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.