Question

D is the midpoint of side BC of a triangle ABC and E is the midpoint of BD. if O is the midpoint of AE. prove that ar(BOE) = 18ar(ABC)

Answer 1

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$d \: is \: the \: midpoint \: of \: side \: bc \:$
$ad \: median \: divides \: triangle \: abc \: in \: two \: equal \: parts \:$
$area \: of \: tiangle \: abd = \frac{1}{2} \times area \: of \: triangle \: abc \:$
$as \: e \: is \: the \: midpoint \: of \: bd \:$
$ae \: median \: divides \: triangle \: abd \: in \: two \: equal \: parts \:$
$area \: of \: tringle \: abe = \frac{1}{2} \times area \: of \: tringle \: abd \:$
$= \frac{1}{2} \times \frac{1}{2} \times area \: of \: triangle \: abc$
$= \frac{1}{4} \times area \: of \: triangle \: abc$
$as \: o \: is \: the \: midpoint \: of \: ae$
$ao \: median \: \: divides \: triangle \: abe \: in \: 2 \: equal \: parts \:$
$area \: of \: triangle \: boe = area \: of \: triangle \: abe \:$
$= \frac{1}{2} \times \frac{1}{4} \times area \: of \: tringle \: abc$
$\frac{1}{8} \times area \: of \: triangle \: abc \:$
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Answer 2

Answer:

ar(BOE)= 1/8ar(ABC).

Please see attached answer for the same.