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.