$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.

(N/A) Given: $D, E, F$ are mid-points of sides $BC, CA, AB$ respectively in $\Delta ABC$.
To prove: $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.
Proof:
$1$. Since $D$ and $F$ are mid-points of $BC$ and $AB$ respectively,by the Mid-point Theorem,$DF \parallel AC$ and $DF = \frac{1}{2} AC = AE$. Thus,$AFDE$ is a parallelogram.
$2$. Similarly,$BDEF$ and $CDFE$ are parallelograms.
$3$. In parallelogram $BDEF$,the diagonal $DF$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
$4$. In parallelogram $AFDE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$.
$5$. In parallelogram $CDFE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta CDE) = \operatorname{ar}(\Delta DEF)$.
$6$. Therefore,$\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta AFE) + \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta CDE) + \operatorname{ar}(\Delta DEF)$.
$7$. Substituting the equal areas: $\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) = 4 \operatorname{ar}(\Delta DEF)$.
$8$. Hence,$\operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.