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Answer :

∵ Lengths of two tangents drawn from an external point of circle are equal,

\(AP \ = \ AS \) , \(BP \ = \ BQ \) , \(DR \ = \ DS \) , and \(CR \ = \ CQ \)

Adding these all, we get

\( (AP \ + \ BP) \ + \ (CR \ + \ RD) \ \)

\( = \ (BQ \ + \ QC) \ + \ (DS \ + \ SA) \)

\( => \ AB \ + \ CD \ = \ BC \ + \ DA \)

Hence proved

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