Question

in the rhombus ABCD prove that ac²+BD²=4AB²​

Answer 1

15



 Hey there !! 

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

To Prove :- 4AB² = ( AC² + BD² ).

Proof :- 

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





From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]





=> 4AB² = ( AC² + BD² ).

✔✔ Hence, it is proved ✅✅.