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

In a regular hexagon ABCDEF,vec AE

In a regular hexagon ABCDEF, vec(AE)

In a regular hexagon ABCDEF, vec(AE)

In a regular hexagon ABCDEF, vec(AB) +vec(AC)+vec(AD)+ vec(AE) + vec(AF)=k vec(AD) then k is equal to

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.

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

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

In a regular hexagon ABCDEF, vec(AB)=a, vec(BC)=b and vec(CD) = c. Then, vec(AE) =