In the adjoining figure, D, E and F are midpoints of the sides BC, CA and AB respectively of ∆ABC. Prove that BCEF is a trapezium and area of the trap. BCEF = $\dfrac{3}{4}$ area of ∆ABC.

We know that D and E are the mid-points of BC and CA, respectively.

By mid-point theorem,

DE || AB and DE = $\dfrac{1}{2}$ AB = BF (As F is mid-point of AB)

From figure,

BF || DE and BF = DE.

Hence, BDEF is a || gm.

Similarly,

F and E are the mid-points of AB and CA.

By mid-point theorem,

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

From figure,

EF || DC and EF = DC.

Hence, EFDC is a || gm.

Since, FE || BC and FB, EC are not parallel.

Hence, EFBC is a trapezium.

F and D are the mid-points of AB and BC.

By mid-point theorem,

FD || AC and FD = $\dfrac{1}{2}$ AC = AE (As E is mid-point of AC)

From figure,

FD || AE and FD = AE.

Hence, AFDE is a || gm.

We know that a diagonal divides a || gm in two triangles of equal area.

In || gm BDEF, FD is the diagonal,

∴ area of △DEF = area of △BDF ………(i)

In || gm EFDC, DE is the diagonal,

∴ area of △DEF = area of △EDC ………(ii)

In || gm AFDE, FE is the diagonal,

∴ area of △DEF = area of △AFE ………(iii)

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

area of △DEF = area of △BDF = area of △EDC = area of △AFE …….(iv)

From figure,

area of △ABC = area of △DEF + area of △BDF + area of △EDC + area of △AFE

= area of △DEF + area of △DEF + area of △DEF + area of △DEF

= 4 x area of △DEF

∴ area of △ABC = 4 x area of △DEF

⇒ area of △DEF = $\dfrac{1}{4}$ area of △ABC ……(v)

From figure,

area of trapezium BCEF = area of △DEF + area of △BDF + area of △EDC

= area of △DEF + area of △DEF + area of △DEF

= 3 x area of △DEF

∴ area of trapezium BCEF = 3 x area of △DEF

⇒ area of △DEF = $\dfrac{1}{3}$ x area of trapezium BCEF …….(vi)

From (v) and (vi) we get,

⇒ $\dfrac{1}{4}$ area of △ABC = $\dfrac{1}{3}$ area of trapezium BCEF

⇒ area of trapezium BCEF = $\dfrac{3}{4}$ area of △ABC.

Hence, proved that area of trapezium BCEF = $\dfrac{3}{4}$ area of △ABC.