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# In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = $$\frac{1}{4}$$ ar (ABC).

ar(BED) = $$\frac{1}{2}$$×BD×DE
(Since, E is the mid-point of AD,)
AE = DE
(Since, AD is the median on side BC of triangle ABC,)
BD = DC

DE = $$\frac{1}{2}$$ AD — (i)
BD = $$\frac{1}{2}$$BC — (ii)
From (i) and (ii), we get,
ar(BED) = $$\frac{1}{2}$$×$$\frac{1}{2}$$BC × $$\frac{1}{2}$$AD
$$\Rightarrow$$ ar(BED) = $$\frac{1}{2}$$×$$\frac{1}{2}$$ar(ABC)
$$\Rightarrow$$ ar(BED) = $$\frac{1}{4}$$ ar(ABC)

Hence, proved.