If $O$ is any point inside a rectangle $A B C D$ . Prove that $O B^{2} + O D^{2} = O A^{2} + O C^{2}$ .

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Solution

$G i v e n : R e c t a n g l e A B C D w i t h c e n t e r O T o P r o v e : {O B}^{2} + {O D}^{2} = {O C}^{2} + {O A}^{2} T h r o u g h O, d r a w P Q ∥ B C s o t h a t P l i e s o n A B a n d Q l i e s o n D C . N o w, P Q ∥ B C T h e r e f o r e, P Q ⊥ A B a n d P Q ⊥ D C [∠ B = 90 a n d ∠ C = 90) S o, ∠ B P Q = 90 a n d ∠ C Q P = 90 T h e r e f o r e, B P Q C a n d A P Q D a r e b o t h r e c t a n g l e s . N o w, f r o m R i g h t a n g l e d t r i a n g l e △ O P B, {O B}^{2} = {B P}^{2} + {O P}^{2} . . . (1) S i m i l a r l y, f r o m R i g h t a n g l e d t r i a n g l e △ O D Q, {O D}^{2} = {O Q}^{2} + {D Q}^{2} . . . (2) F r o m O P A, w e h a v e {O A}^{2} = {A P}^{2} + {O P}^{2} . . . (3) F r o m △ O Q C {C Q}^{2} + {O Q}^{2} = {O C}^{2} A d d i n g (1) n d (2) {O B}^{2} + {O D}^{2} = {B P}^{2} + {O P}^{2} + {O Q}^{2} + {D Q}^{2} = {C Q}^{2} + {O P}^{2} + {O Q}^{2} + {A P}^{2} [A s B P = C Q a n d D Q = A P] = {C Q}^{2} + {O Q}^{2} + {O P}^{2} + {A P}^{2} = {O C}^{2} + {O A}^{2} [F r o m (3) a n d (4))] H e n c e P r o v e d$

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