Q. 55.0( 3 Votes )

Let us prove that a parallelogram circumscribed by a circle is a rhombus.

Answer :



Let ABCD be the parallelogram


AP and AS are tangents drawn from point A


BP and BQ are tangents drawn from point B


CR and CQ are tangents drawn from point C


DR and DS are tangents drawn from point D


Each of the above pairs of tangents are equal to each other in length since tangents drawn from an external point are always equal.


Using this concept we can say


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


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


Adding each length segment we can say


AB + CD = AD + BC


Since Opposite sides of a parallelogram are equal so


2AB = 2AD


AB = AD


Since adjacent sides of a parallelogram are equal, so it’s a rhombus.


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