A B C D is a rhombus, whose diagonals intersect at E. Then E A+E … - Vidyadip

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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