Question

proof that sum of the square of the sides of a rhombus is equal to the sum of the square of the diagonal​

Answer 1

Answer:

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 )