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ABCD is a parallelogram and P is the point of intersection of the diagonals. If $\mathrm{O}$ is the origin, then $\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$ is equal to
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$\quad 4 \overrightarrow{\mathrm{OP}}$
$\because \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{AP}} ; \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PB}}$
$\overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{OP}}+\overrightarrow{\mathrm{PC}} \& \overrightarrow{\mathrm{OD}}=\overrightarrow{\mathrm{OP}}-\overrightarrow{\mathrm{DP}}$
Also, $\overrightarrow{\mathrm{AP}}=\overrightarrow{\mathrm{PC}} \& \overrightarrow{\mathrm{DP}}=\overrightarrow{\mathrm{PB}}$
$\therefore \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$
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