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Consider the points $A',B',C'$ on the sides $BC,CA,AB$ of a triangle $ABC$ respectively, such that $BA'/A'C=CB'/B'A=AC'/C'B$.

Show that the triangles $ABC$ and $A'B'C'$ share a common centroid.

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    $\begingroup$ What techniques do you have to use? What have you tried? $\endgroup$ – Ted Shifrin Mar 24 '14 at 22:26
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Here is the Pure geometry method:( I just list main points)

$A'J$// $AB \implies AJ=CB'$

$F$ is mid point of $B'J \implies 2FD'=JA'$

$D'$ is mid point of $A'B'$ , $BF$ cross $C'D'$ at $G$

$\dfrac{FD'}{C'B}=\dfrac{D'G}{GC'}=\dfrac{FG}{GB}$

now if you can prove $\dfrac{JA'}{AB}=\dfrac{C'B}{AB}$,then $G$ is the common centroids.

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Using vectors, let $\vec A$ denote the vertex $A$, etc, and let $r := A' B / B C$. Then the centroid of $\Delta ABC$ is the arithmetic mean of the vertices: $$ \vec G = \dfrac{ \vec A + \vec B + \vec C } 3. $$ We also have by definition $$ \vec A' = (1-r) \vec B + r \vec C, $$ etc., so $$ \dfrac{ \vec A' + \vec B' + \vec C'} 3 = \dfrac{ (1-r) \vec B + r \vec C + (1-r) \vec C + r \vec A + (1-r) \vec A + r \vec B}3 = \vec G $$ so the centroids coincide.

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  • $\begingroup$ Thank you for your answer. Due to the fact that I'm not very familiar with analytic geometry, I haven't fully understood your proof. Could you please explain why $$\vec A' = (1-r) \vec B + r \vec C,$$ ? $\endgroup$ – Matheo Mar 24 '14 at 23:13

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