prove that the parallelogram circumscribing in a circle is rhombus.
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Given: ABCD be a parallelogram circumscribing a circle with centre O.
To prove: ABCD is a rhombus.
We know that the tangents drawn to a circle from an exterior point are equal in length.
Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.
Adding the above equations,
AP + BP + CR + DR = AS + BQ + CQ + DS.
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