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Question
If the sides of a parallelogram touch a circle in following figure, prove that the parallelogram is a rhombus.
Solution
From A, AP and AS are tangents to the circle.
Therefore, 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.
But AB = CD and BC = AD ...(v) (Opposite sides of a ||gm)
Therefore, AB + AB = BC + BC
2AB = 2BC
AB = BC ...(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.
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