Question

prove that the parallelogram inscribed in a circle is a rhombus

Answer 1

we know that ABCD is a parallelogram.
AB=CD......1
BC=AD......2
It is observed that,
DR=DS (tangent on the circle from point D).....3
CR=CQ (tangent on the circle from point C).....4
BP=BQ (tangent on the circle from point B).....5
AP=AS (tangent on the circle from point A)....6
Adding equation 3, 4, 5 & 6
we get,
DR+CR+BP+AP=DS+CQ+BQ+AS
CD+AB=AD+BC......7
On putting values of equation 1, 2 and 7
we get,
2AB=2BC
Compairing equation 1, 2 and 7
we get,
AB=BC=CD=AD
Hence, ABCD is a rhombus