Question

If $D,E,F$ are midpoint of the sides $BC,CA$ an $AB$ of triangle $ABC$, then
Area of $△DEF=1/4$ (area of $△ABC$)

Answer 1

In $△ABC,$

$F$ is the mid-point of $AB$,

$E$ is the mid-point of $AC$,

$\therefore FE\parallel BC$

$\Rightarrow FE\parallel BD$ since parts of parallel lines are parallel .......$\left(1\right)$

$D$ is the mid-point of $BC$,

$E$ is the mid-point of $AC$,

$\therefore DE\parallel AB$

$\Rightarrow DE\parallel FB$ since parts of parallel lines are parallel .......$\left(2\right)$

Now, $FE\parallel BD$ and $DE\parallel FB$

In $BDEF$, both pairs of opposite sides are parallel,

$\therefore BDEF$ is a parallelogram.

In $△ABC,$

$F$ is the mid-point of $AB$,

$E$ is the mid-point of $AC$,

$\therefore FD\parallel EC$

$\Rightarrow FE\parallel DC$ since parts of parallel lines are parallel .......$\left(3\right)$

$D$ is the mid-point of $BC$,

$E$ is the mid-point of $AC$,

$\therefore DF\parallel AC$

$\Rightarrow DF\parallel EC$ since parts of parallel lines are parallel .......$\left(2\right)$

Now, $FE\parallel DC$ and $DF\parallel FE$

In $CEFD$, both pairs of opposite sides are parallel,

$\therefore FDCE$ is a parallelogram.

$\therefore △DBF\cong △DEF$ since diagonals of a parallelogram divides it into two congruent triangles.

$\Rightarrow$ area$\left(DBF\right)=$area$\left(DEF\right)$ since area of congruent triangles are equal .......$\left(4\right)$

$\therefore △AFE\cong △DEF$ since diagonals of a parallelogram divides it into two congruent triangles.

$\Rightarrow$ area$\left(AFE\right)=$area$\left(DEF\right)$ since area of congruent triangles are equal .......$\left(4\right)$

$\therefore$ area$\left(FBD\right)=$area$\left(DEC\right)=$area$\left(AFE\right)=$area$\left(ABC\right)$ .........$\left(5\right)$

Now, area$\left(FBD\right)+$area$\left(DEC\right)+$area$\left(AFE\right)+$area$\left(DEF\right)=$area$\left(ABC\right)$

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

$\Rightarrow$area$\left(DEF\right)=\frac{1}{4}$area$\left(ABC\right)$

Hence proved.