Question

ABCD is a parallelogram in which E,F,G,H are the midpoints of sides. Prove that 2ar(EFGH)=ar(ABCD)
Answer fast in detail.

Answer 1

Area of EFGH=EFG+FGH
                     =height*GF/2+ height*GF/2
                     =AFGD/2  + BCGH/2
                     =ABCD/2
⇒  2EFGH=ABCD

Answer 2

G is the midpoint of DC
H is the midpoint of AD

By midpoint theorem
      HG // AC     and
      HG = 1/2 AC             eq1
Similarly
      EF // AC
      EF = 1/2 AC              eq2 
From eq 1 and eq 2
  HG // EF and HG =EF
 therefore EFGH is a parallelogram

In ΔHEF and //gm HABF lie on the same base HF and between same
p//s HF and AB
therefore ar(ΔHEF) = 1/2 ar(HABF)      eq3
similarly ar(ΔHGF) = 1/2 ar(HFCD)      eq4
 adding eq3 and eq 4
ar (ΔHEF) + ar(ΔHGF) = 1/2 ar (HABF+HFCD)
ar(EFGH) = 1/2 ar (ABCD)

⇒2ar(EFGH) = ar(ABCD)