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.

Answer 1

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 )