The mid-points of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

1. $\dfrac{1}{2}$ area of △ABC

2. $\dfrac{1}{3}$ area of △ABC

3. $\dfrac{1}{4}$ area of △ABC

4. area of △ABC

Let CDEF be a parallelogram.

So, diagonal DF divides it into two triangles of equal area.

∴ area of △CDF = area of △EDF ……..(i)

E and F are mid-points of side AB and AC respectively.

By mid-point theorem,

EF = $\dfrac{1}{2}$ BC = BD (As D is mid-point of BC) and EF || BC.

Since, FE || BC so from figure,

EF || BD and EF = BD.

So, EBDF is a paralleogram with diagonal ED dividing it into two triangles of equal area.

∴ area of △EBD = area of △EDF ……..(ii)

E and D are mid-points of side AB and BC respectively.

By mid-point theorem,

ED = $\dfrac{1}{2}$ AC = AF (As F is mid-point of AC) and ED || AC.

Since, ED || AC so from figure,

ED || AF.

So, AEDF is a paralleogram with diagonal EF dividing it into two triangles of equal area.

∴ area of △AEF = area of △EDF ……..(iii)

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

area of △EDF = area of △CDF = area of △EBD = area of △AEF = x

From figure,

area of △ABC = area of △EDF + area of △CDF + area of △EBD + area of △AEF = 4x.

area of || gm EDCF = area of △EDF + area of △CDF = 2x.

So,

$\Rightarrow \dfrac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \dfrac{2x}{4x} \\[1em] \Rightarrow \dfrac{\text{area of || gm EDCF}}{\text{area of △ABC}} = \dfrac{1}{2} \\[1em]$

area of || gm EDCF = $\dfrac{1}{2}$ area of △ABC.

Hence, Option 1 is the correct option.