1.

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

Answer»

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.


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