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Question

If the sides of a parallelogram touch a circle (refer figure of Q. 7), 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 = 2 BC
AB = BC ........(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.

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