`bar(AC) " and " bar(BD)` are the diagonals of the parallelogram ABCD. Prove that,
`vec(AC)+vec(BD)=2vec(BC) " and " vec(AC)-vec(BD)=2vec(AB)`

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

In a parallelogram ABCD. Prove that vec(AC)+ vec (BD) = 2 vec(BC)

vec(AC) and vec(BD) are the diagonals of a parallelogram ABCD. Prove that (i) vec(AC) + vec(BD) - 2 vec(BC) (ii) vec(AC) - vec(BD) - 2vec(AB)

ABCD is a parallelogram and AC, BD are its diagonals. Show that : vec(AC)+vec(BD)=2vec(BC), vec(AC)-vec(BD)=2vec(AB) .

ABCD is a parallelogram and AC, BD are its diagonals. Show that : vec(AC)+vec(BD)=2vec(BC),vec(AC)-vec(BD)=2vec(AB) .

If ABCD is a parallelogram, then vec(AC) - vec(BD) =

If ABCD is a parallelogram, then vec(AC) - vec(BD) =

In the parallelogram ABCD,vec AC^(2)-vec BD^(2)=

In pentagon ABCDE, prove that : vec(AB)+vec(BC)+vec(CD)+vec(DE)+vec(EA)=vec0 .

ABCD is a parallelogram Fig. 2 (c ) .64. AC and (BD) are its diagonals. Show that (a) vec (AC) +vec (BD) =2 vec (BC) (b) vec (AC) - vec (BD) =2 vec (AB) .