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If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/ 2 ar (ABCD).

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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.

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