**Given :** A parallelogram ABCD · E, F, G, H are mid-points of sides AB, BC, CD, DA respectively

**To Porve :** ar (EFGH) = 1/ 2 ar (ABCD)

**Construction : **Join AC and HF.

**Proof :** In ∆ABC,

E is the mid-point of AB.

F is the mid-point of BC.

⇒ EF is parallel to AC and EF = 1 2 AC ... (i)

Similarly, in ∆ADC, we can show that

HG || AC and HG = 1 2 AC ... (ii)

From (i) and (ii)

EF || HG and EF = HG

∴ EFGH is a parallelogram. [One pour of opposite sides is equal and parallel]

In quadrilateral ABFH, we have

HA = FB and HA || FB [AD = BC ⇒ 1/ 2 AD = 1/ 2 BC ⇒ HA = FB]

∴ ABFH is a parallelogram. [One pair of opposite sides is equal and parallel]

Now, triangle HEF and parallelogram HABF are on the same base HF and between the same parallels HF and AB.

∴ Area of ∆HEF = 1/ 2 area of HABF ... (iii)

Similarly, area of ∆HGF = 1/ 2 area of HFCD ... (iv)

Adding (iii) and (iv),

Area of ∆HEF + area of ∆HGF = 1/ 2 (area of HABF + area of HFCD)

⇒ ar (EFGH) = 1/ 2 ar (ABCD)** Proved.**