Question

If all sides a parallelogram touch a circle, show that the parallelogram 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.