question_answer

A point 0 in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Then,

A)  \[OB+OD=OC+OA\]

B)  \[O{{B}^{2}}+O{{A}^{2}}=O{{C}^{2}}+O{{D}^{2}}\]

C)  \[OB\cdot OD=OC\cdot OA\]

D)  \[O{{B}^{2}}+O{{D}^{2}}=O{{C}^{2}}+O{{A}^{2}}\]

Correct Answer: D

Solution :

Draw EF || AB

In right angled \[\Delta EOA\] and \[\Delta OCF,\]

\[O{{A}^{2}}=O{{E}^{2}}+A{{E}^{2}}\]and \[O{{C}^{2}}=O{{F}^{2}}+C{{F}^{2}}\]

\[\therefore \]\[=O{{A}^{2}}+O{{C}^{2}}=O{{E}^{2}}+A{{E}^{2}}+O{{F}^{2}}+C{{F}^{2}}\]?(i)

In right angled \[\Delta DEO\]and \[\Delta OBF,\]

\[O{{D}^{2}}=O{{E}^{2}}+D{{E}^{2}},\]\[O{{B}^{2}}=O{{F}^{2}}+B{{F}^{2}}\]

\[\therefore \] \[O{{D}^{2}}+O{{B}^{2}}=O{{E}^{2}}+O{{F}^{2}}+D{{E}^{2}}+B{{F}^{2}}\]            ...(ii)

As         FB = EA and DE = CF

Here, from Eqs. (i) and (ii), we get

\[O{{A}^{2}}+O{{C}^{2}}=O{{D}^{2}}+O{{B}^{2}}\]