$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $\operatorname{ar}( BDEF ) = \frac{1}{2} \operatorname{ar}( ABC )$

(N/A) Given that $D, E$ and $F$ are the mid-points of sides $BC, CA$ and $AB$ respectively.
By the Mid-point Theorem,$FE \parallel BC$ and $FE = \frac{1}{2} BC = BD$.
Since $FE \parallel BD$ and $FE = BD$,$BDEF$ is a parallelogram.
Similarly,$AFDE$ and $FDCE$ are also parallelograms.
In parallelogram $BDEF$,the diagonal $DE$ divides it into two triangles of equal area,so $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
Similarly,in parallelogram $AFDE$,$\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$,and in parallelogram $FDCE$,$\operatorname{ar}(\Delta DEF) = \operatorname{ar}(\Delta EDC)$.
Thus,$\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF) = \operatorname{ar}(\Delta EDC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
Now,$\operatorname{ar}(BDEF) = \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC) + \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{2}{4} \operatorname{ar}(\Delta ABC) = \frac{1}{2} \operatorname{ar}(\Delta ABC)$.