Question

If the sides of a parallelogram touch a circle in following figure, prove that the parallelogram is a rhombus.

Answer 1

From A, AP and AS are tangents to the circle.

Therefore, AP = AS   ...(i)

Similarly, we can prove that:

BP = BQ  ...(ii)

CR = CQ  ...(iii)

DR = DS   ...(iv)

Adding,

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence, AB + CD = AD + BC.

But AB = CD and BC = AD   ...(v) (Opposite sides of a ||gm)

Therefore, AB + AB = BC + BC

2AB = 2BC

AB = BC  ...(vi)

From (v) and (vi)

AB = BC = CD = DA

Hence, ABCD is a rhombus.