Question

E,F,G,H are respectively, the midpoints of the sides AB,BC,CD and DA of parallelogram. ABCD. Show that the quadrilateral EFGH is a parallelogram and that its area is half the area of the parallelogram ABCD.

Answer 1

Area of HFG$=\frac{1}{2}\times$ Area of HDCF (Parallelogram and triangle on same base)

Multiplying by $2$

$2\times$ Area of HGF$=2\times [\frac{1}{2}\times AreaHDCF]$

Area of HGFE $=\frac{1}{2}\times$ABCD

Hence proved.