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PROVE THAT THE SUM OF THE SQUARES OF THE SIDES OF THE RHOMBUS IS EQUAL TO THE SUM OF THE SQUARES OF ITS DIAGONALS.​

Answer 1

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Answer 2

Answer:

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).

Proof :- 

➡ We know that the diagonals of a rhombus bisect each other at right angles.

==>∠AOB = ∠BOC = ∠COD = ∠DOA = 90°,

$OA = \frac{1}{2} AC \: and \: OB = \frac{1}{2} BD$

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

$= = > {AB}^{2} = \frac{1}{2} {AC}^{2} + \frac{1}{2} {BD}^{2}$

$= = > {AB}^{2} = \frac{1}{2} ( {AC}^{2} + {BD}^{2} )$

==> 4AB² = ( AC² + BD² ) .

==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .

•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .

[ In a rhombus , all sides are equal ] .

Hence, it is proved.