CBSE Class 10 Answered
PLEASE ANSWER
Asked by Rishabh1 | 25 Jun, 2009, 09:42: PM
Expert Answer
Let ABCD be the parallelogram which touches a circle with sides AB,BC,CD,DA touching the circle at P,Q,R,S respectively.
So, using the fact that lengths of tangent segments drawn from an external point to a circle are equal , we get,
AP=AS..(I)
PB=BQ..(II)
CR=CQ..(III)
DR=DS(IV)
Adding, we get
AP+PB+CR+DR=AS+BQ+CQ+DS
So we have,
AB+CD=AD+BC...(v)
but AB=CD and AD=BC..(opp sides of a paralelogram)
So,(v) becomes,
2AB=2BC
so,
AB=BC
thus we have proved that ABCD is a parallelogram with one pair of adjacent sides equal , so ABCD is a rhombus by definition.
Answered by | 25 Jun, 2009, 10:17: PM
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