Question

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

Answer 1

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Let ABCD be a parallelogram which circumscribes the circle.

AP = AS. [Tangents drawn from an external point to a circle are equal in length]

BP = BQ. [Tangents drawn from an external point to a circle are equal in length]

CR = CQ. [Tangents drawn from an external point to a circle are equal in length]

DR = DS. [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  [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.

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Answer 2

given
ABCD is a parallelogram
and circle PQRS touches it's side resp.
prove
AP=AS (tangents of circle ) (1)
PB=BQ ( ") (2)
DR=D'S (") (3)
RC=cq (") (4)
adding equation 1,2,3,4
we get
AP+PB+ DR +DC =AS +DS +BQ +QC
AB +DC= AD+ BC
given
AB =DC
AD= BC
so, AB = BC
Hence, AB = BC = CD = DA
ABCD is rhombus
Hence proved