Parallelogram ABCD of which midpoints are E, F, G and H of sides AB, BC, CD and DA respectively. Let’s prove that

(i) EFGH is a parallelogram.

(ii) Area of parallelogram shaped region EFGH is half of the area of parallelogram shaped region ABCD.

Given.

Parallelogram ABCD of which midpoints are E, F, G and H of sides AB, BC, CD and DA respectively

Formula used.

SAS congruency rule = If 2 sides and angle between them are equal in both triangles then both triangles are congruent

In triangle DGH and triangle BEF

DG= EB ∵ AB= CD then their half also be equal

DH= FB ∵ BC= DA then their half also be equal

∠ D=∠ B ∵ opposite angles of parallelogram are equal

triangle DGH ≅ triangle BEF [By SAS congruency]

∴ GH = EF

In triangle AEH and triangle CGF

GC= AE ∵ AB= CD then their half also be equal

AH= FC ∵ BC= DA then their half also be equal

∠ C=∠ A ∵ opposite angles of parallelogram are equal

triangle DGH ≅ triangle BEF [By SAS congruency]

∴ HE = GF

If both pair of sides are equal in quadrilateral then this is called parallelogram

∴ EFGH is a parallelogram

Join the mid-points FH

Parallelogram divides in 2 equal parallelograms

In parallelogram ABFH and triangle EFH

Both lies on same base FH

And both lies on parallel lines AB and FH

AB || FH ∵ opposite sides of parallelogram are parallel

Then

Area of triangle EFH = × Area of parallelogram ABFH

In parallelogram FHDC and triangle EFG

Both lies on same base FH

And both lies on parallel lines CD and FH

CD || FH ∵ opposite sides of parallelogram are parallel

Then

Area of triangle GFH = × Area of parallelogram FHDC

Adding both we get;

Area of [Δ EFG+Δ GFH]=×Area of parallelogram [ABFH+FHDC]

Area of parallelogram EFGH = ×Area of parallelogram ABCD