Question

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

Answer 1

Let the sides of parallelogram ABCD touch the circle at points P, Q, R and S. 

AP = AS - (1) 

PB = BQ - {2} {Length of tangents drawn from an external point to a circle a equal) 

DR = DS - {3} 

RC = CQ - (4) 

Adding (1), {2}, {3} and (4) 

AP + PB + DR + RC = AS + BQ + DS + CQ 

AB + CD = AD + BC 

2 AB = 2 BC => AB = BC {Opposite sides of a parallelogram are equal) 

:. AB = BC = CD = DA, 

Hence , ABCD is a rhombus.