prove that the sum of the squares of the side of the rohmbus is equal to sum of square of diagonals
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Given :- A rhombus ABCD whose diagonals AC and BD intersect at O. To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ). ➡ We know that the diagonals of a rhombus bisect each other at right angles. ==> 4AB² = ( AC² + BD² )
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To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ). ➡ We know that the diagonals of a rhombus bisect each other at right angles. ==> 4AB² = ( AC² + BD² ) . ==> AB² + AB² + AB² + AB² = ( AC² + BD² )
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