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If O is any point inside a rectangle ABCD then is the following relation true?

$(OB)^{2} + (OD)^{2} = (OC)^{2} + (OA)^{2}$

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True

False

answer is A.

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Detailed Solution

Given, ABCD is rectangle.
Here, APRD and BPRC are rectangles.
Hence,

AP = DR \dots \dots

.(i)

BP = CR \dots \dots

…(ii)
Using the Pythagoras theorem in right angle triangle OPB,

O B^{2} = {PB}^{2} + O P^{2}

Similarly in right angle triangles ORD, OPA and ORC,

{OD}^{2} = {OR}^{2} + {RD}^{2}

{OA}^{2} = {AP}^{2} + {OP}^{2}

{OC}^{2} = {OR}^{2} + {RC}^{2}

Now the left hand side given in the problem is,

{OB}^{2} + {OD}^{2}

Put the values,

= {BP}^{2} + {OP}^{2} + {OR}^{2} + {DR}^{2}

= {OP}^{2} + {OR}^{2} + {BP}^{2} + {AP}^{2}

Now the right hand side given in the problem is,

{OA}^{2} + {OC}^{2}

= {AP}^{2} + {OP}^{2} + {OR}^{2} + {CR}^{2}

= O P^{2} + O R^{2} + A P^{2} + B P^{2}

As both sides are equal. Thus,

(OB)^{2} + (OD)^{2} = (OC)^{2} + (OA)^{2}

Hence, given relation is true.
Option 1 is correct.

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