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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