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


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

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