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