Question

If ABCDEF be a regular hexagon with centre 'O' then show that $\overline{AB}+\overline{AC}+\overline{AD}+\overline{AE}+\overline{AF}=3\overline{AD}=6\overline{AO}$

Answer 1

$AB+AC+AD+AE+AF=AB+(AC+AF)+AD+DE$

$=AB+CAC+\left(D\right)+AD+DE$

$AF=CD$

$=AB+AD+AD+AE$

$=2AD+(AB+AE)$

$=2AD+(ED+AE)(AB=ED)$

$=2AD+AD$

$AF+AB+AC+AD+AE=3AD$

(As "$O$" is the centre and $AO=OD$)

$=3\left(2AO\right)$

$=6AO$

$AB+AC+AD+AE+AF+6AD$