Question

Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.

Answer 1

Given: A regular octagon of eight sides with centre O.
To show: \[\overrightarrow{OA} +\overrightarrow{OB} + \overrightarrow{OC} +\overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} = \overrightarrow{0} .\]
Proof: We know centre of the regular octagon bisects all the diagonals passing through it.
\[\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} +\overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} =\overrightarrow{0} .\] and \[\overrightarrow{OC} = -\overrightarrow{OG} .\]
\[\overrightarrow{OA} + \overrightarrow{OE} = \overrightarrow{0} , \overrightarrow{OB} + \overrightarrow{OF} = \overrightarrow{0} , \overrightarrow{OD} + \overrightarrow{OH} = \vec{0}\] 

\[\overrightarrow{OA} = - \overrightarrow{OE} , \overrightarrow{OB} = - \overrightarrow{OF} , \overrightarrow{OD} = - \overrightarrow{OH}\] and \[\overrightarrow{OC} +\overrightarrow{OG} = \overrightarrow{0} .\]    ............(i)
Now, 

\[\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} +\overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} \]

\[ = \left(\overrightarrow{OA} +\overrightarrow{OE} \right) + \left( \overrightarrow{OB} + \overrightarrow{OF} \right) + \left( \overrightarrow{OC} + \overrightarrow{OG} \right) + \left( \overrightarrow{OD} + \overrightarrow{OH} \right)\]

\[ = \overrightarrow{0} +\overrightarrow{0} +\overrightarrow{0} + \overrightarrow{0} \]

\[ = \overrightarrow{0}\]
Hence proved.\]