D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that

 (i) BDEF is a parallelogram.        

(ii) ar(DEF) = ¼ ar(ABC)

(iii) ar (BDEF) = ½ ar(ABC)

Solution:

(i) In ΔABC,

EF || BC and EF = ½ BC (by mid point theorem)

also,

BD = ½ BC (D is the mid point)

So, BD = EF

also,

BF and DE are parallel and equal to each other.

∴, the pair opposite sides are equal in length and parallel to each other.

∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),

BDEF, DCEF, AFDE are parallelograms.

Diagonal of a parallelogram divides it into two triangles of equal area.

∴ar(ΔBFD) = ar(ΔDEF) (For parallelogram BDEF) — (i)

also,

ar(ΔAFE) = ar(ΔDEF) (For parallelogram DCEF) — (ii)

ar(ΔCDE) = ar(ΔDEF) (For parallelogram AFDE) — (iii)

From (i), (ii) and (iii)

ar(ΔBFD) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)

⇒ ar(ΔBFD) +ar(ΔAFE) +ar(ΔCDE) +ar(ΔDEF) = ar(ΔABC)

⇒ 4 ar(ΔDEF) = ar(ΔABC)

⇒ ar(DEF) = ¼ ar(ABC)

(iii) Area (parallelogram BDEF) = ar(ΔDEF) +ar(ΔBDE)

⇒ ar(parallelogram BDEF) = ar(ΔDEF) +ar(ΔDEF)

⇒ ar(parallelogram BDEF) = 2× ar(ΔDEF)

⇒ ar(parallelogram BDEF) = 2× ¼ ar(ΔABC)

⇒ ar(parallelogram BDEF) = ½ ar(ΔABC)