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Proof that the parallelogram cricumscribing a …

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Proof that the parallelogram cricumscribing a circle is a rhombus

Posted by Ayush Saxena 7 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Miru Mahesh 7 years, 3 months ago

Let ABCD be the llgm circumscribing a circle wid centre O. As ABCD is a llgm. AB= CD AND BC =AD -----> Eq.1 Also, 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.2 From Eq 1 and 2, we get, AB=BC=CD=AD Hence, ABCD is a rhombus.

2Thank You

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