Question

ABCD is a parallelogram and P is the point of intersection of its diagonals. If O is the origin of reference, show that $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}$.

Answer 1

Given a parallelogram ABCD and P is the point of intersection of its diagonals. We know the diagonals of a parallelogram, bisect each other. Therefore,
$$\begin{gathered} \frac{\overrightarrow{OA}+\overrightarrow{OC}}{2}=\overrightarrow{OP} \\ \overrightarrow{OA}+\overrightarrow{OC}=2\overrightarrow{OP}.....\left(1\right) \\ and\frac{\overrightarrow{OB}+\overrightarrow{OD}}{2}=\overrightarrow{OP} \\ \overrightarrow{OB}+\overrightarrow{OD}=2\overrightarrow{OP}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), We get,
$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OP}$