Geometry - Proving a common centroid.

$PQCR$ and $PQRB$ are parallelograms by construction, hence: $PQ=BR=RC$ and $R$ is the midpoint of $BC$.

$PRQA$ is by construction a parallelogram, hence its diagonals $PQ$ and $AR$ meet at their common midpoint $M$.

It follows that median $AR$ of triangle $ABC$ lies on the same line as median $RM$ of triangle $PQR$. The same reasoning holds for the other medians of those triangles.

Hence the medians of triangle $ABC$ meet at the same point as the medians of triangle $PQR$.

Solution with vectors. The centroid $G$ of $ABC$ is $G = {1\over 3}(A+B+C)$

Since $R$ is a midpoint of $BC$ we have $$R = {1\over 2}(B+C)$$ and similar for $P$ and $Q$. Now the centroid $G'$ of $PRQ$ is $$G' = {1\over 3}(P+Q+R) = {1\over 3}\Big({1\over 2}(B+C)+{1\over 2}(A+C)+ {1\over 2}(B+A)\Big) $$$$= {1\over 3}(A+B+C) = G$$

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 $$\fbox{$S_{\triangle PQR}R=\frac23 RM=\frac 13 AR=S_{\triangle ABC}R\implies S_{\triangle PQR}=S_{\triangle ABC}$}$$