Question

Prove that the sum of the squares on the sides of a rhombus is equal
to the sum of the squares on its diagonals. ​

Answer 1

Answer :-

▶ Step-by-step explanation :-

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°,

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

==> 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.