P is the point of intersection of the diagona | vedclass

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(D) We know that in a parallelogram,the diagonals bisect each other. Therefore,$P$ is the midpoint of both diagonals $AC$ and $BD$.Using the section f

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$4\,\overrightarrow{OP}$

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(D) We know that in a parallelogram,the diagonals bisect each other. Therefore,$P$ is the midpoint of both diagonals $AC$ and $BD$.Using the section formula for the midpoint,for diagonal $AC$ with midpoint $P$,we have:$\overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OP}$ ......$(i)$Similarly,for diagonal $BD$ with midpoint $P$,we have:$\overrightarrow{OB} + \overrightarrow{OD} = 2\overrightarrow{OP}$ ......$(ii)$Adding equations $(i)$ and $(ii)$,we get:$(\overrightarrow{OA} + \overrightarrow{OC}) + (\overrightarrow{OB} + \overrightarrow{OD}) = 2\overrightarrow{OP} + 2\overrightarrow{OP}$$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 4\overrightarrow{OP}$

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $O$ is any point,then $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = $

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