# In a rhombus ABCD, prove that: AB2 + BC2 + CD2 + DA2 = AC2 + BD2

In rhombus ABCD, AB = BC = CD = DA

We know that diagonals bisect each other at 90°

And

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

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