(i) In ΔABC,
EF || BC and EF = ½ BC(by mid point theorem)
also,
BD = ½ BC (D is themid point)
So, BD = EF
also,
BF and DE are paralleland equal to each other.
∴, the pair oppositesides are equal in length and parallel to each other.
∴ BDEF is aparallelogram.
(ii) Proceeding fromthe result of (i),
BDEF, DCEF, AFDE are parallelograms.
Diagonal of aparallelogram 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) = ¼ar(ABC)
(iii) Area(parallelogram BDEF) = ar(ΔDEF) +ar(ΔBDE)
⇒ ar(parallelogramBDEF) = ar(ΔDEF) +ar(ΔDEF)
⇒ ar(parallelogramBDEF) = 2× ar(ΔDEF)
⇒ ar(parallelogramBDEF) = 2× ¼ ar(ΔABC)
⇒ ar(parallelogramBDEF) = ½ ar(ΔABC)