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

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

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BC

Δ ABC

E

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ar Δ BOE = \frac{1}{8} ar Δ ABC

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Δ B O E = \frac{1}{8} a r (Δ A B C)

ar (∆ B O E) = \frac{1}{8} ar (∆ A B C)

\frac{1}{8}

D is the mid point of side BC of ABC and E is the mid point of BD. If O is the mid point of AE, then

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