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)