$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$

(N/A) In $\Delta ABC, F$ and $E$ are the midpoints of $AB$ and $AC$ respectively.
By the Midpoint Theorem,$FE \parallel BC$ and $FE = \frac{1}{2} BC$.
Since $D$ is the midpoint of $BC$,$BD = \frac{1}{2} BC$.
Therefore,$FE \parallel BD$ and $FE = BD$.
Since one pair of opposite sides is equal and parallel,quadrilateral $BDEF$ is a parallelogram. (Result $i$)
Similarly,quadrilaterals $AFDE$ and $FDCE$ are parallelograms.
In parallelogram $BDEF, FD$ is a diagonal,so $ar(BDF) = ar(DEF)$. $(1)$
In parallelogram $AFDE, EF$ is a diagonal,so $ar(AFE) = ar(DEF)$. $(2)$
In parallelogram $FDCE, ED$ is a diagonal,so $ar(DCE) = ar(DEF)$. $(3)$
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\Delta ABC$ is composed of four non-overlapping triangles: $\Delta BDF, \Delta AFE, \Delta DCE$ and $\Delta DEF$.
$ar(ABC) = ar(BDF) + ar(AFE) + ar(DCE) + ar(DEF)$
Using $(1), (2)$ and $(3)$:
$ar(ABC) = ar(DEF) + ar(DEF) + ar(DEF) + ar(DEF) = 4 ar(DEF)$
Therefore,$ar(DEF) = \frac{1}{4} ar(ABC)$. (Result $ii$)
Now,$ar(BDEF) = ar(BDF) + ar(DEF) = ar(DEF) + ar(DEF) = 2 ar(DEF)$.
Substituting $ar(DEF) = \frac{1}{4} ar(ABC)$:
$ar(BDEF) = 2 \times \frac{1}{4} ar(ABC) = \frac{1}{2} ar(ABC)$. (Result $iii$)