Geometry - Proving a common centroid.

Question

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.

Answer 1

$PQCR$ 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 $PQR$.

Answer 2

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 $$G' = {1\over 3}(P+Q+R) = {1\over 3}\Big({1\over 2}(B+C)+{1\over 2}(A+C)+ {1\over 2}(B+A)\Big) $$$$= {1\over 3}(A+B+C) = G$$

Answer 3

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

Answer 4

Notice that $\angle CQR=\angle QRP=\angle CAB$. With a same argument, it must be easy for you to derive that $AQPR$ is a parallellogram. Similarly, you can show that $BLQR, QPRC$ are parallellograms. So $BR=PQ=CR \implies R$ is the midpoint of $BC$. Similarly, $Q,P$ are the respective midpoints. And now by symmetry, you can take half turn of the triangle $PQR$, and enlarge each of its sides by a scale factor of $2$ with respect to the centroid of $\Delta PQR$, and now this triangle will completely coincide with $ABC$, indeed implying it shares its centroid with the other triangle.

Answer 5

Here’s is another method. Let $h$ be the homothecy of ratio $-\frac12$ centered at $G$ centroid of $\triangle ABC$. Then $h(A)=R$, $h(B)=Q$ and $h(C)=P$ thus $h(\triangle ABC)=\triangle PQR$. Since $h(G)=G$ these two triangle have the same centroid.