Question

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

Answer 1

Let ABC be a triangle and AD is the median of $\Delta ABC$

E is the mid point of AD

To prove : $ar\left(BED\right)=\frac{1}{4}.ar\left(ABC\right)$

In $\Delta ABC$,

$ar\left(ABD\right)=ar\left(ACD\right)$ ___________ (1)

In $\Delta ABD$, BE is the median

$ar\left(ABE\right)=ar\left(BED\right)$ __________ (2)

Now, $ar\left(ABD\right)=ar\left(BED\right)$

= 2. ar (BED) ______ (3)

ar(ABC)=ar(ABD)+ar(ACD)

ar(ABD)=2.ar(ABD) using (1)

ar(ABC)=2.2.ar(BED) - (2)

ar(ABC)= 4.ar (BED)

$\therefore ar\left(BED\right)=\frac{1}{4}ar\left(ABC\right)$

Hence it is proved.