ABCD एक समांतर चतुर्भुज है ।AC और BD इसके विकर्ण है । सिद्ध कीजिए -
`vec(AC) + vec(BD) = 2vec(BC)`
`vec(AC) - vec(BD) = 2vec(AB)`.

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) .

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)

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

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) =

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)