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Question

# D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE, prove that $\mathrm{ar}\left(∆BOE\right)=\frac{1}{8}\mathrm{ar}\left(∆ABC\right)$.

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Solution

## Given: D is the midpoint of BC; E is the midpoint of BD; O is the mid point of AE. To prove: ar(∆BOE) = $\frac{1}{8}$ ⨯ ar(∆ABC) Proof: D is the midpoint of BC, so AD is the median of ∆ABC. E is the midpoint of BD, so AE is the median of ∆ABD. O is the mid point of AE, so BO is median of ∆ABE. We know that a median of a triangle divides it into two triangles of equal areas. So, we have: ar(∆ABD ) = $\frac{1}{2}$ ar(∆ABC) ...(i) ar(∆ABE ) =$\frac{1}{2}$​ ar (∆ ABD) ...(ii) ar(∆BOE ) =$\frac{1}{2}$​ ar (∆ ABE) ...(iii) From (i), (ii) and (iii), we have: ar(∆BOE) =$\frac{1}{2}$​ ar(∆ABE) ar(∆BOE) = $\frac{1}{2}$ ⨯​ $\frac{1}{2}$ ⨯​ $\frac{1}{2}$ ⨯​ ar(∆ABC)​ ∴​​ ar(∆BOE)​ = $\left(\frac{1}{8}\right)$ ⨯ ar(∆ABC)

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