D is the midpoint f side BC of $△ A B C$ and E is the midpoint of BD. IF O is the midpoint of AE, prove that ar $(△ B O E) = \frac{1}{8} a r (△ A B C)$

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Solution

In $△ A B C,$ D is mid point of BC, E is mid point BE and O is the midpoint of AE. BO, AE, AD are joined.

To prove : ar $(△ B O E) = \frac{1}{8} a r (△ A B C)$
Proof: In $△ A B C,$
$∴$ D is the midpoint of BC
$a r (△ A B D) = a r (△ A D C) = \frac{1}{2} a r (△ A B C) I n △ A B D, E i s t h e m i d p o i n t o f B D ∴ a r (△ A B E) = \frac{1}{2} a r (△ A B D) = \frac{1}{2} (\frac{1}{2} a r (△ A B C)) = \frac{1}{4} a r (△ A B C) ∵ O i s t h e m i d p o i n t o f A E ∴ a r (△ B O E) = \frac{1}{2} a r (△ A B E) = \frac{1}{2} \times (\frac{1}{4} a r (△ A B C)) = \frac{1}{8} a r (△ A B C) H e n c e a r (△ B O E) = \frac{1}{8} a r (△ A B C)$

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Similar questions

D is midpoint of side BC of $△ A B C$ and E is the midpoint of BD.If O is the midpoint of AE,prove that $a r (△ B O E) = \frac{1}{8} a r (△ A B C)$

In $△ A B C$ ,it is given that D is the midpoint of BC;E is the midpoint of BD and O is the midpoint of AE.Then,ar( $△ B O E$ )=?

(a) $\frac{1}{3} a r (△ A B C)$

(b) $\frac{1}{4} a r (△ A B C)$

(d) $\frac{1}{8} a r (△ A B C)$

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

Δ A B C

Δ B O E

△ A B C

△ B O E

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