Question

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?​

Answer 1

Answer:

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Answer 2

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