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$AA_1$, $BB_1$, $CC_1$ are the medians of triangle $ABC$ whose centroid is $G$. If points $A, C_1, G, B_1$ are concylic then prove that $2a^2= b^2 + c^2$.

Thanks

My try:- $ar(GBC)=1/3ar(ABC)$

$\frac{1}{2}(GB.GC.\sin(\pi-A))=\frac{1}{3}(\frac{1}{2}bc\sin A)$

$GB.GC=\frac{1}{3}bc$

Now I can't think any further. Here is my diagram:

My diagram

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  • $\begingroup$ Hi and welcome to M.S.E what have you tried on the problem $\endgroup$ – Shobhit Oct 4 '13 at 13:39
  • $\begingroup$ i couldn't think of anything, what I have done is in the post now...I know that wouldn't be correct. $\endgroup$ – dknight Oct 4 '13 at 13:45
  • $\begingroup$ please help me solve this $\endgroup$ – dknight Oct 4 '13 at 14:15
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Draw the line $B_1C_1$, which is parallel to $BC$. Hence we have $\angle GC_1B_1=\angle GCB$. But $AB_1C_1G$ lie on a circle. Therefore we have: $$ \angle GC_1B_1=\angle GAB_1\implies \angle GCB=\angle GAB_1. $$ Therefore the triangles $GA_1C$ and $AA_1C$ are similar and hence we have: $$ \frac{A_1C}{GA_1}=\frac{AA_1}{A_1C}. $$ Call the length of $AA_1$, $m_a$. We know that $GA_1=\frac{m_a}{3}$. Therefore we get: $$ a^2=\frac{4}{3}m_a^2 $$ Now it is enough to replace $m_a^2$ by its length in terms of $a,b,c$: $$ m_a=\sqrt{\frac{2b^2+2c^2-a^2}{4}} $$ and the result follows.

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  • $\begingroup$ thanks Arash for the answer:) $\endgroup$ – dknight Oct 4 '13 at 14:35
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It is really very simple. Using power of point property

$$BC_{1}.BA=BG.BB_{1}$$

$$\frac{c}{2}.c=\frac{2}{3}m_{b}.m_{b}$$

$$3c^2=4m^2_{b}$$

$$3c^2=2a^2+2c^2-b^2$$

$$2a^2=b^2+c^2$$

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