If all sides of a parallelogram touch a circle,prove that the parallelogram is a rhombus

Asked by  | 29th Feb, 2008, 03:08: PM

Expert Answer:







Parallelogram ABCD touches a circle with centre O.
To prove
ABCD is a rhombus.
 Since the length of the tangents from an external point to a given circle are equal
AP=AS           (i)
BP=BQ           (ii)
CR=CQ and  (iii)
DR=DS          (iv)
Adding (i), (ii),(iii),(iv) we get
Since ABCD is a Parallelogram CD=AB and BC=AD
This implies, AB+AB=AD+AD
This implies AB=AD
But AB=CD and AD=BC as opposite sides of a Parallelogram are equal.
Therefore , AB=BC=CD=AD.
Hence ABCD is a rhombus.



Answered by  | 29th Feb, 2008, 04:09: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.