If `E ,\ F ,\ G\ a n d\ H` are respectively the mid-points of the sides of a parallelogram `A B C D ,\ ` Show that `a r(E F G H)=1/2A R\ (A B C D)`

Explanation

Given:
`E,F,G and H` are respectively the mid-points of the sides of a parallelogram `ABCD`.
To Prove:
`ar(EFGH)= 1/2 ar(ABCD)`Construction:
`H and F` are joined.
Proof:
`AD |\ | BC and AD=BC` (Opposite sides of a parallelogram)
`⇒ 1/2AD=1/2BC`
Also,
`AH |\ | BF` and and `DH |\ | CF`
`⇒AH=BF` and `DH=CF ∣ H` and `F` are mid points
Thus,
`ABFH` and `HFCD` are parallelograms.
Now,
`/_\ EFH and |\ | gm ABFH` lie on the same base `FH` and between the same parallel lines `AB` and `HF`.
`∴ Area of EFH= 1/2 ar(ABFH) --- (i)`
Also,
`Area of GHF= 1/2ar(HFCD) --- (ii)`
Adding (i) and (ii),


`Area of /_\ EFH+ area of /_\ GHF = 1/2ar(ABFH)+ 1/2 ar(HFCD)`

⇒ Area of `EFGH=` Area of `ABFH`

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