Let ABCDE be a regular pentagon. If `vecOD = vecOA +lambdavecOB + muvecOC`, where O is the origin, then `mu-lambda`

If E is the intersection point of diagonals of parallelogram ABCD and vec( OA) + vec (OB) + vec (OC) + vec (OD) = xvec (OE) where O is origin then x=

ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find angle AMC

ABCDE is a regular pentagon.The bisector of angle A meets side CD at P, then prove that APC=90^(@).

Show that if P, A, B are any three points, then lambda vec(PA)+mu vec(PB)=(lambda + mu)vec(PC , where C divides [AB] in the ratio mu : lambda .

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: