In rhombus ABCD, AB = BC = CD = DA
We know that diagonals bisect each other at 90°
Consider right triangle AOB
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (OA)2 + (OB)2 = (AB)2
⇒ AC2 + BD2 = 4AB2
⇒ AC2 + BD2 = AB2 + AB2 + AB2 + AB2
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + DA2
Hence Proved