# Prove that the intersection of medians is the same point in both triangles.

Consider triangle $ABC$ We construct $A',B',C'$ like the picture below:

(Here the name of points are different.)$A,B,C$ are mid points.Prove that the intersection of medians is the same point in both triangles.

What should we do to prove that?The medians are not the same so working with the definition of all medians seems to be useless so I considered one median and found a point on it that divide it into two part that one part is two times bigger than the other but this definition also didn't work.What should I do now?

We have $D=2A-B$, $E=2C-A$ and $F=2B-C$. Therefore, the intersections of medians of the triangle $DEF$ is $$\frac{D+E+F}{3} = \frac{2A-B+2C-A+2B-C}{3} = \frac{A+B+C}{3}$$ which is the intersection of medians of the triangle $ABC$
$$A+B+C=\frac{D+B}{2}+\frac{F+C}{2}+\frac{A+E}{2}$$ or $$A+B+C=D+F+E$$ or $$\frac{A+B+C}{3}=\frac{D+F+E}{3}$$ and we are done!