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Question
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.
Solution
ABCD is a quadrilateral. Suppose a circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at P, Q, R and S, respectively.
We know that the length of tangents drawn from an external point to a circle are equal.
DR = DS ...(1)
CR = CQ ...(2)
BP = BQ ...(3)
AP = AS ...(4)
Adding (1), (2), (3) and (4), we get
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
Or, AB + CD = BC + DA
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