Answers

2014-12-06T16:13:15+05:30
Area of EFGH=EFG+FGH
                     =height*GF/2+ height*GF/2
                     =AFGD/2  + BCGH/2
                     =ABCD/2
⇒  2EFGH=ABCD
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2014-12-07T15:28:35+05:30
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)
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