Krdytkyu

Suppose that M,N, and P are the midpoints of sides AB, AC, and BC of triangle ABC. Show that the centroid of triangle MNP is the same point as the centroid of triangle ABC.

1.
Consider the homothety which maps $∆ABC$ to its medial triangle $∆PNM$.

HINT:

What's the homothety center in such a construction?

2. Note that by the inversion of Thale's Theorem (not to be confused with another theorem with the same name), since
   $$frac{overline{AM}}{overline{AB}}=frac{overline{AN}}{overline{AC}}$$
   the segment $overline{MN}$ is parallel to $overline{BC}$.

Let $K=overline{AP}cap{overline{MN}}$. Again, by Thale's theorem
$$frac{overline{AM}}{overline{AB}}=frac{overline{MK}}{overline{BP}}$$
$$frac{overline{AN}}{overline{AC}}=frac{overline{KN}}{overline{PC}}$$
$$Rightarrowfrac{overline{MK}}{overline{BP}}=frac{overline{KN}}{overline{PC}}Rightarrowfrac{overline{MK}}{overline{KN}}=frac{overline{BP}}{overline{PC}}=1$$
This implies that K is also the midpoint of $overline{MN}$. Since $K$ lies on $overline{AP}$, $overline{AP}$ is one median in the triangle $∆PNM$. You can analugously prove this statement for $overline{BN}$ and $overline{CM}$. The centorid of $∆ABC$ is therefore the centroid of the triangle $∆PNM$