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Question
If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
Solution
Let the sides of parallelogram ABCD touch the circle at points P, Q, R and S.
AP = AS - (1)
PB = BQ - {2} {Length of tangents drawn from an external point to a circle a equal)
DR = DS - {3}
RC = CQ - (4)
Adding (1), {2}, {3} and (4)
AP + PB + DR + RC = AS + BQ + DS + CQ
AB + CD = AD + BC
2 AB = 2 BC => AB = BC {Opposite sides of a parallelogram are equal)
:. AB = BC = CD = DA,
Hence , ABCD is a rhombus.
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