Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of it's diagonals

Answer 1

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°,

$\bf { OA = \frac{1}{2} AC \: and \: OB = \frac{1}{2} BD . }OA=21ACandOB=21BD.$

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

$\bf { \implies {AB}^{2} = (\frac{1}{2} AC}^{2}) + ( { \frac{1}{2}BD) }^{2} .⟹AB2=(21AC2)+(21BD)2.$

$\bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }⟹AB2=41(AC2+BD2)$

==> 4AB² = ( AC² + BD² ) .

==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .

•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .

[ In a rhombus , all sides are equal ] .