Question

prove that same of the squares of the sides of rhombus is equal to the sum of the squares of its diagonal

AB^2+BC^2+CD^2+AD^2=AC^2+BD^2

Answer 1

In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB
AB2 = OA2 + OB2   [By Pythagoras theorem]

⇒  4AB2 = AC2+ BD2
⇒  AB2 + AB2 + AB2 + AB2 = AC2+ BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2
Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.