Question

If $D,E,F$ are the mid-points of the sides $BC,CA$ and $AB$ respectively of $△ABC$, prove that $BDEF$ is a parallelogram. And also show that $ar\left(△DEF\right)=\frac{1}{4}ar\left(△ABC\right)$.

Answer 1

We have,

$Giventhat:\Delta ABC$

Where $D,E,F$ are mid points of $BC,CA$ and $AB$ respectively.

To prove:-$BDEF$ is a parallelogram

Proof:- In $\Delta ABC$,

$F$ is mid point of $AB$,

$D$ is the midpoint of $BC$

$E$ is the midpoint of $CA$

$\therefore FE\parallel BC\Rightarrow \therefore FE\parallel BD$

$\therefore DE\parallel AB\Rightarrow \therefore DE\parallel FB$

$\therefore FE\parallel BDandDE\parallel FB$

So, in $BDEF$,

Both pairs of opposite sides are parallel,

$\therefore BDEF$ is a parallelogram.

Hence proved.

Now,

$BDEF$ is a parallelogram

$\therefore \Delta DBF\cong \Delta DEF$

$ar\left(\Delta DBF\right)=ar\left(\Delta DEF\right)......\left(1\right)$

Similarly, we can prove that,

$FDCE$ is a parallelogram

$\therefore \Delta DEC\cong \Delta DEF$

$ar\left(\Delta DEC\right)=ar\left(\Delta DEF\right)......\left(2\right)$

Similarly, we can prove that,

$AFDE$ is a parallelogram

$\therefore \Delta AFE\cong \Delta DEF$

$ar\left(\Delta AFE\right)=ar\left(\Delta DEF\right)......\left(3\right)$

From equation (1), (2), (3) to, and we get,

$ar\left(\Delta DBF\right)=ar\left(\Delta DEC\right)=ar\left(\Delta AFE\right)=ar\left(\Delta DEF\right)$

Now,

$ar\left(\Delta DBF\right)+ar\left(\Delta DEC\right)+ar\left(\Delta AFE\right)+ar\left(\Delta DEF\right)=ar\left(\Delta ABC\right)$

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

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

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

Hence proved.