D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC.

Solution:



(i) $\ln \triangle A B C$,

$E$ and $F$ are the mid-points of side $A C$ and $A B$ respectively.

Therefore, $E F \| B C$ and $E F=\frac{1}{2} B C$ (Mid-point theorem)

However, $B D=\frac{1}{2} B C(D$ is the mid-point of $B C)$

Therefore, $B D=E F$ and $B D \| E F$

Therefore, BDEF is a parallelogram.

(ii) Using the result obtained above, it can be said that quadrilaterals BDEF, DCEF, AFDE are parallelograms.

We know that diagonal of a parallelogram divides it into two triangles of equal area.

$\therefore$ Area $(\Delta B F D)=$ Area $(\Delta D E F)($ For parallelogram BD $)$

Area (ΔCDE) = Area (ΔDEF) (For parallelogram DCEF)

Area (ΔAFE) = Area (ΔDEF) (For parallelogram AFDE)

$\therefore$ Area $(\triangle \mathrm{AFE})=$ Area $(\triangle \mathrm{BFD})=$ Area $(\triangle \mathrm{CDE})=$ Area $(\triangle \mathrm{DEF})$

Also,

Area $(\Delta \mathrm{AFE})+$ Area $(\Delta \mathrm{BDF})+$ Area $(\Delta \mathrm{CDE})+$ Area $(\Delta \mathrm{DEF})=$ Area $(\triangle \mathrm{ABC})$

$\Rightarrow$ Area $(\Delta \mathrm{DEF})+$ Area $(\Delta \mathrm{DEF})+$ Area $(\Delta \mathrm{DEF})+$ Area $(\Delta \mathrm{DEF})=$ Area $(\Delta \mathrm{ABC})$

$\Rightarrow 4$ Area $(\triangle \mathrm{DEF})=$ Area $(\triangle \mathrm{ABC})$

$\Rightarrow$ Area $(\Delta \mathrm{DEF})=\frac{1}{4}$ Area $(\triangle \mathrm{ABC})$

(iii) Area (parallelogram BDEF) = Area ( $\triangle \mathrm{DEF})+$ Area ( $\triangle \mathrm{BDF}$ )

$\Rightarrow$ Area $($ parallelogram BDEF $)=$ Area $(\triangle D E F)+$ Area $(\triangle D E F)$

$\Rightarrow$ Area (parallelogram BDEF) $=2$ Area ( $\triangle \mathrm{DEF}$ )

$\Rightarrow$ Area (parallelogram BDEF) $=2 \times \frac{1}{4}$ Area ( $\triangle \mathrm{ABC}$ )

$\Rightarrow$ Area (parallelogram BDEF) $=\frac{1}{2}$ Area ( $\left.\triangle \mathrm{ABC}\right)$