Question

If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}$.

Answer 1

Let D, E and F are the midpoints of BC, CA and AB respectively.
Therefore,
$\frac{\overrightarrow{OB}+\overrightarrow{OC}}{2}=\overrightarrow{OD}$
$$\begin{gathered} \overrightarrow{OB}+\overrightarrow{OC}=2\overrightarrow{OD}.....\left(1\right) \\ Similarly, \\ \overrightarrow{OC}+\overrightarrow{OA}=2\overrightarrow{OE}.....\left(2\right) \\ \overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OF}.....\left(3\right) \end{gathered}$$

Adding (1), (2) and (3). We get,

$$\begin{gathered} 2(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})=2(\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}). \\ \Rightarrow \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}. \end{gathered}$$
Hence Proved.