TITLE: Geometry - Proving a common centroid.
QUESTION [5 upvotes]: Triangle PQR is drawn. Through it's vertices are lines drawn which are
parallel to the opposite sides of the triangle. The new triangle formed
is ABC. Prove that these two triangles have a common centroid.

I started by letting $M$ be the median of $[PQ]$, and then prove that bisector of $[BC]$ from $A$ occurs at $R$ while $M$ is collinear to $[AP]$. Firstly, I'm not sure whether or not this will suffice in proving they share a common centroid, and secondly I'm not sure where to start with the proof as setting up similar triangles from the parallel lines identity has lead to nothing.

REPLY [2 votes]: You can alternatively use this very well-known property (look at the very end of the answer linked).

Lemma 1
In any given triangle $\triangle ABC$, the medians $AM_a, BM_b, CM_c$ concur at the centroid $S$ such that $$\frac{AS}{SM_a}=\frac{BS}{SM_b}=\frac{CS}{SM_c}=2$$

Observe now that - back to your picture -  $APRQ$ is a parallelogram and hence $M$ is the midpoint of $PQ$. Thus $S_{\triangle PQR}\in MR$ satisfies $$\fbox{$S_{\triangle PQR}R=\frac23 RM=\frac 13 AR=S_{\triangle ABC}R\implies S_{\triangle PQR}=S_{\triangle ABC}$}$$