The vertex A of ∆ABC is joined to a point D on BC. If E is the midpoint of AD, then ar(∆BEC) = ?
(a) $\frac{1}{2} \operatorname{ar}(\Delta A B C)$
(b) $\frac{1}{3} \operatorname{ar}(\Delta A B C)$
(c) $\frac{1}{4} \operatorname{ar}(\Delta A B C)$
(d) $\frac{1}{6} \operatorname{ar}(\Delta A B C)$
(a) $\frac{1}{2} \operatorname{ar}(\Delta A B C)$
Since E is the midpoint of AD, BE is a median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., $\operatorname{ar}(\Delta B E D)=\frac{1}{2} \times \operatorname{ar}(\triangle A B D)$ ...(i)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., $\operatorname{ar}(\triangle C E D)=\frac{1}{2} \times \operatorname{ar}(\triangle A D C)$ ...(ii)
Adding (i) and (ii), we have:
$\operatorname{ar}(\Delta B E D)+\operatorname{ar}(\Delta C E D)=\frac{1}{2} \times \operatorname{ar}(\Delta A B D)+\frac{1}{2} \times \operatorname{ar}(\Delta A D C)$
$\Rightarrow \operatorname{ar}(\Delta B E C)=\frac{1}{2}(\Delta A B D+A D C)=\frac{1}{2} \Delta A B C$
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