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

Solution :

Let the diagonals BD and AC of the rhombus ABCD intersect each other at O.rhombus

Since the diagonals of the rhombus bisect each other at right angles.

\(\therefore\)  \(\angle\) AOB = \(\angle\) BOC = \(\angle\) COD = \(\angle\) DOA = 90

and  AO = CO, BO = OD

Since AOB is a right triangle angled at O, therefore,

\({AB}^2\) = \({OA}^2\) + \({OB}^2\)

Since AO = CO, BO = OD,

\(\implies\) \({AB}^2\) = \(({1\over 2}{AB})^2\) + \(({1\over 2}{BD})^2\)

\(\implies\)  \(4{AB}^2\) = \({AC}^2\) + \({BD}^2\)            ……….(1)

Similarly, we have :

\(4{BC}^2\) = \({AC}^2\) + \({BD}^2\)            ……..(2)

\(4{CD}^2\) = \({AC}^2\) + \({BD}^2\)            ……..(3)

\(4{AD}^2\) = \({AC}^2\) + \({BD}^2\)             ………(4)

Adding all the equations above, we get

\({AB}^2\) + \({BC}^2\) + \({CD}^2\) + \({DA}^2\) = \({AC}^2\) + \({BD}^2\)

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