Question

if all sides of a parallelogram touches a circle show that the parallelogram is a rhombus​

Answer 1

Answer:

Let ABCD be a parallelogram.

Parallelogram touches the circle at P, Q, R, S.

AP=AS .... (1)

PB=BQ .... (2)

CR=CQ .... (3)

DR=DS .... (4)

Now add (1), (2), (3) and (4),

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

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

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

AB + CD = AD + BC

AB+CD = AD+BC

AB+AB = BC+BD (∵ In parallelogram AB = CD, BC = AD)

2AB = 2BC

2AB=2BC

AB=BC

∴ the parallelogram becomes a rhombus.