Question

prove that it's a rhombus

Answer 1

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