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The parallelogram circumscribing a circle is a
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(c): Let ABCD be a parallelogram circumscribing a circle. Using the property that the tangents to a circle from an exterior point are equal in length. and CO=CNDO=DP(AM+BM)+(CO+DO)⎧⎨⎪⎪⎪BC=BN+NC,AD=AP+PD⎭⎪⎪⎪ AB+CD=(AP+PD)+(BN+NC) 2AB=2BC∵⇒AB=CDBC=AD⇒∴AB=BC=CD=DA\displaystyle{H}{e}{n}{c}{e},{A}{B}{C}{D}{i}{s}{a}\rho{m}{b}{u}{s}.
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Question Text | The parallelogram circumscribing a circle is a |
Answer Type | Text solution:1 |
Upvotes | 150 |