Ratios of triangles Consider a triangle $ABC$. Let 3 points $D,E$ and $F$ divide the sides $BC,CA,AB$ respectively in the same ratio. Prove that the centroids of both triangles namely $ABC$ and $DEF$ coincide.
(The problem would be easy if $D,E$ and $F$ are the midpoints of the sides of the triangle. However I do not know if this is the case.)
 A: When points are viewed as vectors, the centroid of a triangle is the average of the points. Denoting the vector associated with a point by the lowercase of the letter used for the point, we have
$$d=(1-t)b+tc\qquad e=(1-t)c+ta\qquad f=(1-t)a+tb$$
where $t\in[0,1]$. Thus the centroid of $DEF$ is
$$\frac{d+e+f}3=\frac{(1-t)b+tc+(1-t)c+ta+(1-t)a+tb}3=\frac{a+b+c}3$$
which is also the centroid of $ABC$.
A: Broad hints: 
You may use analytic geometry. It's easy to show that a centroid has coordinates $(x_1+x_2+x_3)/3$ and $(y_1+y_2+y_3)/3$ $\;$ where $(x_1,y_1)$ and etc. are the coordinates of the triangle vertices. You just need to use the fact that medians intersect at the centroid which divides the medians in ratio $1/3$. 
The line segment AB with $A(x_1,y_1)$ and $B(x_2,y_2)\;$ divided by some point on it with ratio $k$ yields the coordinates (the basics of analytic geometry): 
$$\frac{x_1+k\,x_2}{1+k}\;;\;\;\frac{y_1+k\,y_2}{1+k}$$
Now it easy to find the coordinates of the centroid, and I'm leaving it out for you giving only the answer here $(x_1+x_2+x_3)/3\;$and $(y_1+y_2+y_3)/3\,.$
Just as easy is to find the centroid of a new triagle with points D, E, F which divides the sides of the original triangle ABC in the same ratio, say, $\;p.\;$ We use the formulas above to find the vertices or our new triangle:
$$\bigg(\frac{x_1+p\,x_2}{1+p}\;;\;\;\frac{y_1+p\, y_2}{1+p}\bigg);\quad \bigg(\frac{x_2+p\,x_3}{1+p}\;;\;\;\frac{y_2+p\, y_3}{1+p}\bigg);\quad \bigg(\frac{x_3+p\,x_1}{1+p}\;;\;\;\frac{y_3+p\, y_1}{1+p}\bigg)$$
Now we find its centroid. Make sure you can derive all the formulas above using the properties of medians and the formulas for the line segment. I'm talking about deriving $(x_1+x_2+x_3)/3.\;$ The rest is easy. It's the same procedure:
$$\bigg(\frac{x_1+p\,x_2}{1+p}+\frac{x_2+p\,x_3}{1+p}+\frac{x_3+p\,x_1}{1+p}\bigg)/3=(x_1+x_2+x_3)/3$$
