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# If O and O' are circumcentre and orthocentre of ∆ ABC, then $\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}$ equals (a) 2 $\stackrel{\to }{OO}$' (b) $O\stackrel{\to }{O\text{'}}$ (c) $\stackrel{\to }{O\text{'}O}$ (d) $2\stackrel{\to }{O\text{'}O}$

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Solution

## Option (b). Given: $O$ be the circumcentre and $O\text{'}$ be the orthocentre of $△ABC$. Let $G$ be the centroid of the triangle. We know that $O,G$ and $H$ are collinear and by geometry $\stackrel{\to }{O\text{'}G}=2\stackrel{\to }{OG}.$ This yields, $\stackrel{\to }{O\text{'}O}=\stackrel{\to }{O\text{'}G}+\stackrel{\to }{GO}=2\stackrel{\to }{GO}+\stackrel{\to }{GO}=3\stackrel{\to }{GO}.\phantom{\rule{0ex}{0ex}}$ In other words $\stackrel{\to }{OO\text{'}}=3\stackrel{\to }{OG}.$ Since, $\stackrel{\to }{OG}=\frac{\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}}{3}$. ∴ $\stackrel{\to }{OO\text{'}}=3×\frac{\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}}{3}=\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}.$

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