A) \[~AB+CD=AD+BC\]
B) \[AB+AC=BC+BD\]
C) \[AB+BC=CD+DA\]
D) \[AB+BD=BC+CA\]
Correct Answer: A
Solution :
(a): Using theorem, the lengths of tangents drawn from an external point to a circle are equal. A is an external point, then AP = AS .... (i) B is an external point, then BP = BQ .... (ii) C is an external point, then CQ = RC .... (iii) D is an external point, then SD = RD .... (iv) On adding Eq. (i), (ii), (iii) and (iv), we get \[\left( AP+BP \right)+\left( RC+RD \right)\] \[=\left( AS+BQ \right)+\left( CQ+SD \right)\] \[\Rightarrow \]\[AB+CD=\left( AS+SD \right)+\left( BQ+CQ \right)\] \[\Rightarrow \]\[AB+CD=AD+BC\]
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