Prove that the parallelogram of service centre in the Rhombus

1.	Prove that the parallelogram of service centre in the Rhombus
Answer» Given : ABCD is a parallelogram circumscribing a circle.To prove: ABCD is a rhombus\xa0Proof:\xa0{tex}\\because{/tex} The tangent segments from an external point to a circle are equal{tex}\\therefore{/tex}\xa0AP = ASBP = BQCR = CQand DR = DS{tex}\\Rightarrow{/tex}\xa0(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ){tex}\\Rightarrow{/tex}\xa0AB + CD = AD + BC{tex}\\Rightarrow{/tex}\xa0AB + AB = AD + AD[{tex}\\because{/tex} Opposite sides of parallelogram are equal]{tex}\\Rightarrow{/tex}\xa02AB = 2AD{tex}\\Rightarrow{/tex}\xa0AB = ADBut\xa0AB = CD And AD = BC [Opposite sides of parallelogram]{tex}\\therefore{/tex}\xa0AB = BC = CD = AD{tex}\\therefore{/tex}\xa0Parallelogram ABCD is a rhombus.

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