PLEASE ANSWER

Asked by Rishabh1 | 25th 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  | 25th Jun, 2009, 10:17: PM

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