ABCD एक चतुर्भुज है, तो सिद्ध कीजिये कि `vec(BA)+vec(BC)+vec(CD)+vec(DA)=2vec(BA)`

The value of vec(AB)+vec(BC)+vec(DA)+vec(CD) is

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:

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:

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

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

If ABCD is a quadrilateral, then vec(BA) + vec(BC)+vec(CD) + vec(DA)=

Assertion A : If A, B, C, D are four points on a semi-circular arc with centre at 'O' such that |vec(AB)| = |vec(BC)|=|vec(CD)| , then vec(AB) +vec(AC) +vec(AD) =4 vec(AO) +vec(OB) +vec(OC) Reason R : Polygon law of vector addition yields vec(AB) +vec(BC) +vec(CD) +vec(AD)=2vec(AO) In the light of the above statements, choose the most appropriate answer from the options given below :