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# If $ABCD$ is a Rhombus, with position vectors of vertices,$\overrightarrow {OA},\:\overrightarrow {OB},\:\overrightarrow {OC}\:and\:\overrightarrow {OD}$, then $\overrightarrow {OA}+\overrightarrow {OB}+\overrightarrow {OC}+\overrightarrow {OD}=?$

Can you answer this question?

In the Rhombus $ABCD$, the opp sides are equal and parallel.
$\therefore\:\overrightarrow {AB}=\overrightarrow {DC}$
$\Rightarrow\:\overrightarrow {OB}-\overrightarrow {OA}=\overrightarrow {OC}-\overrightarrow {OD}$
$\Rightarrow\:\overrightarrow {OB}+\overrightarrow {OD}=\overrightarrow {OC}+\overrightarrow {OA}$
Now
$\overrightarrow {OA}+\overrightarrow {OB}+\overrightarrow {OC}+\overrightarrow {OD}=2(\overrightarrow {OA}+\overrightarrow {OC})$
$=2(\overrightarrow {OA}+\overrightarrow {OB}-\overrightarrow {OB}+\overrightarrow {OC})$
$(i.e.,$ Adding and subtracting $\overrightarrow {OB}$)
$=2(\overrightarrow {AB}+\overrightarrow {BC})$
answered Nov 7, 2013 by
edited Nov 12, 2013