|In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that \[AB+CD=BC+DA\].|
|We have, AB, BC, CD and DA are the tangents touching the circle at P, Q, R and S respectively|
|Now, \[AP=AS,BP=BQ,CR=CQ\] and \[DR=DS\].|
|On adding we get|
|\[AB+CD=AD+BC\] Hence Proved.|
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