If ABCDEF is a regular hexagon, prove that `vec(AC)+vec(AD)+vec(EA)+vec(FA)=3vec(AB)`

If ABCDEF is a regualr hexagon, then vec(AC) + vec(AD) + vec(EA) + vec(FA)=

In Fig. ABCDEF is a ragular hexagon. Prove that vec(AB) +vec(AC) +vec(AD) +vec(AE) +vec(AF) = 6 vec(AO) .

ABCDEF is a regular hexagon. Show that : vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=6vec(AO) . Where O is the centre of the hexagon.

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon with vec(AB) = vec(a) ,vec(BC ) = vec(b) then vec(CE) equals :