hoping someone can help.

Answer 1

Consider one of the medians of $ABC$, say from $A$ to the midpoint $M$ of $BC$. As a vector, this is $M - A$. Now, $M = \frac12(B+C)$, so the median is $$\frac12(B+C) - A.$$ Find similar expressions for the other two medians, and add everything together.

Answer 2

The medians of any triangle is preserved by any affine transformation. In particular, any triangle is affinely equivalent to an equilateral triangle, but the sum of the vectors from the vertices to the midpoint of the opposite side of an equilateral triangle is zero. This is true, in general, for any regular polygon with an odd number of sides including a triangle. Prove it using the fact that the sum is invariant under rotation around the centroid (=center) and hence must be zero. That is, the vectors from the vertices to the midpont of the opposite sides sum to zero. It is also true for a regular polygon with any number of sides that the sum of the vectors from the vertices to the centroid is zero.

Note that this proof does not imply that it is true for general polygons since not all polygons with the same number of sides are affinely equivalent. However the result still holds in general using vector algebra. For example, suppose that $\,v_1,v_2,v_3,v_4,v_5\,$ are all the vertices of a polygon, $\,c \!:=\! (v_1\!+\!v_2\!+\!v_3\!+\!v_4\!+\!v_5)/5\,$ is the centroid, and define $\;a_i := v_i - c\,$ is the vector from the centroid to $\,v_i,\,$ then $\,a_1 + \cdots + a_5 = v_1 + \cdots + v_5 - 5c = 0.\,$ A closely related result is that if $\,b_1 := (v_3+v_4)/2 - v_1\,$ is the vector from $\,v_1\,$ to the midpoint of its opposite side, and similarly for the other vertices, then $\, b_1 + \cdots + b_5 = (v_1 + \cdots + v_5) - (v_1 + \cdots + v_5) = 0.\,$

Answer 3

Pick some origin in the plane and let $A,B,C$ be vectors to the vertices of the triangle. Note that the midpoint of the side BC is $\frac 12(B+C)$. The vector that represents the median from $A$ is then $\frac 12(B+C)-A$ You can cyclically permute the vertices to get the other two medians. Now add them together and note you get $0$.