Q. 104.0( 20 Votes )

Prove that a parallelogram circumscribing a circle is a rhombus.

Answer :

Given: ABCD is a ||gm.


To Prove: ABCD is a rhombus.


Theorem Used:


The length of two tangents drawn from an external point are equal.


Explanation:


Let ABCD is a ||gm and its sides touches the circle with centre O.



As A is external point and AS and AP are tangents,


By theorem stated above,


AP = AS …. (1)


Similarly, for point B,


BP = BQ …. (2)


For point C,


CR = CQ …. (3)


For point D,


DR = DS …. (4)


Add 1,2,3 and 4 to get


AP + BP + CR + DR = AS + BQ +CQ +DS


(AP + BP) + (CR + DR) = (AS + DS) +(BQ + CQ)


AB + CD = AD + BC


As ABCD is a ||gm,


AB = CD


AD = BC


2 AB = 2AD


AB = AD


AB = AD = BC = CD


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