Let ABCD is a rhombus. Let O is a point where diagonals are intersect to each other.

So AB = BC = CD = DA

    AC is perpendicular to BD and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90

and OA = OC = AC/2

OB= OD = BD/2

Now from triangle AOB

AB² = OA² + OB²

=> AB² = (AC/2)² + (BD/2)²

            = AC² /4 + BD² /4

=> 4AB² = AC² + BD²

=>  AB² + AB² +AB² + AB² = AC² + BD²

=> AB² + BC² + CD² + DA² = AC² + BD²

So the sum of squares of the sides of a rhombus is equal to the sum of the square of its diagonal