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If $G$ be the centroid of $\triangle ABC$, prove that $AB^{2} + BC^{2} + CA^{2} = 3(GA^{2}+GB^{2}+GC^{2}).$

Please help me. I couldn't get even to the first step. However, i guess centroid is the point of intersection of medians of triangle and median is the the line that joins the vertex to the mid point of opposition side, not the perpendicular. Am I right?

Please help me to solve this.

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Hint

$1)$ If $M$ is the midpoint of $BC$ then $GA=(2/3)AM$.

$2)$ Use Stewart's theorem to find $AM$: enter image description here

$$AB^2\cdot CM+AC^2\cdot BM=BC(AM^2+BM\cdot CM)\quad (1)$$

Use that $$BM=CM=BC/2$$

and then $(1)$ becomes:

$$ AM^2=\frac{1}{4}\left(2AB^2+2AC^2-BC^2\right)$$

So,

$$GA^2=\frac{1}{9}\left(2AB^2+2AC^2-BC^2\right)$$

Use the same idea to find $GB^2$ and $GC^2$.

Can you finish?

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  • $\begingroup$ What is Stewart theorem? $\endgroup$ – pi-π Jan 10 '17 at 16:32
  • $\begingroup$ see here:en.wikipedia.org/wiki/Stewart's_theorem $\endgroup$ – Arnaldo Jan 10 '17 at 16:32
  • $\begingroup$ Could you please elaborate a bit more? I can't get all. $\endgroup$ – pi-π Jan 10 '17 at 16:34
  • $\begingroup$ Ok, I will write in more details. $\endgroup$ – Arnaldo Jan 10 '17 at 16:34
  • $\begingroup$ Is that the Stewart theorem you have used? $\endgroup$ – pi-π Jan 10 '17 at 16:43

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