D, E and F are mid-points of the sides BC, CA and AB respectively of a ∆ABC. Prove that

(i) FDCE is a parallelogram

(ii) area of ∆DEF = $\dfrac{1}{4}$ area of ∆ABC

(iii) area of || gm FDCE = $\dfrac{1}{2}$ area of ∆ABC

∆ABC with D, E and F as mid-points of the sides BC, CA and AB, respectively is shown below:

(i) F and E are midpoints of AB and AC respectively.

So, by mid-point theorem,

FE || BC and FE = $\dfrac{1}{2}$ BC ………(1)

Also, D is the mid-point of BC

CD = $\dfrac{1}{2}$ BC …….. (2)

From (1) and (2),

FE || BC and FE = CD

Since, FE || BC so,

FE || CD and FE = CD ……… (3)

Similarly,

D and F are the midpoints of BC and AB.

So, by mid-point theorem,

DF || AC and DF = $\dfrac{1}{2}$ AC ………(4)

Also, E is the mid-point of AC

EC = $\dfrac{1}{2}$ AC …….. (5)

From (4) and (5),

DF || AC and DF = EC

Since, DF || AC so,

DF || EC and DF = EC ……… (6)

From 3 and 6,

FE || CD, FE = CD, DF || EC and DF = EC.

Since, opposite sides of FDCE are parallel and equal.

Hence proved FDCE is a parallelogram.

(ii) Since,

BD = CD (As D is mid-point of BC), FE = CD (Proved above) and FE || CD

So, BD = FE and BD || FE.

Hence, BDEF is a || gm.

Since,

FD = EC (Proved above) and AE = EC (As E is mid-point of AC) and FD || EC.

So, FD = AE and FD || AE.

Hence, AFDE is a || gm.

We know that, FDCE is a parallelogram and DE is a diagonal of || gm FDCE.

So, area of ∆DEF = area of ∆DEC (As diagonal divides || gm into two triangles with equal area) ……..(1)

We know that, BDEF is a parallelogram and FD is a diagonal of || gm BDEF.

So, area of ∆DEF = area of ∆FBD (As diagonal divides || gm into two triangles with equal area) ……..(2)

We know that, AFDE is a parallelogram and FE is a diagonal of || gm AFDE.

So, area of ∆DEF = area of ∆AFE (As diagonal divides || gm into two triangles with equal area) ……..(3)

From 1, 2 and 3 we get,

area of ∆DEF = area of ∆DEC = area of ∆FBD = area of ∆AFE ………(4)

From figure,

area of ∆ABC = area of ∆DEF + area of ∆DEC + area of ∆FBD + area of ∆AFE

= area of ∆DEF + area of ∆DEF + area of ∆DEF + area of ∆DEF (From 4)

= 4 x area of ∆DEF.

∴ area of ∆ABC = 4 x area of ∆DEF

⇒ area of ∆DEF = $\dfrac{1}{4}$ area of ∆ABC

Hence, proved that area of ∆DEF = $\dfrac{1}{4}$ area of ∆ABC.

(iii) From figure,

Area of || gm FDCE = Area of ∆DEF + Area of ∆DEC

= Area of ∆DEF + Area of ∆DEF (From part (ii) eqn. 4)

= 2 area of ∆DEF

= $2 \times \dfrac{1}{4}$ area of ∆ABC [As area of ∆DEF = $\dfrac{1}{4}$ area of ∆ABC]

= $\dfrac{1}{2}$ area of ∆ABC.

Hence, proved that area of || gm FDCE = $\dfrac{1}{2}$ area of ∆ABC.