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.

Question

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) = \(\frac{1}{4}\) ar(ABC)

(iii) ar (BDEF) = \(\frac{1}{2}\) ar(ABC)

Samar shah · Answer

(i) In ΔABC,

EF || BC and EF = \(\frac{1}{2}\) BC (by mid point theorem)

BD = \(\frac{1}{2}\) BC (D is the mid point)

So, BD = EF

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)

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) = \(\frac{1}{4}\) 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 × \(\frac{1}{4}\) ar(ΔABC)

⇒ ar(parallelogram BDEF) = \(\frac{1}{2}\) ar(ΔABC)

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