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

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Area of EFGH=EFG+FGH

=height*GF/2+ height*GF/2

=AFGD/2 + BCGH/2

=ABCD/2

⇒ 2EFGH=ABCD

=height*GF/2+ height*GF/2

=AFGD/2 + BCGH/2

=ABCD/2

⇒ 2EFGH=ABCD

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)