Question
If $A B C D$ is a rhombus, whose diagonals intersect at $E$, then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals

Answer

(a) : $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$
$
=\overrightarrow{E A}+\overrightarrow{E B}-\overrightarrow{E A}-\overrightarrow{E B}
$ [As diagonals of a rhombus bisect each other]
$=\overrightarrow{0}$

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