Prove that parallelogram circumscribing a circle is rombus.

1.	Prove that parallelogram circumscribing a circle is rombus.
Answer» Given ABCD is a parallelogram in which all the sides touch a given circleTo prove:- ABCD is a rhombusProof:-{tex}\\because{/tex}\xa0ABCD is a parallelogram{tex}\\therefore{/tex}\xa0AB = DC and AD = BCAgain AP, AQ are tangents to the circle from the point A{tex}\\therefore{/tex}\xa0AP = AQSimilarly, BR = BQCR = CSDP = DS{tex}\\therefore{/tex}(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS){tex}\\Rightarrow{/tex}\xa0AD + BC = AB + DC{tex}\\Rightarrow{/tex}\xa0BC + BC = AB + AB [{tex}\\because{/tex} AB = DC, AD = BC]{tex}\\Rightarrow{/tex}\xa02BC = 2AB{tex}\\Rightarrow{/tex}\xa0BC = ABHence, parallelogram ABCD is a rhombus

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