Question

PROVE THAT A PARALLELOGRAM CIRCUSCRIBING A CIRCLE IS A RHOMBUS

Answer 1

answer for the question..

Answer 2

Let ABCD be a parallelogram which circumscribes the circle.
AP = AS               ...(1)
BP = BQ               ...(2)
CR = CQ               ...(3)
DR = DS               ...(4)
For (1) (2) (3) (4) [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 = CD
Thus ABCD is a rhombus.

:) Hope this Helps!!!