' $O$ ' 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

LHS $= {O B}^{2} + {O D}^{2}$
$= ({B Q}^{2} + {O Q}^{2}) + ({O S}^{2} + {D S}^{2})$ (Considering right angled triangles $B Q O$ and $O S D$ )
But, $B Q = A S$ and $Q C = D S$
$({B Q}^{2} + {O Q}^{2}) + ({O S}^{2} + {D S}^{2})$
$= ({A S}^{2} + {O Q}^{2}) + ({O S}^{2} + {Q C}^{2})$
$= ({A S}^{2} + {O S}^{2}) + ({O Q}^{2} + {Q C}^{2})$
$= {O A}^{2} + {O C}^{2}$ (Considering right angled triangles $A S O$ and $C Q O$ )
$=$ RHS
So, ${O B}^{2} + {O D}^{2} = {O A}^{2} + {O C}^{2}$ proved.

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LHS =OB2+OD2=(BQ2+OQ2)+(OS2+DS2) (Considering right angled triangles BQO and OSD)But, BQ=AS and QC=DS(BQ2+OQ2)+(OS2+DS2) =(AS2+OQ2)+(OS2+QC2) =(AS2+OS2)+(OQ2+QC2) =OA2+OC2 (Considering right angled triangles ASO and CQO)= RHSSo, OB2+OD2=OA2+OC2 proved.

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