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

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, 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

ABCDEF is a regular hexagon with point O as centre. The value of vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF) is

If ABCDEF is a regular hexagon , then A vec D + E vec B + F vec C equals

If ABCDE is a pentagon, then vec(AB) + vec(AE) + vec(BC) + vec(DC) + vec(ED) + vec(AC) is equal to

ABCDEF is a regular hexagon, Fig. 2 (c ) .65. What is the value of (vec (AB) + vec (AC) + vec (AD) + vec (AE) + vec (AF) ? .

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

In a regular hexagon ABCDEF,vec AE