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Let $\mathrm{ABCD}$ be a parallelogram whose diagonals intersect at $\mathrm{P}$ and let $\mathrm{O}$ be the origin, then what is
$\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$ equal to $?$
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$\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}$ equal to $?$
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2234 Upvotes
Verified Answer
The correct answer is:
$4 \overline{\mathrm{OP}}$
Diagonals of a parallelogram bisect each other. Therefore, $\mathrm{P}$ is the mid point of $\mathrm{AC}$ and $\mathrm{BD}$ both.
So, in $\Delta \mathrm{OAC}, \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$
and in $\Delta \mathrm{ODB}, \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$
So, in $\Delta \mathrm{OAC}, \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$
and in $\Delta \mathrm{ODB}, \overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OC}}=2 \overrightarrow{\mathrm{OP}}$ and $\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OD}}=2 \overrightarrow{\mathrm{OP}}$
$\Rightarrow \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}=4 \overrightarrow{\mathrm{OP}}$
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