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Problem: Let $ABC$ be a triangle and a median $AD$ is drawn from $A$ to $B$C. $O$ is any point on $AD$. Now joining $BO$ and $CO$ extend it. $BO$ and $CO$ cut $AC$ and $AB$ at $P$ and $Q$ respectively. Prove that $PQ$ is parallel to $BC$.

I understood that proving that $PQ$ is bisected by the median can solve this. But I can't prove that. Any help will be appreciated.

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Hint: Use Ceva's theorem to conclude that $\dfrac{AP}{PC} = \dfrac{AQ}{QB}$.

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