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

Asked by 29th February 2008, 3:08 PM
Answered by Expert
Answer:

 

 

 

 

Solution

 Given

Parallelogram ABCD touches a circle with centre O.
To prove
ABCD is a rhombus.
 
Proof
 Since the length of the tangents from an external point to a given circle are equal
So,
AP=AS           (i)
BP=BQ           (ii)
CR=CQ and  (iii)
DR=DS          (iv)
 
Adding (i), (ii),(iii),(iv) we get
(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)
AB+CD=AD+BC
Since ABCD is a Parallelogram CD=AB and BC=AD
This implies, AB+AB=AD+AD
2AB=2AD
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 Expert 29th February 2008, 4:09 PM
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