**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)