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Question
E, F, G, and H are the midpoints of the sides of a parallelogram ABCD.
Show that the area of quadrilateral EFGH is half of the area of parallelogram ABCD.
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Solution
Join HF.
Since H and F are mid-points of AD and BC respectively,
∴ AH = `1/2 "AD and BF" = 1/2 "BC"`
Now, ABCD is a parallelogram.
⇒ AD = BC and AD ∥ BC
⇒ `1/2 "AD" = 1/2`BC and AD || BC
⇒ AH = BF and AH ∥ BF
⇒ ABFH is a parallelogram.
Since parallelogram FHAB and ΔFHE are on the same base FH and between the same parallels HF and AB,
A( ΔFHE ) = `1/2`A ( ||m FHAB ) .....(i)
Similarly,
A( ΔFHG ) = `1/2`A ( ||m FHDC ) .......(ii)
Adding (i) and (ii), We get,
A( ΔFHE ) + A( ΔFHG ) = `1/2 "A"( ||^m "FHAB" ) + 1/2`A ( ||m FHDC )
⇒ A( EFGH ) = `1/2`[ A ( ||m FHAB ) + A ( ||m FHDC ) ]
⇒ A( EFGH ) = `1/2`A( ||m ABCD )
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