Q. 154.9( 8 Votes )

Prove that the parallelogram circumscribing a circle, is a rhombus.

Answer :


Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.


To Proof : ABCD is a rhombus.


As ABCD is a parallelogram


AB = CD and BC = AD …[1]


[opposite sides of a parallelogram are equal]


Now, As tangents drawn from an external point are equal.


We have


AP = AS


[tangents from point A]


BP = BQ


[tangents from point B]


CR = CQ


[tangents from point C]


DR = DS


[tangents from point D]


Add the above equations


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


AB + CD = AS + DS + BQ + CQ


AB + CD = AD + BC


AB + AB = BC + BC [From 1]


AB = BC …[2]


From [1] and [2]


AB = BC = CD = AD


And we know,


A parallelogram with all sides equal is a rhombus


So, ABCD is a rhombus.


Hence Proved.


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