# 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.)

• And by the way ,do these 3 points namely D,E and F are the mid points of the sides of the triangle? – ssk Oct 29 '18 at 7:14
• Are you asking me what your question means?! – José Carlos Santos Oct 29 '18 at 7:16
• I can't understand what u are saying – ssk Oct 29 '18 at 7:17

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: $$\bigg(\frac{x_1+p\,x_2}{1+p}+\frac{x_2+p\,x_3}{1+p}+\frac{x_3+p\,x_1}{1+p}\bigg)/3=(x_1+x_2+x_3)/3$$
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 $$ABC$$.