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Question

# D is the mid-point of side BC of ΔABC and E is the mid-point BD. If O is the mid point of AE, then prove that ar(ΔBOE) = $\frac{1}{8}$ ar(ΔABC).

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Solution

## $\mathrm{Given}:\mathrm{In}∆\mathrm{ABC},\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{BC},\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{BD}\mathrm{and}\mathrm{O}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{AE}.\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{prove}:\mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{8}\mathrm{ar}\left(∆\mathrm{ABC}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}\mathrm{In}∆\mathrm{ABC},\phantom{\rule{0ex}{0ex}}\because \mathrm{AD}\mathrm{is}\mathrm{median}\left(\mathrm{As},\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{BC}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{ar}\left(∆\mathrm{ABD}\right)=\frac{1}{2}\mathrm{ar}\left(∆\mathrm{ABC}\right).....\left(\mathrm{i}\right)\left(\mathrm{Median}\mathrm{divides}\mathrm{the}\mathrm{triangle}\mathrm{into}\mathrm{two}\mathrm{parts}\mathrm{equal}\mathrm{in}\mathrm{area}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{in}∆\mathrm{ABD},\phantom{\rule{0ex}{0ex}}\because \mathrm{AE}\mathrm{is}\mathrm{median}\left(\mathrm{As},\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{BD}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{ar}\left(∆\mathrm{ABE}\right)=\frac{1}{2}\mathrm{ar}\left(∆\mathrm{ABD}\right).....\left(\mathrm{ii}\right)\left(\mathrm{Median}\mathrm{divides}\mathrm{the}\mathrm{triangle}\mathrm{into}\mathrm{two}\mathrm{parts}\mathrm{equal}\mathrm{in}\mathrm{area}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{in}∆\mathrm{ABE},\phantom{\rule{0ex}{0ex}}\because \mathrm{BO}\mathrm{is}\mathrm{median}\left(\mathrm{As},\mathrm{O}\mathrm{is}\mathrm{the}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{AE}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{2}\mathrm{ar}\left(∆\mathrm{ABE}\right)\left(\mathrm{Median}\mathrm{divides}\mathrm{the}\mathrm{triangle}\mathrm{into}\mathrm{two}\mathrm{parts}\mathrm{equal}\mathrm{in}\mathrm{area}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{2}\left[\frac{1}{2}\mathrm{ar}\left(∆\mathrm{ABD}\right)\right]\left[\mathrm{Using}\left(\mathrm{ii}\right)\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{4}\mathrm{ar}\left(∆\mathrm{ABD}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{4}\left[\frac{1}{2}\mathrm{ar}\left(∆\mathrm{ABC}\right)\right]\left[\mathrm{Using}\left(\mathrm{i}\right)\right]\phantom{\rule{0ex}{0ex}}\therefore \mathrm{ar}\left(∆\mathrm{BOE}\right)=\frac{1}{8}\mathrm{ar}\left(∆\mathrm{ABC}\right)$

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