Why is the Centroid of a Triangle Shared with Its Midpoint Triangle

Question

The question is as follows:

Let $A'$, $B'$ and $C'$ be the midpoint triangle of triangle $ABC$. In other words, $A'$, $B'$ and $C'$ are the midpoints of segments $BC$, $CA$ and $AB$, respectively. Show that triangles $A'B'C'$ and $ABC$ have the same centroid.

My reasoning was through the means of the Midline Theorem. After drawing the diagram, we can see, for example, $C'A'$ is a midline of $\triangle ABC$, which means that $C'A'$ is parallel to the base $AC$ and half the length of it as well. Therefore, the median $B'B$ would pass through the midpoint of $C'A'$, including the midpoint of $AC$.

Is my reasoning valid or, at least, is it on the right track? Any advice on solving this problem will be greatly appreciated.

Answer 1

Let $A(a)$, $B(b)$ and $C(c)$ in the Gauss's plane.

Thus, $M\left(\frac{a+b+c}{3}\right)$ is a centroid of $\Delta ABC$ or $$M\left(\frac{\frac{b+c}{2}+\frac{a+c}{2}+\frac{a+b}{2}}{3}\right),$$ which is a centroid of $\Delta A'B'C'$.

Done!

Answer 2

Observe a homothety with center at the centroid $T$ of the triangle $ABC$ and $k=-{1\over 2}$. This homothety takes $A$ to $A'$ and so on. Thus it takes centroid $T$ to the centroid $T'$ of $A'B'C'$ (beacuse a homothety is a similarity transformation). But the center of homothethy is fixed point for a homothety, so $T'=T$.