Question

If $ABCDEF$ is a regular hexagon, show that $\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}=6\overline{AO}$, where $O$ is the centre of the hexagon.

Answer 1

$ABCDEF$ is a regular hexagon.
$\therefore \overline{AB}=\overline{ED}$ and $\overline{AF}=\overline{CD}$
$\therefore$ by the triangle law of addition of vectors,
$\overline{AC}+\overline{AF}=\overline{AC}+\overline{CD}=\overline{AD}$
$\overline{AE}+\overline{AB}=\overline{AE}+\overline{ED}=\overline{AD}$
$\therefore LHS=\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}$
$=\overline{AD}+(\overline{AC}+\overline{AF})+(\overline{AE}+\overline{AB})$
$=\overline{AD}+\overline{AD}+\overline{AD}$
$=3\overline{AD}=3\left(2\overline{AO}\right)$......[$O$ is midpoint of $AD$]
$6\overline{AO}$.