prove that the diagonals of a rhombus are perpendicular to each other.

Explanation

In Fig. 2.54 , ABCD is a rhombus. `vec(AB) =vec(DC)=veca and vec(AD)=vec(BC)=vecb`, the values of `veca and vecb` are the same for a rhombus.

Let a=b=m
`vec(AC)=vec(AB)+vec(BC)=veca+vecb`
and `vec(BD)=vec(BA)+vec(AD)=-veca+vecb`
`therefore vecAC*vecBD=(veca+vecb)*(-veca+vecb) =-veca*veca+veca*vecb-vecb*veca+vecb*vecb`
`=-a^2+veca*vecb-veca*vecb+b^2=-m^2+m^2=0`
As the dot product is 0, the angle between the vector =`90^@`, i.e., `vec(AC) and vec(BD)` are perpendicular to each other.