MCQ
$A B C D$ is a rhombus, whose diagonals intersect at $E$. Then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equals
  • A
    $\overrightarrow{0}$
  • B
    $\overrightarrow{A D}$
  • C
    $2 \overrightarrow{B C}$
  • D
    $2 \overrightarrow{A D}$

Answer

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

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