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Prove that parallelogram circumscribing a circle …

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Prove that parallelogram circumscribing a circle is rombus.

    • Posted by Tripti Sahu 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago


    Given ABCD is a parallelogram in which all the sides touch a given circle
    To prove:- ABCD is a rhombus
    Proof:-
    {tex}\because{/tex} ABCD is a parallelogram
    {tex}\therefore{/tex} AB = DC and AD = BC
    Again AP, AQ are tangents to the circle from the point A
    {tex}\therefore{/tex} AP = AQ
    Similarly,  BR = BQ
    CR = CS
    DP = DS
    {tex}\therefore{/tex}(AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS)
    {tex}\Rightarrow{/tex} AD + BC = AB + DC
    {tex}\Rightarrow{/tex} BC + BC = AB + AB [{tex}\because{/tex} AB = DC, AD = BC]
    {tex}\Rightarrow{/tex} 2BC = 2AB
    {tex}\Rightarrow{/tex} BC = AB
    Hence, parallelogram ABCD is a rhombus

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