# Parallelogram $=A F E D$.6. If $E, E, G$ and $H$ are respectively the midpoint of the sides of

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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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by (42.4k points) 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)