Question

$ar\left(△DEF\right)=\frac{1}{4}ar\left(△ABC\right)$

Answer 1

Consider $△ABC,D,E$ and $F$ are mid points of $BC,AC$ and $AB$.

$\therefore F,E$ and $D$ are mid points of $AB,AC$ and $BC$.

Line segments joining the mid points of two sides of a triangle is parallel to the third side.

$\therefore FE||BC\Rightarrow FE||BD$and $DE||AB\Rightarrow DE||FB$ [parts of parallel lines are parallel]

$\therefore BDEF$ is a parallelogram.

Diagonals of a parallelogram divides it into two congruent triangles.

$\therefore △DBF\cong △DEF$

$\Rightarrow ar\left(DBF\right)=ar\left(DEF\right)$ [Area of congruent triangle is equal]

Similarly, we can prove $FDCE$ is a parallelogram

$\therefore △DEC\cong △DEF$

$\Rightarrow ar\left(DEC\right)=ar\left(DEF\right)$
Similarly, $AFDE$ is also a parallelogram.

$\therefore △AFE\cong △DEF$

$\Rightarrow △AFE\cong △DEF$

$\Rightarrow ar\left(AFE\right)=ar\left(DEF\right)$ [Area of congruent triangles is equal]

$\therefore ar\left(FBD\right)=ar\left(DEC\right)=ar\left(AFE\right)=ar\left(DEF\right)$

$\therefore ar\left(FBD\right)+ar\left(DEC\right)+ar\left(AFE\right)+ar\left(DEF\right)=ar\left(ABC\right)$

$ar\left(DEF\right)+ar\left(DEF\right)+ar\left(DEF\right)+ar\left(DEF\right)=ar\left(ABC\right)$

$\Rightarrow 4ar\left(DEF\right)=ar\left(ABC\right)$

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