Question

$D,E$ and $F$ are respectively the mid-point of the sides $BC,CA$ and $AB$ of a $△ABC$
Show that
$ar\left(BDEF\right)=\frac{1}{2}ar\left(ABC\right)$

Answer 1

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

$2\times ar\left(DEF\right)=\frac{2}{4}ar\left(ABC\right)$

$2ar\left(DEF\right)=\frac{1}{2}ar\left(ABC\right)$

$ar\left(DEF\right)+ar\left(DEF\right)=\frac{1}{2}ar\left(ABC\right)$

$ar\left(DEF\right)+ar\left(FBD\right)=\frac{1}{2}ar\left(ABC\right)$

$ar\left(BDEF\right)=\frac{1}{2}ar\left(ABC\right)$

$\therefore$ Proved .