Prove that in a square diagnols are equal and bisect each other at right angles

let ABCD be a square of side 'a'. and AD and BC are its diagonals.

then we know that AC² = AB²+BC² = 2a²

similarly BD² = BC²+CD² = 2a²

so the length of diagonals are same and equal to 2a

now in vector we can write AC = AB + BC  and  BD = BA + AD

now AC.BD = (AB + BC) . (BA + AD) = AB.BA + AB.AD + BC.BA + BC.AD

AB.BA =-AB.AB = -|AB|² = -a²   and  AB.AD = 0(since AB AD )

similarly BC.BA = 0 and BC.AD = a² (since BC AD )

so AC.BD = -a² +0+0+a² =0

which means AC and BD are perpendicular to each other.