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

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

A B C D E is pentagon, prove that vec A B + vec B C + vec C D + vec D E+ vec E A = vec0 vec A B+ vec A E+ vec B C+ vec D C+ vec E D+ vec A C=3 vec A C

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

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 A,B,C,D are any four points in space prove that vec(AB)xxvec(CD)+vec(BC)xvec(AD)+vec(CA)xxvec(BD)=2vec(AB)xxvec(CA)

If D E and F be the mid ponts of the sides BC, CA and AB respectively of the /_\ABC and O be any point, then prove that vec(OA)+vec(OB)+vec(OC)=vec(OD)+vec(OE)+vec(OF)

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals:

Let O be the centre of a regular pentagon ABCDE and vec(OA) = veca , then vec(AB) +vec(2BC) + vec(3CD) + vec(4DE) + vec(5EA) is equals:

ABCDE is a pentagon.prove that the resultant of force vec AB,vec AE,vec BC,vec DC,vec ED and vec AC is 3vec AC