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Question
If the sides of a quadrilateral ABCD touch a circle, prove that : AB + CD = BC + AD.
Solution
Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.
Since AP and AS are tangents to the circle from external point A
AP = AS ...(i)
Similarly, we can prove that:
BP = BQ ...(ii)
CR = CQ ...(iii)
DR = DS ...(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
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