If ABCDEF is a regular hexagon, prove that `AD+EB+FC=4AB`.

If ABCDEF is a regular hexagon, then Delta ACE is

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

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

If ABCDEF is a regular hexagon,them vec AD+vec EB+vec FC equals 2vec AB b.vec 0 c.3vec AB d.4vec AB

In the adjoining figure, ABCDEF is a regular hexagon. Prove that squareABDE, square ACDF and squareAGDH are parallelograms.

In the adjoining figure, ABCDEF is a regular hexagon. Prove that squareABDE, square ACDF and squareAGDH are parallelograms.

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :

If ABCDEF is a regular hexagon then vec(AD)+vec(EB)+vec(FC) equals :