Proof that the parallelogram cricumscribing a circle is a rhombus

1.	Proof that the parallelogram cricumscribing a circle is a rhombus
Answer» Let ABCD be the llgm circumscribing a circle wid centre O. As ABCD is a llgm.AB= CD AND BC =AD -----> Eq.1Also, the tangents to a circle frm an external point are equal in length.Therefore, AM= AP, BM= BN, CO= CN, DO= DP.=> (AM+ BM) +(CO+ DO)= AP+BN+CN+DP=> AB+CD=(AP+BC) +( BN+ NC) =AD+BC ---->Eq.2From Eq 1 and 2, we get,AB=BC=CD=ADHence, ABCD is a rhombus.

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