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Parallelogram \( =A F E D \). 6. If \( E, E, G \) and \( H \) are respectively the midpoints of the sides of a Parallelogram \( A B C D \), show that [wr (E FCAH) \( =\frac{1}{2} \operatorname{ar}(A B C D) \) \[ \operatorname{arc}\left(E F(G H)=\frac{1}{\theta} \operatorname{ar}(A B C D)\right. \]

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Construction: Join E & G.

\(\because \) AB = CD (opposite sides of parallelogram)

\(\frac{AB}2 = \frac{CD}2\) 

⇒ BE = GC (\(\because \) E is mid-point of AB & G is mid-point of CD)

Also BE || GC (\(\because \) AB || CD)

\(\therefore\) EBCG is a parallelogram.

Now, since, parallelogram EBCG & triangle △EFG are on the same base EG & between same parallels EG & BC.

\(\therefore\) ar(△EFG) = \(\frac12\)ar(EBCG)-----(i)

Similarly, ar(△EHG) = \(\frac12\)ar(EGDA)----(ii)

Adding equations (i) & (ii), we get

ar(△EFG) + ar(△EHG) = \(\frac12\)ar(EBCG) + \(\frac12\)ar(EGDA)

⇒ ar(EFGH) = \(\frac12\)ar(ABCD)

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