Prove that the co-ordinates of the centroid of a triangle is an average of that of vertices For a given triangle [ABC], how do I prove that the co-ordinates of the Centroid $O_{xy}$ (intersection of the medians) is the average of the individual vertices?
$O_x =  \left(\frac {A_x + B_x + C_x}{3}\right)$
$O_y =  \left(\frac {A_y + B_y + C_y}{3}\right)$
I tried proving the height of the $O_y$ coordinate - which should be equal to the sum total of heights of the points A, B, C - divided by 3 - but to no avail.
 A: I'd calculate the centroid coordinates directly from the definition.  Place $A$ at the origin, $B$ at $(b,0)$ and $C$ somewhere in the upper half-plane.  Then the coordinates of the centroid are:
$$C_x = \frac{1}{A} \int_A x f(x) dx,$$
$$C_y = \frac{1}{2A} \int_A [f(x)]^2 dx,$$
where $A$ is the area of the triangle.  The lower bounding function, $g(x)$ is zero on the interval of interest because of our choice of positions.
Consider both a triangle with all acute angles, and one with an obtuse angle.
The function $f(x)$ will be split up into two sections for the acute triangle.  Calculating the centroid for the obtuse triangle involves calculating the integral for two right triangles, and subtracting one from the other.
Then, once you have these, calculate the average of the coordinates of the vertices, and see that they match what you just calculated.
A: The median through $C$ passes through the mid point of the opposite side ${1 \over 2} (A+B)$. Find an equation of this line.
Find the equation of another median and see where they intersect. Show that
this is the average of the points.
A: What is the average of the vertices ? $\dfrac{A+B+C}3~.~$ And where do the medians intersect ? A third above the middle, and two thirds below the vertex, or $\dfrac23\cdot\dfrac{X+Y}2+\dfrac13~Z,~$ where X, Y, Z form a permutation of A, B, C.
A: The median drawn from $A$ and the median drawn from $B$ intersect where $t$ and $u$ solve
$\displaystyle tA+(1-t)\frac{B+C}{2}=uB+(1-u)\frac{A+C}{2}$
Now rearrange terms so you get something that looks like $xA+yB+zC=0$, and $x$, $y$, and $z$ will all depend on $t$ and $u$.  Since $A$, $B$, and $C$ form a triangle, they are linearly independent.  That means that $x$, $y$, and
$z$ must all equal zero.  This gives you three equations for $t$ and $u$.  Solve and you have your intersection.  Verify that the median from $C$ intersects in the same place.
