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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.

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    $\begingroup$ $\exists$ a homothety mapping the sides of $\Delta PQR$ to the sides of $\Delta ABC$. $\endgroup$ – user665856 Apr 29 at 13:54
  • $\begingroup$ I am not talking about similarity, homothety $\endgroup$ – user665856 Apr 29 at 13:58
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    $\begingroup$ So what if you are in high school? I am in middle school $\endgroup$ – user665856 Apr 29 at 14:03
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    $\begingroup$ Okay i am trying to help you without homothety $\endgroup$ – user665856 Apr 29 at 14:03
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    $\begingroup$ @ShamimAkhtar rather a lot of us have managed to get thru a math major without learning that term. It's a bit specialized. $\endgroup$ – Carl Witthoft Apr 29 at 18:27
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$PQCR$ 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 $PQR$.

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  • $\begingroup$ +1 Thank you, the main problem I have with geometry is not knowing whether or not the proof is complete, and if this is complete then I suppose my answer is too. $\endgroup$ – John Miller Apr 29 at 14:16
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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 $$G' = {1\over 3}(P+Q+R) = {1\over 3}\Big({1\over 2}(B+C)+{1\over 2}(A+C)+ {1\over 2}(B+A)\Big) $$$$= {1\over 3}(A+B+C) = G$$

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    $\begingroup$ Thank you, I do feel as if the proof using a parallelogram is sufficient. $\endgroup$ – John Miller Apr 29 at 14:31
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    $\begingroup$ It always amazes me how a simple idea like vectors - a magnitude with a direction after all - can lead to these conclusions. [+1]! $\endgroup$ – Dr. Mathva Apr 29 at 18:39
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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 $$\fbox{$S_{\triangle PQR}R=\frac23 RM=\frac 13 AR=S_{\triangle ABC}R\implies S_{\triangle PQR}=S_{\triangle ABC}$}$$

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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.

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  • $\begingroup$ Thank you, I'm not sure which answer to give the check to as they both answer the question. $\endgroup$ – John Miller Apr 29 at 14:17
  • $\begingroup$ You can check the other answer, it is obviouslu better than mine, it uses elementary techniques. $\endgroup$ – user665856 Apr 29 at 14:22
  • $\begingroup$ Okay, thank you for your patience and sorry I assumed complexity in homothety but reading about it more, it seems to be very useful in projective geometry and the idea is quite simple. $\endgroup$ – John Miller Apr 29 at 14:28
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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.

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