Prove that the parallelogram circumscribing a circle is a rhombus.

Solution

$Given : ABCD be a parallelogram circumscribing a circle with centre O . To prove : ABCD is a rhombus . We know that the tangents drawn to a circle from an exterior point are equal in length Therefore, AP = AS, BP = BQ, CR = CQ, and DR = DS . Adding the above equations, AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) AB + CD = AD + BC 2 AB = 2 BC (Since, ABCD is a parallelogram so AB = DC and AD = BC) AB = BC Therefore, AB = BC = DC = AD Hence, ABCD is a rhombus .$

For Daily Free Study Material Join wiredfaculty

Circles

Prove that the parallelogram circumscribing a circle is a rhombus.

Some More Questions From Circles Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction