Question

OR
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares
of its diagonals, ie AB+BC+CD+AD = ACBD (see figure)
28
3​

Answer 1

Answer:

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 ) .