CBSE Class 10 Answered
If all sides of a parallelogram touch a circle,prove that the parallelogram is a rhombus
Asked by | 29 Feb, 2008, 03:08: PM
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 | 29 Feb, 2008, 04:09: PM
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