prove that the parallelogram circumscribing a circle is a rhombus.
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Given:-)A parallelogram ABCD circumscribing a circle with O centre.
TO PROVE:-)AB=BC=CD=AD
Proof:-)We know that the length of tangents drawn from an exterior point to a circle are equal.
so,
AP=AS
BP=BQ
CR=CQ
DR=DS
so,
AB+CD=AP+BP+CR+DR
=AS+BQ+CQ+DS
=(AS+DS)+[BQ+CQ]
Hence,
AB+CD=AD+BC
So,
2AB=2AD
cancel 2 And 2
then,
AB=AD
so,
Now,
CD=AB=AD=BC
Hence,
ABCD Is a rhombus.
that's all
By Sujeet Yaduvanshi.