Question

$D,E$ and $F$ are respectively the mid-point of the sides $BC,CA$ and $AB$ of a $△ABC$
Show that $ar\left(DEF\right)=\frac{1}{4}ar\left(ABC\right)$

Answer 1

R.E.F image

As BDEF is a $\parallel gm$

$\therefore △DEF\cong △DBF$

$\Rightarrow ar\left(DBF\right)=ar\left(DEF\right)$

similarly, we can prove FDCE is $\parallel gm$

$\therefore △DEC\cong △DEF$

$\Rightarrow ar\left(DEC\right)=ar\left(DEF\right)$

similarly, we have prove AFDE is$\parallel gm$

$\therefore △AFE\cong △DEF$

$\Rightarrow ar\left(AFE\right)=ar\left(DEF\right)$

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

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

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

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

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

$\therefore$ Proved.