Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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Question

Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

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Accepted Answer

Step 1: $AB^2 + BC^2 + CD^2 + AD^2$
Step 2: $= 4AB^2$ ($\because$ ABCD is a rhombus)
Step 3: $= 4(OA^2 + OB^2)$
Step 4: $= 4\left(\frac{AC^2}{4} + \frac{BD^2}{4}\right) = AC^2 + BD^2$

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