If quadrilateral ABCD is drawn to circmscribe a circle then prove that ab+cd=ad+bc
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Since, tangents drawn from an exterior point to a circle are equal in length.
Therefore,
AP = AS------(i)
BP = BQ-----(ii)
CR = CQ-----(iii)
And,
DR = DS------(iv)
Adding (i), (ii), (iii) & (iv)
We get,
AP + BP + CR + DR = AS + BQ + CQ + DS
=> (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
=> AB+CD=AD+BC
Hence,
AB+CD=BC+DA
Therefore,
AP = AS------(i)
BP = BQ-----(ii)
CR = CQ-----(iii)
And,
DR = DS------(iv)
Adding (i), (ii), (iii) & (iv)
We get,
AP + BP + CR + DR = AS + BQ + CQ + DS
=> (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ)
=> AB+CD=AD+BC
Hence,
AB+CD=BC+DA
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