D, E, and F are respectively the mid-points of the sides BC, CA, and AB of a triangle ABC. Show that (i) BDEF is a parallelogram (ii) ar(DEF) = 1/4 ar(ABC) (iii) ar(BDEF) = 1/2 ar(ABC)

In ΔABC, E and F are midpoints of side AC and AB
Therefore, EF || BC and EF = 1/2 BC
Similarly, ED || AB and ED = 1/2 AB
Since, EF || BC ⇒ EF || BD and ED || FB
∴ ΔBDEF is a parallelogram
(ii) Using above concept , we will have BDEF, CDFE, DFAE, are parallelogram ,
And as we know that diagonal of parallelogram divides it into two triangles of equal area
∴A(ΔBDF) = A(ΔDFE) = A(ΔCDE) = A(ΔAFE)
A(ΔBDF) + A(ΔDFE) + A(ΔCDE) + A(ΔAFE) = A(ΔABC)
A(ΔDFE) = 1/4 A(ΔABC)
(iii) Using the relation in (ii), we have
A(ΔBDEF) = A(ΔDBF) + A(ΔDFE)
and A(ΔBDF) = A(ΔDFE) = A(ΔCDE) = A(ΔAFE) = 1/4 A(ΔABC)
A(ΔBDEF) = A(ΔDBF) + A(ΔDFE) = 1/4 A(ΔABC) + 1/4 A(ΔABC) = 1/2 A(ΔABC)