Question

If all the sides of a parallelogram touch a circle, show that it is a rhombus

Answer 1

Let ABCD be a parallelogram which circumscribes the circle.
AP = AS [Tangents drawn from an external point to a circle are equal in length]
BP = BQ [Tangents drawn from an external point to a circle are equal in length]
CR = CQ [Tangents drawn from an external point to a circle are equal in length]
DR = DS [Tangents drawn from an external point to a circle are equal in length]
Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
But AB = CD and BC = AD  [Opposite sides of parallelogram ABCD]
AB + CD = AD + BC
Hence 2AB = 2BC
Therefore, AB = BC  
Similarly, we get AB = DA and  DA = CD
Thus, ABCD is a rhombus.