if a circle touches all the four side of a quadrilateral of ABCD at the point p,q,r, and s then prove that AB+CD=BC+DA?
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Step-by-step explanation:
Given: Quadrilateral ABCD touches a circle at 4 points P, Q, R, S.
AP = AS --------(1) ---------
BP = BQ --------(2) | --------- tangents from an external
CR = CQ --------(3) | --------- point are equal
DR = DS --------(4) ----------
Adding eqns (1), (2), (3), (4)
AP + BP + CR + DR = AS + BQ + CQ + DS
AP + BP + CR + DR = AS + DS + BQ + CQ
∴ AB + CD = AD + BC
Hence, proved
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