1.

If a parallelogram circumscribes a circle then prove that it must be a rhombus.

Answer»

Solution :Given a parallelogram, say ABCD, Let this parallelogram TOUCH the circle at the point P, Q, R and S.
As AP and AS are tangents to the circle drawn from an EXTERNAL point A.
`AP=AS""…(1)`
SIMILARLY,`""BP=BQ""…(2)`
`CR=CQ""...(3)`
`DR=DS""...(4)`
Circles
Adding (1), (2), (3) and (4), we get
`implies""AP+BP+CR+DR=AS+BQ+CQ+DS`
`implies""(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)`
`implies""AB+CD=AD+BC`
`implies""AB+AB=AD+AD`
(CD=AB, DB=AD, opposite sides of a parallelogram)
`implies""2AB=2AD`
`implies""AB=AD`
Hence, ABCD is a RHOMBUS. (`because` adjacent sides of a parallelogram are equal) Hence Proved.


Discussion

No Comment Found

Related InterviewSolutions