Let triangle $ABC, BC=a,AB=c,AC=b$. Know that the bisector of $ACB$ perpendicular with the line $OG$, with $O$ is inscribed circle center and $G$ is center of $ABC$. Prove that: $$\frac{2ab}{a+b}=\frac{a+b+c}{3}$$

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    $\begingroup$ What do you mean by "$G$ being the center of (triangle) $ABC$"? $\endgroup$ – Quintofron Sep 22 '12 at 13:47
  • $\begingroup$ Let M,N,P is Midpoint of BC,AC,AB. G is intersection of AM,BN and CP. Sorry because of my bad English $\endgroup$ – LevanDokite Sep 23 '12 at 6:02
  • $\begingroup$ G is the centroid(jimloy.com/geometry/centers.htm) $\endgroup$ – lab bhattacharjee Sep 23 '12 at 6:09

I suppose the home work is finished and this answer is for reference.

the answer


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