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

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

ABCDEF is a regular hexagon. Show that : vec(OA)+vec(OB)+vec(OC)+vec(OD)+vec(OE)+vec(OF)=vec(0)

ABCDE is a pentagon prove that vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EA)=vec0

Let O be the centre of the regular hexagon ABCDEF then find vec(OA)+vec(OB)+vec(OD)+vec(OC)+vec(OE)+vec(OF)

In the regular hexagon shown in figure vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EF)+vec(AF) can be expressed as :

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

A parallelogram ABCD. Prove that vec(AC)+ vec (BD) = 2 vec(BC) '