Question

Prove that parallelogram circumscribing a circle is a rombhus.

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 eequal in length]
CR = CQ [Since tangents drawn from an external point to a circle are eequal in length]
DR = DS [Since tangents drawn from an external point to a circle are eequal in length]
Consider, (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
Hence 2AB = 2BC
Therefore, AB = BC
Similarly, we get AB = DA and DA = CD
Thus ABCD is a rhombus.