Show that the barycenters of ABC and DEF coincide.

Question

Given a triangle $ABC$, consider the triangle $DEF$ where $B$ is the midpoint of $AD$, $A$ is the midpoint of $CF$ and $C$ is the midpoint of $BE$. Show that the barycenters of $ABC$ and $DEF$ coincide.

Obs: Use plan geometry

I have by vectors. I need by plan geometry

The midpoint is given by $\vec{X}=\dfrac{\vec{X}+\vec{Y}}{2}$

$\vec{A}=\dfrac{\vec{C}+\vec{F}}{2}$

$ \vec{B}=\dfrac{\vec{A}+\vec{D}}{2}$

$\vec{C}=\dfrac{\vec{B}+\vec{E}}{2}$

$ \therefore \dfrac{\vec{A}+\vec{B}+\vec{C}}{3}= \dfrac{\vec{D}+\vec{E}+\vec{F}}{3} \implies$

$\vec{G}_{ABC}=\vec{G}_{DEF}$

Answer 1

Comment: It can be shown that resulting triangle by extending each side of a triangle equal to itself is similar to original one. As shown in figure if you continue this you will get third triangle similar to original one. The position of all points like orthocenter, circumcircle and incenter vary due to scale of triangle, but centroid remains fixed at its original position. In fact we have a kind of affine transformation(https://mathworld.wolfram.com/AffineTransformation.html) by rotating about centroid and enlargement.

Answer 2

If you wonder why the solution you found on the Internet, is working, this and this one will probably help you understand the solution. If you still cannot convince yourself that the solution is working, you may adopt a new analytic approach explained below.

WLOG, we may assume that $A=(-1,0), C=(1,0),$ and $B=(a,b)$, where $a$ and $b$ are real numbers. Of course, at this step, you must persuade yourself why defining a triangle in such a way covers all of the possible cases.

Since the centroid divides in $2:1$ ratio, the centroid of $\triangle ABC$ is $(\frac{a}{3}, \frac{b}{3}).$

Now, we also have:

$$F=(-3,0),\\ E=(2-a,-b), \\ D=(1+2a,2b).$$

So, the midpoint of $FE$ is:

$$(\frac{-1-a}{2},\frac{-b}{2});$$

and the centroid of $\triangle FED$ is:

$$(\frac{2(\frac{-1-a}{2})+(1+2a)}{3}, \frac{2(\frac{-b}{2})+(2b)}{3})=(\frac{a}{3}, \frac{b}{3}).$$

Note that in the last line, we calculated a third of the segment between the midpoint of $FE$ and $D$.

We are done.